KC Sinha: Exercise 4.4 - Mathematics Solution Class 10 Chapter 4 Trigonometric Ratios and Identities

[mathjax]

Question 1

Fill in the blanks

(i) sin² θ cosec² θ = ……..
(ii) 1 + tan² θ = ……
(iii) Reciprocal sin θ. cot θ = ……
(iv) 1–.......= cos²θ
(v) $\tan A=\frac{\cdots \cdots \cdot}{\cos A}$
(vi) $\ldots \ldots=\frac{\cos A}{\sin A}$
(vii) cos θ is reciprocal of .........
(viii) Reciprocal of sin θ is.........
(ix) Value of sin θ in terms of cos θ is
(x) Value of cos θ in terms of sin θ is
Sol :
(i) Given: sin² θ cosec² θ

$\Rightarrow \sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\csc \theta}\right]$
= 1

(ii) Given: 1 + tan² θ

$=1+\left(\frac{\sin \theta}{\cos \theta}\right)^{2}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos ^{2} \theta}$[∵ cos² θ + sin² θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec² θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$

(iii) Given : sin θ cot θ
Firstly, we simplify the given trigonometry

$=\sin \theta \times \frac{\cos \theta}{\sin \theta}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
= cos θ
Now, the reciprocal of cos θ is
$=\frac{1}{\cos \theta}$
=sec θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$

(iv) Given: 1 – x = cos²θ
Subtracting 1 to both the sides, we get
1 – x –1 = cos²θ – 1
⇒ –x = – sin²θ
⇒ x =sin² θ

(v) $\tan A=\frac{\sin A}{\cos A}$

(vi) $\cot A=\frac{\cos A}{\sin A}$

(vii) $\cos \theta=\frac{1}{\sec \theta}$

(viii) $\sin \theta=\frac{1}{\cos \theta}$

(ix) We know that
cos² θ + sin² θ = 1
⇒ sin² θ = 1 – cos² θ
⇒ sin θ = √(1 – cos² θ)
(x) We know that
cos² θ + sin² θ = 1
⇒ cos ² θ = 1 – sin² θ
⇒ cos θ = √(1 – sin² θ)

---

Question 2

If sin θ= p and cos θ = q, what is the relation between p and q ?

Sol :
We know that,
cos² θ + sin² θ = 1 …(i)
Given : sin θ = p and cos θ = q
Putting the values of sin θ and cos θ in eq. (i), we get
(q)² + (p)² =1
⇒ p² + q² =1

---

Question 3

If cos A = x, express sin A in terms of x

Sol :
We know that
cos² θ + sin² θ = 1
⇒ sin² θ = 1 – cos² θ
⇒ sin θ = √(1 – cos² θ)
And Given that cos θ = x
⇒ sin θ = √(1– x²)

---

Question 4

If x cos θ = 1 and y sin θ = 1 find the value of tan θ.

Sol :
Given x cosθ = 1 and y sinθ = 1

$\Rightarrow \cos \theta=\frac{1}{x}$ and $\sin \theta=\frac{1}{\mathrm{y}}$

Now, we know that

$\tan \theta=\frac{\sin \theta}{\cos \theta}$

Putting the value of sin θ and cos θ, we get
$\tan \theta=\frac{\frac{1}{y}}{\frac{1}{x}}$
$\Rightarrow \tan \theta=\frac{x}{y}$

---

Question 5

If cos40^o = p, then write the value of sin 40^o in terms of p.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 40° + sin² 40° = 1
⇒ sin² 40° = 1 – cos² 40°
⇒ sin 40° = √(1 – cos² 40°)
And Given that cos 40° = p
⇒ sin 40° = √(1– p²)

---

Question 6

If sin 77° = x, then write the value of cos 77^o in terms of x.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 77° + sin² 77° = 1
⇒ cos ² 77° = 1 – sin² 77°
⇒ cos 77° = √(1 – sin² 77°)
And Given that sin 77° = x
⇒ cos 77° = √(1 – x² )

---

Question 7

If cos55° = x², then write the value of sin 55^o in terms of x.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 55° + sin² 55° = 1
⇒ sin² 55° = 1 – cos² 55°
⇒ sin 55° = √(1 – cos² 55°)
And Given that cos 55° = x²
⇒ sin 55° = √{1– (x²)²}
⇒ sin 55° = √(1 – x⁴)

---

Question 8

If, sin 50° = α then write the value of cos 50° in terms of α.

Sol :
We know that
cos² θ + sin² θ = 1
⇒ cos² 50° + sin² 50° = 1
⇒ cos ² 50° = 1 – sin² 50°
⇒ cos 50° = √(1 – sin² 50°)
And Given that sin 50° = a
⇒ cos 50° = √(1 – a² )

---

Question 9

If x cos A = 1 and tan A = y, then what is the value of x² – y².

Sol :
Given x cos A = 1 and tan A =y
$\Rightarrow \mathrm{x}=\frac{1}{\cos \mathrm{A}}$ and $\frac{\sin A}{\cos A}=y$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$ 
To find: x² – y²

Putting the values of x and y , we get
$\left(\frac{1}{\cos A}\right)^{2}-\left(\frac{\sin A}{\cos A}\right)^{2}$
$=\frac{1}{\cos ^{2} A}-\frac{\sin ^{2} A}{\cos ^{2} A}$
$=\frac{1-\sin ^{2} A}{\cos ^{2} A}$
$=\frac{\cos ^{2} A}{\cos ^{2} A}$ [∵ cos² θ + sin² θ = 1]
= 1

---

Question 10

Prove the followings identities:

(1 – sin θ)(1 + sin θ) = cso² θ
Sol :
Taking LHS = (1 – sinθ)(1+ sinθ)
Using identity , (a + b) (a – b) = (a² – b²) , we get
= (1)² – (sinθ)²
= 1 – sin² θ
= cos² θ [∵ cos² θ + sin² θ = 1]
= RHS
Hence Proved

---

Question 11

Prove the followings identities:

(1 + cos θ)(1 – cos θ) = sin²θ
Sol :
Taking LHS =(1 – cosθ)(1+ cosθ)
Using identity , (a + b) (a – b) = (a² – b²) , we get
= (1)² – (cosθ)²
= 1 – cos² θ
= sin² θ [∵ cos² θ + sin² θ = 1]
= RHS
Hence Proved

---

Question 12

Prove the followings identities:

$\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}=\tan ^{2} \theta$

Sol :
Taking LHS $\frac{(1-\cos \theta)(1+\cos \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{(1)^{2}-(\cos \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$ [Using identity , (a + b) (a – b) = (a² – b²)]
$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
= tan² θ  $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= RHS
Hence Proved

---

Question 13

Prove the followings identities:

