Class 10: Maths Chapter 10 solutions. Complete Class 10 Maths Chapter 10 Notes.
Contents
- 1 RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios
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- 2 RS Aggarwal Solutions for Class 10 Maths Chapter 10: Download PDF
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RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios
RS Aggarwal 10th Maths Chapter 1, Class 10 Maths Chapter 10 solutions
Page No 546:
Question 1:
If sin , find the value of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that sin = = = .
So, if AB = , then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2 = (2k)2 ()2
⇒ BC2 = 4k2 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos = =
tan =
∴ cot = , cosec = and sec =
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Question 2:
If cos find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cos = = = .
So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2 = (25k)2 (7k)2.
⇒ AB2 = 625k2 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin = =
tan =
∴ cot = , cosec = and sec =
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Question 3:
If tan find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that tan = = = .
So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k
Now, finding the other T-ratios using their definitions, we get:
sin = =
cos =
∴ cot = , cosec = and sec =
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Question 4:
If cot θ = 2, find the value of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cot = = = 2.
So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = k
Now, finding the other T-ratios using their definitions, we get:
sin = =
cos =
∴ tan = , cosec = and sec =
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Question 5:
If cosec θ = , the find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cosec = = = .
So, if AC = ()k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2 = 10k2 k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan = =
cos =
∴ , cot = and sec =
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Question 6:
If , find the values of all T-ratios of .
Answer:
We have ,
As,
Also,
Now,
Also,
And,
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Question 7:
If , where d > 0 then find the values of cos θ and tan θ.
Answer:
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Question 8:
If then evaluate (cos2θ – sin2θ).
Answer:
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Question 9:
If 4tan θ = 3 then prove that .
Answer:
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Question 10:
If , show that .
Answer:
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Question 11:
If tan θ = ab, show that (a sinθ−b cosθa sinθ+b cosθ)=(a2−b2)(a2+b2).
Answer:
It is given that tan θ = ab.
LHS = a sinθ − b cosθa sinθ + b cosθ
Dividing the numerator and denominator by cos θ, we get:
a tan θ − ba tan θ + b (∵ tan θ = sin θcos θ)
Now, substituting the value of tan θ in the above expression, we get:
a(ab) − ba(ab) + b= a2b − ba2b + b= a2 − b2a2 + b2 = RHS
i.e., LHS = RHS
Hence proved.
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Question 12:
If sin θ=1213 then evaluate (2sin θ−3cos θ4sin θ−9cos θ).
Answer:
Given: sinθ=1213Since, sinθ=PH⇒P=12 and H=13Using Pythagoras theorem,P2+B2=H2⇒122+B2=132⇒B2=169−144⇒B2=25⇒B=5Therefore,cosθ=BH=513Now,(2sinθ−3cosθ4sinθ−9cosθ)=(2(1213)−3(513)4(1213)−9(513)) =(2413−15134813−4513) =(24−1548−45) =93 =3Hence, (2sinθ−3cosθ4sinθ−9cosθ)=3.
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Question 13:
If tan θ=12 then evaluate (cos θsin θ+sin θ1+cos θ)
Answer:
Given: tanθ=12Since, tanθ=PB⇒P=1 and B=2Using Pythagoras theorem,P2+B2=H2⇒12+22=H2⇒H2=1+4⇒H2=5⇒H=5–√Therefore,sinθ=PH=15√cosθ=BH=25√Now,(cosθsinθ+sinθ1+cosθ)=(25√15√+15√1+25√) =⎛⎝21+15√5√+25√⎞⎠ =(21+15√+2) =(2+(15√+2×5√−25√−2)) =(2+(5√−25−4)) =(2+5–√−2) =5–√Hence, (cosθsinθ+sinθ1+cosθ)=5–√.
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Question 14:
If sinα=12, prove that (3cosα−4cos3α)=0.
Answer:
LHS=(3cosα−4cos3α)=cosα(3−4cos2α)=1−sin2α−−−−−−−−√[3−4(1−sin2α)]=1−(12)2−−−−−−−√[3−4(1−(12)2)]=11−14−−−−−√[3−4(11−14)]=34−−√[3−4(34)]=34−−√[3−3]=34−−√[0]=0=RHS
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Question 15:
If 3 cot θ = 2, show that (4sinθ−3cosθ2sinθ+6cosθ)=13.
Answer:
It is given that cot θ = 23.
LHS = 4 sinθ − 3 cosθ2 sinθ + 6 cosθ
Dividing the above expression by sin θ, we get:
4 − 3 cot θ2 + 6 cot θ [∵ cot θ = cosθsinθ]
Now, substituting the values of cot θ in the above expression, we get:
4 − 3(23)2 + 6(23)= 4 − 22 + 4 = 26=13
i.e., LHS = RHS
Hence proved.