$\frac{1}{\sec \theta+\tan \theta}=\sec \theta-\tan \theta$

Sol :
Taking LHS $=\frac{1}{\sec \theta+\tan \theta}$
Multiplying and divide by the conjugate of sec θ + tan θ
$=\frac{1}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{\sec \theta-\tan \theta}{\sec ^{2} \theta-\tan ^{2} \theta}$ [Using identity , (a + b) (a – b) = (a² – b²)]
= sec θ – tan θ [∵ 1+ tan² θ = sec² θ ]
= RHS
Hence Proved

---

Question 14 A 

Prove the following identities :

sinθ. cotθ = cos θ
Sol :
Taking LHS = sin θ cot θ
$=\sin \theta \times \frac{\cos \theta}{\sin \theta}\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
= cos θ
=RHS
Hence Proved

---

Question 14 B 

Prove the following identities :

sin² θ(1+ cot² θ) = 1
Sol :

Taking LHS = sin² θ(1+ cot² θ)
= sin² θ (cosec² θ) [∵ cot² θ +1= cosec² θ ]

$=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\operatorname{cscec} \theta}\right]$
= 1
=RHS
Hence Proved

---

Question 14 C 

Prove the following identities :

cos² A (tan² A+1) = 1
Sol :
Taking LHS = cos² A (tan² A+1)
= cos² θ (sec² θ) [∵ 1+ tan² θ = sec² θ ]

$=\cos ^{2} \theta \times \frac{1}{\cos ^{2} \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
= 1
=RHS
Hence Proved

---

Question 14 D 

Prove the following identities :

tan⁴θ + tan²θ = sec⁴θ – sec²θ
Sol :
Taking LHS = tan⁴ θ + tan² θ
= (tan² θ)² + tan² θ
= ( sec² θ – 1)² + (sec² θ – 1) [∵ 1+ tan² θ = sec² θ ]
= sec⁴ θ + 1 – 2 sec² θ + sec² θ – 1 [∵ (a – b)² = (a² + b² – 2ab)]
= sec⁴ θ – sec² θ
=RHS
Hence Proved

---

Question 14 E 

Prove the following identities :

$\frac{\left(1+\tan ^{2} \theta\right) \sin ^{2} \theta}{\tan \theta}=\tan \theta$
Sol :
Taking LHS $=\frac{\left(1+\tan ^{2} \theta\right) \sin ^{2} \theta}{\tan \theta}$
$=\frac{\left(\sec ^{2} \theta\right) \sin ^{2} \theta}{\frac{\sin \theta}{\cos \theta}}$ [∵ 1+ tan² θ = sec² θ]
$=\frac{1 \times \sin \theta \times \cos \theta}{\cos ^{2} \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin \theta}{\cos \theta}$
= tan θ
=RHS
Hence Proved

---

Question 14 F 

Prove the following identities :

$\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+1=\frac{\tan ^{2} \theta}{\sin ^{2} \theta}$
Sol :
Taking LHS$=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+1$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1}{\cos ^{2} \theta}$
= sec²θ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$

Now, RHS $=\frac{\tan ^{2} \theta}{\sin ^{2} \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\sin ^{2} \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

$=\frac{\sin ^{2} \theta}{\sin ^{2} \theta \times \cos ^{2} \theta}$

$=\frac{1}{\cos ^{2} \theta}$
= sec²θ  $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
∴ LHS = RHS
Hence Proved

---

Question 14 G 

Prove the following identities :

$\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}=3-\tan ^{2} \theta$
Sol :
Taking LHS $=\frac{3-4 \sin ^{2} \theta}{\cos ^{2} \theta}$
$=\frac{3}{\cos ^{2} \theta}-\frac{4 \sin ^{2} \theta}{\cos ^{2} \theta}$
= 3 sec² θ – 4 tan² θ
We know that,
1+ tan² θ = sec² θ
= 3(1+ tan² θ) – 4 tan² θ
= 3 + 3 tan² θ – 4 tan² θ
= 3 – tan² θ
=RHS
Hence Proved

---

Question 14 H 

Prove the following identities :

(1+ tan² θ) cos θ. sin θ = tan θ
Sol :
Taking LHS = (1+ tan² θ) cos θ sin θ
= (sec² θ) cos θ sin θ [∵ 1+ tan² θ = sec² θ]

$=\frac{1}{\cos ^{2} \theta} \times \cos \theta \times \sin \theta$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$
$=\frac{\sin \theta}{\cos \theta}$
= tan θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved

---

Question 14 I 

Prove the following identities :

sin²θ – cos² ϕ = sin²ϕ – cos²θ
Sol :
Taking LHS = sin² θ – cos² φ
=( 1 – cos² θ) – (1 – sin² φ) [∵ cos² θ + sin² θ = 1] & [∵ cos² φ + sin² φ = 1]
= 1 – cos² θ – 1 + sin² φ
= sin² φ – cos² θ
=RHS
Hence Proved

---

Question 14 J 

Prove the following identities :

$\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}=\tan ^{2} \theta$

Sol :
Taking LHS $=\frac{1-\tan ^{2} \theta}{\cot ^{2} \theta-1}$
$=\frac{1-\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-1}$
$=\frac{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos ^{2} \theta}}{\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\sin ^{2} \theta}}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos ^{2} \theta} \times \frac{\sin ^{2} \theta}{\cos ^{2} \theta-\sin ^{2} \theta}$
= tan² θ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved

---

Question 15 A 

Prove the following identities :

(1 – cosθ)(1+ cosθ)(1+ cot² θ) = 1
Sol :
Taking LHS = (1 – cosθ)(1+ cosθ)(1+ cot² θ)
Using identity , (a + b) (a – b) = (a² – b²) in first two terms , we get
= (1)² – (cosθ)² (cosec² θ) [∵ cot² θ +1= cosec² θ]
= (1 – cos² θ) (cosec² θ)
= (sin² θ) (cosec² θ) [∵ cos² θ + sin² θ = 1]

$=\sin ^{2} \theta \times \frac{1}{\sin ^{2} \theta}$ $\left[\because \sin \theta=\frac{1}{\csc \theta}\right]$
=1
= RHS
Hence Proved

---

Question 15 B 

Prove the following identities :

$\frac{(1+\sin \theta)^{2}+(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}$
Sol :
Taking LHS $=\frac{(1+\sin \theta)^{2}(1-\sin \theta)^{2}}{2 \cos ^{2} \theta}$
$=\frac{1+\sin ^{2} \theta+2 \sin \theta+1+\sin ^{2} \theta-2 \sin \theta}{2 \cos ^{2} \theta}$
[∵ (a + b)² = (a² + b² + 2ab) and (a – b)² = (a² + b² – 2ab)]
$=\frac{2+2 \sin ^{2} \theta}{2 \cos ^{2} \theta}$
$=\frac{2\left(1+\sin ^{2} \theta\right)}{2\left(1-\sin ^{2} \theta\right)}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1+\sin ^{2} \theta}{1-\sin ^{2} \theta}$
=RHS
Hence Proved