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Question 16:
If sec θ = 178 then prove that 3−4sin2θ4cos2θ−3=3−tan2θ1−3tan2θ.
Answer:
It is given that sec θ = 178.
Let us consider a right △ABC right angled at B and ∠C=θ.
We know that cos θ = 1sec θ= 817 = BCAC
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 − BC2 = (17k)2 − (8k)2
⇒ AB2 = 289k2 − 64k2 = 225k2
⇒ AB = 15k.
Now, tan θ = ABBC = 158 and sin θ = ABAC = 15k17k= 1517
The given expression is 3 − 4sin2θ4cos2θ− 3 = 3 − tan2θ1 − 3tan2θ.
Substituting the values in the above expression, we get:
LHS= 3 − 4(1517)24(817)2 − 3 = 3 − 900289256289− 3 = 867−900256−867= −33−611=33611
RHS = 3−(158)21−3(158)2=3−225641−67564=192−22564−675=−33−611=33611
∴ LHS = RHS
Hence proved.
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Question 17:
If tan θ = 2021, show that(1−sinθ+cosθ)(1+sinθ+cosθ)=37.
Answer:
Let us consider a right △ABC right angled at B and ∠C=θ.
Now, we know that tan θ = ABBC = 2021
So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒ AC = 29k
Now, sin θ = ABAC = 2029 and cos θ = BCAC = 2129
Substituting these values in the given expression, we get:
LHS=1 − sinθ + cosθ1 + sinθ + cosθ= 1 − 2029 + 21291 + 2029 + 2129= 29 − 20 + 212929 + 20 + 2129= 3070 = 37 = RHS
∴ LHS = RHS
Hence proved.
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Question 18:
If tan θ = 17√ then prove that (cosec2θ+sec2θcosec2θ−sec2θ)=43.
Answer:
Let us consider a right △△ABC, right-angled at B and .
Now it is given that tan = = .
So, if AB = k, then BC = k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + (k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 2k
Now, finding out the values of the other trigonometric ratios, we have:
sin =
cos =
∴ cosec = and sec =
Substituting the values of cosec and sec in the given expression, we get:
i.e., LHS = RHS
Hence proved.
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Question 19:
If sinθ=34, show that cosec2θ−cot2θsec2θ−1−−−−−−−−−√=7√3.
Answer:
LHS=cosec2θ−cot2θsec2θ−1−−−−−−−−−√=1tan2θ−−−−√=cot2θ−−−−−√=cotθ=cosec2θ−1−−−−−−−−−√=(1sinθ)2−1−−−−−−−−−√=(1(34))2−1−−−−−−−−−−⎷=(43)2−1−−−−−−−√=169−1−−−−−√=16−99−−−−√=79−−√=7√3=RHS
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Question 20:
If 3 tan A = 4 then prove that
(i) sec A−cosec Asec A+cosec A−−−−−−−−−−√=17√
(ii) 1−sin A1+cos A−−−−−−√=122√
Answer:
(i)
LHS=secθ−cosecθsecθ+cosecθ−−−−−−−−√=(1cosθ−1sinθ)(1cosθ+1sinθ)−−−−−−−−−⎷=(sinθ−cosθsinθ cosθ)(sinθ+cosθsinθ cosθ)−−−−−−−−⎷=(sinθ−cosθsinθ)(sinθ+cosθsinθ)−−−−−−−−⎷=(sinθsinθ−cosθsinθ)(sinθsinθ+cosθsinθ)−−−−−−−−−⎷=1−cotθ1+cotθ−−−−−√=(1−34)(1+34)−−−−−⎷
=(14)(74)−−−−⎷=17−−√=17√=RHS
(ii)
Given: 3tanA=4⇒tanA=43Since, tanA=PB⇒P=4 and B=3Using Pythagoras theorem,P2+B2=H2⇒42+32=H2⇒H2=16+9⇒H2=25⇒H=5Therefore,sinA=PH=45cosA=BH=35Now,1−sinA1+cosA−−−−−−√=1−451+35−−−−⎷ =5−455+35−−−⎷ =1585−−⎷ =18−−√ =122√Hence, 1−sinA1+cosA−−−−−−√=122√.
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Question 21:
If cot θ=158 then evaluate (1+sin θ) (1−sin θ)(1+cos θ) (1−cos θ).
Answer:
Given: cotθ=158Since, cotθ=BP⇒P=8 and B=15Using Pythagoras theorem,P2+B2=H2⇒82+152=H2⇒H2=64+225⇒H