---

Question 15 C 

Prove the following identities :

$\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)}=\frac{1+\sin \theta}{1+\cos \theta}$
Sol :
Taking LHS $=\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)}$

Multiplying and divide by the conjugate of (1 – sinθ), we get
$=\frac{\cos ^{2} \theta(1-\cos \theta)}{\sin ^{2} \theta(1-\sin \theta)} \times \frac{(1+\sin \theta)}{(1+\sin \theta)}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta\left[(1)^{2}-(\sin \theta)^{2}\right]}$
$=\frac{\cos ^{2} \theta(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta \times \cos ^{2} \theta}$
$=\frac{(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta}$

Now, multiply and divide by conjugate of 1 – cos θ, we get
$=\frac{(1-\cos \theta)(1+\sin \theta)}{\sin ^{2} \theta} \times \frac{(1+\cos \theta)}{(1+\cos \theta)}$
$=\frac{\left(1^{2}-\cos ^{2} \theta\right)(1+\sin \theta)}{\sin ^{2} \theta(1+\cos \theta)}$
$=\frac{\sin ^{2} \theta(1+\sin \theta)}{\sin ^{2} \theta(1+\cos \theta)}$
$=\frac{1+\sin \theta}{1+\cos \theta}$
=RHS
Hence Proved

---

Question 15 D 

Prove the following identities :

(sin θ – cos θ)² = 1 – 2 sinθ . cos θ
Sol :
Taking LHS = (sin θ – cos θ)²
Using the identity,(a – b)² = (a² + b² – 2ab)
= sin² θ + cos² θ – 2sin θ cos θ
= 1 – 2sin θ cos θ [∵ cos² θ + sin² θ = 1]
=RHS
Hence Proved

---

Question 15 E 

Prove the following identities :

(sin θ + cos θ)² + (sin θ – cos θ)² = 2
Sol :
Taking LHS = (sin θ + cos θ)² + (sin θ – cos θ)²
Using the identity,(a + b)² = (a² + b² + 2ab) and (a – b)² = (a² + b² – 2ab)
= sin² θ + cos² θ + 2sin θ cos θ + sin² θ + cos² θ – 2sin θ cos θ
= 1 +1 [∵ cos² θ + sin² θ = 1]
= 2
=RHS
Hence Proved

---

Question 15 F 

Prove the following identities :

(asin θ + bcos θ)² + (acos θ – bsin θ)² = a² + b²
Sol :
Taking LHS = (asin θ + bcos θ)² + (acos θ – bsin θ )²
Using the identity,(a + b)² = (a² + b² + 2ab) and (a – b)² = (a² + b² – 2ab)
= a² sin² θ + b² cos² θ + 2 ab sin θ cos θ + a² cos² θ + b² sin² θ – 2 ab sin θ cos θ
= a² sin² θ+ a² cos² θ + b² sin² θ + b² cos² θ
= a² (sin² θ + cos² θ) + b² (sin² θ + cos² θ)
= a² + b² [∵ cos² θ + sin² θ = 1]
=RHS
Hence Proved

---

Question 15 G 

Prove the following identities :

cos⁴ A + sin⁴ A + 2 sin² A. cos² A = 1
Sol :
Taking LHS = cos⁴ A + sin⁴ A + 2 sin² A cos² A
Using the identity,(a + b)² = (a² + b² + 2ab)
Here, a = cos² A and b = sin² A
= ( cos² A + sin² A) [∵ cos² θ + sin² θ = 1]
= 1

---

Question 15 H 

Prove the following identities :

sin⁴ A – cos⁴ A = 2 sin² A – 1 = 1 – 2 cos² A = sin² A – cos² A
Sol :
Given:
$\begin{matrix}\sin ^{4} A-\cos ^{4} A&=2 \sin ^{2} A-1&=1-2 \cos ^{2} A&=\sin ^{2} A-\cos ^{2} A\\ \text{I}&\text{II}&\text{III}&\text{IV}\end{matrix}$
Taking I term
= sin⁴ A – cos⁴ A → I term
= (sin² A)² – (cos² A)²
= (sin² A – cos² A)(sin² A+ cos² A )
[∵ (a² – b²) = (a + b) (a – b)]
= (sin² A – cos² A)(1) [∵ cos² θ + sin² θ = 1]
= (sin² A – cos² A) …(i) → IV term
From Eq. (i)
= {sin² A – (1 – sin² A)} [∵ cos² θ + sin² θ = 1]
= sin² A – 1 + sin² A
= 2 sin² A – 1 → II term
Again, From Eq. (i)
= {(1 – cos² A) – cos² A } [∵ cos² θ + sin² θ = 1]
=1 – 2 cos² A → III term
Hence, I = II = III = IV
Hence Proved

---

Question 15 I 

Prove the following identities :

cos⁴ θ – sin⁴ θ = cos² θ – sin² θ = 2 cos² θ – 1
Sol :
Given:

$\begin{matrix}\cos ^{4} \theta-\sin ^{4} \theta&=\cos ^{2} \theta-\sin ^{2} \theta&=2 \cos ^{2} \theta-1\\ \text{I}&\text{II}&\text{III}\end{matrix}$

Taking I term
= cos⁴ θ – sin⁴ θ → I term
= (cos² θ)² – (sin² θ)²
= (cos² θ – sin² θ)(cos² θ+ sin² θ )
[∵ (a² – b²) = (a + b) (a – b)]
= (cos² θ – sin² θ) (1) [∵ cos² θ + sin² θ = 1]
= (cos² θ – sin² θ) …(i) → II term
From Eq. (i)
= {cos² θ – (1 – cos² θ)} [∵ cos² θ + sin² θ = 1]
= 2 cos² θ – 1 → III term
Hence, I = II = III
Hence Proved

---

Question 15 J 

Prove the following identities :

2 cos² θ – cos⁴ θ + sin⁴ θ = 1
Sol :
Taking LHS = 2 cos² θ – cos⁴ θ + sin⁴ θ
= 2 cos² θ – (cos⁴ θ – sin⁴ θ)
= 2 cos² θ – [(cos² θ)² – (sin² θ)²]
Using identity, (a² – b²) = (a + b) (a – b)
= 2 cos² θ – [(cos² θ – sin² θ)(cos² θ+ sin² θ )]
= 2 cos² θ – [(cos² θ – sin² θ)(1)] [∵ cos² θ + sin² θ = 1]
=2 cos² θ – cos² θ + sin² θ
= cos² θ + sin² θ
= 1 [∵ cos² θ + sin² θ = 1]
= RHS
Hence Proved

---

Question 15 K 

Prove the following identities :

1 – 2 cos² θ + cos⁴ θ = sin⁴θ
Sol :
Taking LHS = 1 – 2 cos² θ + cos⁴ θ
We know that,
cos² θ + sin² θ = 1
= 1– 2 cos² θ + (cos² θ)²
= 1 – 2 cos² θ + (1 – sin² θ)²
= 1 – 2 cos² θ +1 + sin⁴ θ – 2sin²θ
= 2 – 2(cos² θ + sin²θ) + sin⁴ θ
= 2 – 2(1) + sin⁴ θ
= sin⁴ θ
=RHS
Hence Proved

---

Question 15 L 

Prove the following identities :

1 – 2 sin² θ + sin⁴ θ = cos⁴θ
Sol :
Taking LHS = 1 – 2 sin² θ + sin⁴ θ
We know that,
cos² θ + sin² θ = 1
= 1– 2 sin² θ + (sin² θ)²
= 1 – 2 sin² θ + (1 – cos² θ)²
= 1 – 2 sin² θ +1 + cos⁴ θ – 2cos²θ
= 2 – 2(cos² θ + sin²θ) + cos⁴ θ
= 2 – 2(1) + cos⁴ θ
= cos⁴ θ
=RHS
Hence Proved

---

Question 16 A 

Prove that the following identities :

sec²θ + cosec²θ = sec²θ.cosec²θ
Sol :
Taking LHS = sec² θ + cosec² θ

$=\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}$ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
$=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\cos ^{2} \theta \sin ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1}{\cos ^{2} \theta \sin ^{2} \theta}$
= sec² θ × cosec² θ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved

---

Question 16 B 

Prove that the following identities :

$\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta=\operatorname{cosec} \theta$
Sol :
Taking LHS $=\frac{\cos ^{2} \theta}{\sin \theta}+\sin \theta$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta}$
$=\frac{1}{\sin \theta}$ [∵ cos² θ + sin² θ = 1]
= cosec θ
=RHS
Hence Proved

---

Question 16 C 

Prove that the following identities :

cotθ + tan θ = cosec θ . sec θ
Sol :
Taking LHS = cot θ + tan θ

$=\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}$

$\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$ and $\left[\tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta}{\cos \theta \sin \theta}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1}{\cos \theta \sin \theta}$
= cosec θ sec θ $\because\left[\cos \theta=\frac{1}{\sec \theta}\right]$ and $\left[\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved

---

Question 17 

Prove that the following identities :

$\frac{1-\sin \theta}{1+\sin \theta}=\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}$
Sol :
Taking LHS $=\frac{1-\sin \theta}{1+\sin \theta}$

Multiplying and divide by the conjugate of 1 + sin θ , we get
$=\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)² and (a + b) (a – b) = (a² – b²)]
$=\frac{(1-\sin \theta)^{2}}{1-\sin ^{2} \theta}$
$=\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}$
=RHS
Hence Proved

---

Question 18 

Prove that the following identities :

$\frac{1-\cos \theta}{1+\cos \theta}=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$

Multiplying and divide by the conjugate of 1 + cos θ , we get
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)² and (a + b) (a – b) = (a² – b²)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
=RHS
Hence Proved

---

Question 19 

Prove that the following identities :

$\frac{1-\cos \theta}{1+\cos \theta}=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$

Multiplying and divide by the conjugate of 1 + cos θ , we get
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)² and (a + b) (a – b) = (a² – b²)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
=RHS
Hence Proved

---

Question 20 

Prove that the following identities :

$\frac{\cos \theta}{1+\sin \theta}=\frac{1-\sin \theta}{\cos \theta}$
Sol :
Taking LHS $=\frac{\cos \theta}{1+\sin \theta}$

Multiplying and divide by the conjugate of 1 + sin θ , we get
$=\frac{\cos \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}$
$=\frac{\cos \theta(1-\sin \theta)}{(1)^{2}-(\sin \theta)^{2}}$ [∵ (a + b) (a – b) = (a² – b²)]
$=\frac{\cos \theta(1-\sin \theta)}{1-\sin ^{2} \theta}$
$=\frac{\cos \theta(1-\sin \theta)}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1-\sin \theta}{\cos \theta}$
=RHS
Hence Proved

---

Question 21 

Prove that the following identities :

(sin⁸θ – cos⁸θ) = (sin²θ – cos²θ)(1 – 2sin²θ .cos²θ)
Sol :
Taking LHS
= sin⁸ θ – cos⁸ θ
= (sin⁴ θ)² – (cos⁴ θ)²
= (sin⁴ θ – cos⁴ θ)(sin⁴ θ+ cos⁴ θ )
[∵ (a² – b²) = (a + b) (a – b)]
= {(sin² θ)² – (cos² θ)²}{(sin² θ)² + (cos² θ)²}
= (sin² θ + cos² θ) (sin² θ – cos² θ) [(sin² θ + cos² θ) – 2 sin² θ cos² θ]
[ ∵ (a² + b²) = (a +b)² – 2ab]
= (1)[ sin² θ –cos² θ][(1) – 2 sin² θ cos² θ]
= (sin² θ – cos² θ)(1 – 2 sin² θ cos² θ)
=RHS
Hence Proved

---

Question 22 

Prove that the following identities :

2(sin⁶ θ – cos⁶ θ) – 3(sin⁴ θ + cos⁴ θ ) + (sin² θ + cos² θ)
Sol :
Taking LHS
= 2(sin⁶ θ – cos⁶ θ) – 3(sin⁴ θ+ cos⁴ θ ) + (sin² θ + cos² θ)
= 2[(sin² θ)³ – (cos² θ)³ ] – 3[(sin² θ)² + (cos² θ)² ]+1 [∵ cos² θ + sin² θ = 1]
Now, we use these identities, (a³ – b³)= (a + b)³ – 3ab(a+b) and (a² + b²) = (a +b)² – 2ab]
= 2[(sin² θ + cos² θ)³ – 3sin²θ cos²θ (sin² θ+ cos² θ)] –3[(sin² θ + cos² θ) – 2 sin² θ cos² θ]+ 1
=2[(1) – 3sin²θ cos²θ (1)] – 3[(1) – 2 sin² θ cos² θ] + 1 [∵ cos² θ + sin² θ = 1]
=2(1 – 3 sin² θ cos² θ )– 3 + 6sin² θ cos² θ+ 1
= 2– 6sin²θ cos²θ – 2 + 6sin² θ cos² θ
=0
=RHS
Hence Proved

---

Question 23 

Prove the following identities

$\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\sin A+\cos A$
Sol :
Taking LHS $=\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}$

$=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\frac{\sin A}{1-\frac{\cos A}{\sin A}}$

$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\frac{\sin A}{\frac{\sin A-\cos A}{\sin A}}$
$=\frac{\cos ^{2} A}{\cos A-\sin A}+\frac{\sin ^{2} A}{\sin A-\cos A}$
$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos A-\sin A}$

Using the identity, (a² – b²)= (a + b) (a – b)
$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{(\cos A-\sin A)}$
= sin A +cos A
=RHS
Hence Proved

---

Question 24 

Prove the following identities

$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=\frac{2}{\sin \theta}$
Sol :
Taking LHS $=\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$
$=\frac{\sin ^{2} \theta+(1+\cos \theta)^{2}}{(1+\cos \theta)(\sin \theta)}$
$=\frac{\sin ^{2} \theta+1+\cos ^{2} \theta+2 \cos \theta}{(1+\cos \theta)(\sin \theta)}$
$=\frac{2+2 \cos \theta}{(1+\operatorname{ses} \theta)(\sin \theta)}$ [∵ cos² θ + sin² θ = 1]
$=\frac{2(1+\cos \theta)}{(1+\cos \theta)(\sin \theta)}$
$=\frac{2}{\sin \theta}$
=RHS
Hence Proved

---

Question 25 

Prove the following identities

$\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=2 \sec ^{2} \theta$
Sol :
Taking LHS $=\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}$
$=\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}$

Using the identity, (a² – b²)= (a + b) (a – b)
$=\frac{2}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2}{1-\sin ^{2} \theta}$
$=\frac{2}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
= 2 sec² θ
=RHS
Hence Proved

---

Question 26 

Prove the following identities
$\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta$
Sol :
Taking LHS $=\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}$
$=\frac{(1+\sin \theta)^{2}+\cos ^{2} \theta}{(1+\sin \theta)(\cos \theta)}$
$=\frac{1+\sin ^{2} \theta+2 \sin \theta+\cos ^{2} \theta}{(\cos \theta)(1+\sin \theta)}$
$=\frac{2+2 \sin \theta}{(\cos \theta)(1+\sin \theta)}$ [∵ cos² θ + sin² θ = 1]
$=\frac{2(1+\sin \theta)}{(\cos \theta)(1+\sin \theta)}$
$=\frac{2}{\cos \theta}$
=2 sec θ
=RHS
Hence Proved

---

Question 27 

Prove the following identities

$\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=\frac{2}{\cos \theta}$
Sol :
Taking LHS $=\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)+\cos \theta(1-\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{\cos \theta+\cos \theta \sin \theta+\cos \theta-\cos \theta \sin \theta}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{2 \cos \theta}{(1-\sin \theta)(1+\sin \theta)}$

Using the identity, (a² – b²)= (a + b) (a – b)
$=\frac{2 \cos \theta}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2 \cos \theta}{1-\sin ^{2} \theta}$
$=\frac{2 \cos \theta}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]

$=\frac{2}{\cos \theta}$
=RHS
Hence Proved

---

Question 28 

Prove the following identities

$\frac{1}{1+\cos \theta}+\frac{1}{1-\cos \theta}=\frac{2}{\sin ^{2} \theta}$
Sol :
Taking LHS $=\frac{1}{1+\cos \theta}+\frac{1}{1-\cos \theta}$
$=\frac{1-\cos \theta+1+\cos \theta}{(1+\cos \theta)(1-\cos \theta)}$

Using the identity, (a² – b²)= (a + b) (a – b)
$=\frac{2}{(1)^{2}-(\cos \theta)^{2}}$
$=\frac{2}{1-\cos ^{2} \theta}$
$=\frac{2}{\sin ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
=RHS
Hence Proved

---

Question 29 

Prove the following identities

$\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}=\frac{2 \tan \theta}{\cos \theta}$
Sol :
Taking LHS $=\frac{1}{1-\sin \theta}-\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta-1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}$

Using the identity, (a² – b²)= (a + b) (a – b)
$=\frac{2 \sin \theta}{(1)^{2}-(\sin \theta)^{2}}$
$=\frac{2 \sin \theta}{1-\sin ^{2} \theta}$
$=\frac{2 \sin \theta}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]

$=\frac{2 \tan \theta}{\cos \theta}$
$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
=RHS
Hence Proved

---

Question 30 

Prove the following identities

cot²θ – cos²θ = cot²θ . cos²θ
Sol :
Taking LHS = cot² θ – cos² θ

$=\frac{\cos ^{2} \theta}{\sin ^{2} \theta}-\cos ^{2} \theta$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta}$
$=\frac{\cos ^{2} \theta\left(1-\sin ^{2} \theta\right)}{\sin ^{2} \theta}$
 [∵ cos² θ + sin² θ = 1]
$=\frac{\cos ^{2} \theta \cos ^{2} \theta}{\sin ^{2} \theta}$
= cot² θ cos² θ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
=RHS
Hence Proved

---

Question 31 

Prove the following identities

tan² φ – sin² φ – tan² φ . sin² φ = 0
Sol :
Taking LHS = tan² φ – sin² φ – tan² φ sin² φ
$=\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi}-\sin ^{2} \varphi-\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi} \sin ^{2} \varphi$
$=\frac{\sin ^{2} \varphi-\sin ^{2} \varphi \cos ^{2} \varphi-\sin ^{4} \varphi}{\cos ^{2} \varphi}$
$=\frac{\sin ^{2} \varphi\left(1-\cos ^{2} \varphi-\sin ^{2} \varphi\right)}{\cos ^{2} \varphi}$
$=\frac{\sin ^{2} \varphi\left\{1-\left(\cos ^{2} \varphi+\sin ^{2} \varphi\right)\right]}{\cos ^{2} \varphi}$ [∵ cos² φ + sin² φ = 1]
$=\frac{\sin ^{2} \varphi\{1-1\}}{\cos ^{2} \varphi}$
= 0
=RHS
Hence Proved

---

Question 32 

Prove the following identities

tan² φ + cot² φ + 2 = sec²ϕ. cosec²ϕ
Sol :
Taking LHS = tan² φ + cot² φ + 2
$=\frac{\sin ^{2} \varphi}{\cos ^{2} \varphi}+\frac{\cos ^{2} \varphi}{\sin ^{2} \varphi}+2$
$=\frac{\sin ^{4} \varphi+\cos ^{4} \varphi+2 \sin ^{2} \varphi \cos ^{2} \varphi}{\cos ^{2} \varphi \sin ^{2} \varphi}$ [∵ (a + b)² = (a² + b² + 2ab)]
$=\frac{\left(\sin ^{2} \varphi+\cos ^{2} \varphi\right)^{2}}{\cos ^{2} \varphi \sin ^{2} \varphi}$ [∵ cos² φ + sin² φ = 1]
$=\frac{1}{\cos ^{2} \varphi \sin ^{2} \varphi}$
= sec² φ cosec² φ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right.$ and $\left.\sin \theta=\frac{1}{\csc \theta}\right]$
=RHS
Hence Proved

---

Question 33 

Prove the following identities

$\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}$
Sol :
Taking LHS $=\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$ [∵ cot² θ – cosec² θ = 1]
$=\frac{(\cot \theta+\operatorname{cosec} \theta)-\{(\operatorname{cosec} \theta+\cot \theta)(\operatorname{cosec} \theta-\cot \theta)\}}{\cot \theta-\operatorname{cosec} \theta+1}$
$=\frac{(\cot \theta+\operatorname{cosec} \theta)(1-\operatorname{cosec} \theta+\cot \theta)}{\cot \theta-\operatorname{cosec} \theta+1}$
= cot θ + cosec θ

$=\frac{\cos \theta}{\sin \theta}+\frac{1}{\sin \theta}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right.$ and $\left.\sin \theta=\frac{1}{\csc \theta}\right]$
$=\frac{1+\cos \theta}{\sin \theta}$
=RHS
Hence Proved

---

Question 34 

Prove the following identities

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$
Sol :
Taking LHS $=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$ $\left[\because \cot \theta=\frac{\cos \theta}{\sin \theta}\right.$ and $\left.\tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
$=\frac{\frac{\sin \theta}{\sin \theta-\cos \theta}}{\sin \theta}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$
$=\frac{\sin \theta}{\cos \theta} \times \frac{\sin \theta}{\sin \theta-\cos \theta}+\frac{\cos \theta}{\sin \theta} \times \frac{\cos \theta}{\cos \theta-\sin \theta}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta\{-(\sin \theta-\cos \theta)\}}$
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{(\cos \theta \sin \theta)(\sin \theta-\cos \theta)}$
$=\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{(\cos \theta \sin \theta)(\sin \theta-\cos \theta)}$ [∵ (a³ – b³)= (a – b)(a² +b² +ab)]
$=\frac{1+\sin \theta \cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta \sin \theta}+\frac{\sin \theta \cos \theta}{\cos \theta \sin \theta}$
= tan θ cot θ + 1
=RHS
Hence Proved

---

Question 35 

Prove the following identities

$\frac{1-\cos \theta}{1+\cos \theta}=(\cot \theta-\operatorname{cosec} \theta)^{2}$
Sol :
Taking LHS $=\frac{1-\cos \theta}{1+\cos \theta}$

Multiplying and divide by the conjugate of 1 + cos θ , we get
$=\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{(1-\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}$
[∵ (a – b) (a – b) =(a – b)² and (a + b) (a – b) = (a² – b²)]
$=\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}$
$=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
$=\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$
$=\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$
= (cosec θ – cot θ)²
= { – (cot θ – cosec θ)}²
= (cot θ – cosec θ)²
=RHS
Hence Proved

---

Question 36 

Prove the following identities

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{1+\cos \theta}{\sin \theta}$
Sol :
Taking LHS $=\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$ [multiplying and divide by conjugate of 1– cosθ]
$=\sqrt{\frac{(1+\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\left(1-\cos ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1+\cos \theta}{\sin \theta}$
=RHS
Hence Proved

---

Question 37 

Prove the following identities

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{\sin \theta}{1-\cos \theta}$
Sol :
Taking LHS $=\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$
$=\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$ [multiplying and divide by conjugate of 1– cosθ]
$=\sqrt{\frac{(1+\cos \theta)^{2}}{(1)^{2}-(\cos \theta)^{2}}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\left(1-\cos ^{2} \theta\right)}}$
$=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$ [∵ cos² θ + sin² θ = 1]
$=\frac{1+\cos \theta}{\sin \theta}$

Multiply and divide by conjugate of 1+ cosθ, we get
$=\frac{1+\cos \theta}{\sin \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$=\frac{1-\cos ^{2} \theta}{\sin \theta \times(1-\cos \theta)}$
$=\frac{\sin ^{2} \theta}{\sin \theta \times(1-\cos \theta)}$
$=\frac{\sin \theta}{1-\cos \theta}$
=RHS
Hence Proved

---

Question 38 

Prove the following identities

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$
Sol :
Taking LHS $=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$
$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}$ [multiplying and divide by conjugate of 1+sinθ]
$=\sqrt{\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}$
$=\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$
$=\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}$ [∵ cos² θ + sin² θ = 1]
=$=\frac{1-\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$
= sec θ – tan θ 

$\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cos \theta=\frac{1}{\sec \theta}\right]$
=RHS
Hence Proved

---

Question 39 

Prove the following identities

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{\cos \theta}{1+\sin \theta}$
Sol :
Taking LHS $=\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$

$=\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}$ [multiplying and divide by of 1+ sin θ]

$=\sqrt{\frac{(1)^{2}-(\sin \theta)^{2}}{(1+\sin \theta)^{2}}}$

$=\sqrt{\frac{1-\sin ^{2} \theta}{(1+\sin \theta)^{2}}}$

$=\sqrt{\frac{\cos ^{2} \theta}{(1+\sin \theta)^{2}}}$ [∵ cos² θ + sin² θ = 1]

$=\frac{\cos \theta}{1+\sin \theta}$
=RHS
Hence Proved

---

Question 40 

Prove the following identities

$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta$
Sol :
Taking LHS $=\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$

[multiplying and divide by conjugate of 1– sinθ in 1^st term and 1+sinθ in 2^nd term]

$=\sqrt{\frac{1+\sin \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1-\sin \theta}{1-\sin \theta}}$

$=\sqrt{\frac{(1+\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}+\sqrt{\frac{(1-\sin \theta)^{2}}{(1)^{2}-(\sin \theta)^{2}}}$

$=\sqrt{\frac{(1+\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\left(1-\sin ^{2} \theta\right)}}$

$=\sqrt{\frac{(1+\sin \theta)^{2}}{\cos ^{2} \theta}}+\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}$ [∵ cos² θ + sin² θ = 1]

=$\frac{1+\sin \theta}{\cos \theta}+\frac{1-\sin \theta}{\cos \theta}$

$=\frac{1+\sin \theta+1-\sin \theta}{\cos \theta}$

$=\frac{2}{\cos \theta}$
= 2 sec θ
=RHS
Hence Proved

---

Question 41 

If secθ + tanθ=m and sec θ – tanθ = n, then prove that $\sqrt{m n}=1$

Sol :
Given : secθ+tanθ=m and sec θ–tanθ=n
To Prove : √mn = 1
Taking LHS = √mn
Putting the value of m and n, we get
$=\sqrt{(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)}$
Using the identity, (a + b) (a – b) = (a² – b²)

$=\sqrt{\sec ^{2} \theta-\tan ^{2} \theta}$

=√(1) [∵ 1+ tan² θ = sec² θ]
=±1
=RHS
Hence Proved

---

Question 42 

If cosθ + sinθ = 1, then prove that cosθ – sin θ = ± 1.

Sol :
Given: cos θ+sinθ=1
On squaring both the sides, we get
(cos θ +sin θ)² =(1)²
⇒ cos² θ + sin² θ + 2sinθ cos θ = 1
⇒ cos² θ + sin² θ = cos² θ + sin² θ – 2sinθ cos θ
[∵ cos² θ + sin² θ = 1]
⇒ cos² θ + sin² θ = (cosθ – sinθ)²
[∵ (a – b)² = (a² + b² – 2ab)]
⇒ 1 = (cos θ – sin θ)²
⇒ (cos θ – sin θ) = ±1
Hence Proved

---

Question 43 

If sinθ + sin²θ = 1, then prove that cos²θ +1 cos⁴θ = 1

Sol :
Given : sin θ + sin² θ = 1
⇒ sin θ = 1 – sin² θ
Taking LHS = cos²θ+ cos⁴θ
= cos² θ + (cos² θ)²
= (1– sin² θ) + (1– sin² θ)² …(i)
Putting sin θ = 1 – sin² θ in Eq. (i), we get
= sin θ + (sin θ)²
= sin θ + sin² θ
= 1 [Given: sin θ + sin² θ = 1]
=RHS
Hence Proved

---

Question 44 

If tanθ + secθ = x, show that sinθ =$\frac{x^{2}-1}{x^{2}+1}$

Sol :
To show : $\sin \theta=\frac{x^{2}-1}{x^{2}+1}$
Taking RHS $=\frac{x^{2}-1}{x^{2}+1}$
Given tanθ+secθ=x

$=\frac{(\tan \theta+\sec \theta)^{2}-1}{(\tan \theta+\sec \theta)^{2}+1}$

$=\frac{\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta-1}{\tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta+1}$

$=\frac{\tan ^{2} \theta+\tan ^{2} \theta+2 \tan \theta \sec \theta}{\sec ^{2} \theta-1+\sec ^{2} \theta+2 \tan \theta \sec \theta+1}$ [∵ 1+ tan² θ = sec² θ]

$=\frac{2 \tan ^{2} \theta+2 \tan \theta \sec \theta}{2 \sec ^{2} \theta+2 \tan \theta \sec \theta}$

$=\frac{\frac{\sin ^{2} \theta}{\cos ^{2} \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}}{\frac{1}{\cos ^{2} \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\cos \theta}}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta} \text { and } \cos \theta=\frac{1}{\sec \theta}\right]$

$=\frac{\sin ^{2} \theta+\sin \theta}{1+\sin \theta}$

$=\frac{\sin \theta(\sin \theta+1)}{1+\sin \theta}$
=sin θ
=LHS
Hence Proved

---

Question 45 

If sinθ + cosθ = p and secθ + cosecθ = q, then show q(p²–1) =2p

Sol :
Given: sin θ + cos θ = p and sec θ + cosec θ = q
To show q(p² – 1) = 2p
Taking LHS = q(p² – 1)
Putting the value of sin θ + cos θ = p and sec θ + cosec θ = q, we get
=(sec θ + cosec θ){( sin θ + cos θ )² – 1)
=(sec θ + cosec θ){(sin² θ + cos² θ + 2sin θ cosθ) – 1)}
[∵ (a + b)² = (a² + b² + 2ab)]
=(sec θ + cosec θ)(1+2sin θ cos θ – 1)
=(sec θ + cosec θ)(2sin θcosθ)

$=\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right) \times(2 \sin \theta \cos \theta)$ $\left[\because \cos \theta=\frac{1}{\sec \theta} \text { and } \sin \theta=\frac{1}{\cos \theta}\right]$

$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta} \times 2 \sin \theta \cos \theta$

= 2(sin θ +cos θ)
= 2p [ given sin θ + cos θ = p]
=RHS
Hence Proved

---

Question 46 

If x cosθ =a and y = a tanθ, then prove that x²–y²=a²

Sol :
Given: x cosθ = a and y = a tanθ

⇒$\mathrm{x}=\frac{\mathrm{a}}{\cos \theta}$ and $y=a \tan \theta$

To Prove : x²–y²=a²
Taking LHS = x²–y²
Putting the values of x and y, we get

$=\left(\frac{a}{\cos \theta}\right)^{2}-(a \tan \theta)^{2}$

$=\frac{a^{2}}{\cos ^{2} \theta}-a^{2} \tan ^{2} \theta$

$=\frac{a^{2}}{\cos ^{2} \theta}-a^{2} \frac{\sin ^{2} \theta}{\cos ^{2} \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$

$=\frac{a^{2}-a^{2} \sin ^{2} \theta}{\cos ^{2} \theta}$

$=\frac{a^{2}\left(1-\sin ^{2} \theta\right)}{\cos ^{2} \theta}$

$=\frac{a^{2} \cos ^{2} \theta}{\cos ^{2} \theta}$ [∵ cos² θ + sin² θ = 1]
= a²
= RHS
Hence Proved

---

Question 47 

If x= r cos α sin β, y = r sin α sin β and z = r cos α then prove that x² + y² + z² = r².

Sol :
Taking LHS = x² + y² + z²
Putting the values of x, y and z , we get
=(r cos α sin β)² + (r sin α sin β)² + (r cos α)²
= r² cos²α sin²β + r² sin²α sin²β + r² cos²α
Taking common r² sin² α , we get
= r² sin²α (cos²β + sin² β) + r²cos² α
= r² sin²α + r² cos² α [∵ cos² β + sin² β = 1]
=r² ( sin² α + cos² α)
= r² [∵ cos² α + sin² α = 1]
=RHS
Hence Proved

---

Question 48 

If secθ – tanθ = x, then prove that

(i)$\cos \theta=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$
(ii) $\sin \theta=\frac{1-x^{2}}{1+x^{2}}$
Sol :
(i) Given sec θ – tan θ = x
Taking RHS $=\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}$
Putting the value of x, we get
$=\frac{2(\sec \theta-\tan \theta)}{1+(\sec \theta-\tan \theta)^{2}}$

$=\frac{2(\sec \theta-\tan \theta)}{1+\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}$

$=\frac{2(\sec \theta-\tan \theta)}{\sec ^{2} \theta+\sec ^{2} \theta-2 \sec \theta \tan \theta}$ [∵ 1+ tan² θ = sec² θ]

$=\frac{2(\sec \theta-\tan \theta)}{2 \sec \theta(\sec \theta-\tan \theta)}$

$=\frac{1}{\sec \theta}$ $\left[\because \cos \theta=\frac{1}{\sec \theta}\right]$

= cos θ
=RHS
Hence Proved

(ii) Given sec θ – tan θ = x

Taking RHS $=\frac{1-x^{2}}{1+x^{2}}$

Putting the value of x, we get

$=\frac{1-(\sec \theta-\tan \theta)^{2}}{1+(\sec \theta-\tan \theta)^{2}}$

$=\frac{1-\sec ^{2} \theta-\tan ^{2} \theta+2 \sec \theta \tan \theta}{1+\sec ^{2} \theta+\tan ^{2} \theta-2 \sec \theta \tan \theta}$

$=\frac{\left.-\tan ^{2} \theta-\tan ^{2} \theta+2 \sec \theta \tan \theta\right)}{\sec ^{2} \theta+\sec ^{2} \theta-2 \sec \theta \tan \theta}$ [∵ 1+ tan² θ = sec² θ]

$=\frac{2 \tan \theta(\sec \theta-\tan \theta)}{2 \sec \theta(\sec \theta-\tan \theta)}$

=$\frac{\sin \theta}{\cos \theta} \times \cos \theta$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right]$
= sin θ
=RHS
Hence Proved

---

Question 49 

If a cos θ + b sinθ = c, then prove that a sinθ – b cos θ = ±  $\sqrt{a^{2}+b^{2}-c^{2}}$

Sol :
Let
(a cos θ + b sin θ)² + (a sin θ – b cos θ)² = a²cos²θ + b² sin²θ + 2ab cos θ sin θ + a²sin²θ
+ b² cos²θ – 2ab cos θ sin θ
⇒ c² + (a sin θ – b cos θ)² = a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ)
⇒ c² + (a sin θ – b cos θ)² = a² + b²
⇒ (a sin θ – b cos θ)² = a² + b² – c²
⇒ (a sin θ – b cos θ) = ±√ (a² + b² – c²)

---

Question 50 

If 1+sin²θ = 3 sinθ . cosθ, then prove that tan θ = 1 or 1/2.

Sol :
Given: 1+sin²θ =3 sin θ cos θ
Divide by cos² θ to both the sides, we get

⇒$\frac{1}{\cos ^{2} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{3 \sin \theta \cos \theta}{\cos ^{2} \theta}$

⇒ sec² θ + tan² θ = 3 tan θ
⇒ 1+ tan² θ+ tan² θ = 3tan θ
⇒ 2 tan² θ –3tan θ +1 = 0
Let tanθ = x
⇒ 2x² – 3x +1 = 0
⇒ 2x² – 2x – x +1 = 0
⇒ 2x ( x – 1) – 1(x – 1) = 0
⇒ (2x – 1)(x – 1) = 0
Putting each of the factor = 0, we get
⇒ x = 1 or $\dfrac{1}{2}$
And above, we let tan θ = x

⇒$\tan \theta=1$ or $\frac{1}{2}$

Hence Proved

---

Question 51 

If a cos θ – b sinθ = x and a sinθ + b cosθ = y that a² + b² = x² + y².

Sol :
Taking RHS =x² + y²
Putting the values of x and y, we get
(a cos θ – b sin θ)² + (a sin θ + b cos θ)²
= a²cos²θ + b² sin²θ – 2ab cos θ sin θ + a²sin²θ + b² cos²θ + 2ab cos θ sin θ
= a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ)
= a² + b² [∵ cos² θ + sin² θ = 1]
=RHS
Hence Proved

---

Question 52 

If x = a sec θ + b tan θ a and y = a tan θ + b sec θ, then prove that x² – y² = a² – b².

Sol :
Taking LHS =x² – y²
Putting the values of x and y, we get
(a sec θ + b tan θ)² – (a tan θ + b sec θ)²
= a² sec²θ + b² tan²θ + 2ab sec θ tan θ – a²tan²θ – b² sec²θ – 2ab sec θ tan θ
= a² (sec²θ – tan²θ) – b² (sec²θ – tan²θ)
= a² – b² [∵ 1+ tan² θ = sec² θ]
=RHS
Hence Proved

---

Question 53 

If (a² – b²) sin θ + 2ab cosθ = a² + b², then prove that .

$\tan \theta=\frac{\mathrm{a}^{2}-\mathrm{b}^{2}}{2 \mathrm{ab}}$
Sol :
Taking (a² – b²) sin θ + 2ab cos θ = a² + b²
We know that

$\sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$ and $\cos \theta=\frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}$

Then, substituting the above values in the given equation, we get

$=\mathrm{a}^{2}-\mathrm{b}^{2} \frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}+2 \mathrm{ab} \frac{1-\tan ^{2} \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}=\mathrm{a}^{2}+\mathrm{b}^{2}$

Now, substituting,$t=\tan \frac{\theta}{2}$ , we have

$a^{2}-b^{2} \frac{2 t}{1+t^{2}}+2 a b \frac{1-t^{2}}{1+t^{2}}=a^{2}+b^{2}$

⇒ ( a² – b²)2t – 2ab(1 – t²) = (a² + b²)(1+t²)
Simplify, we get
(a² + 2ab + b²)t² – 2(a² – b²)t + (a² –2ab +b²)=0
⇒ (a+b)² t² – 2(a² – b²)t + (a – b)² = 0
⇒ (a+b)² t² –2 (a – b)(a+b)t + (a – b)² =0
⇒ [(a+b)t – (a – b)]² = 0 [∵ (a – b)² = (a² + b² – 2ab)]
⇒ [(a+b)t – (a – b)] = 0
⇒ (a+b)t = (a – b)
⇒$ \mathrm{t}=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}$

⇒$ \tan \frac{\theta}{2}=\frac{a-b}{a+b}$

We know that, $\frac{2 t}{1-t^{2}}=\tan \theta$ where $\mathrm{t}=\tan \frac{\theta}{2}$

$\Rightarrow \tan \theta=\frac{2\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)}{1-\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)^{2}}$

$\Rightarrow \tan \theta=\frac{2(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})}{(\mathrm{a}+\mathrm{b})^{2}-(\mathrm{a}-\mathrm{b})^{2}}$

$\Rightarrow \tan \theta=\frac{a^{2}-b^{2}}{2 a b}$

Hence Proved