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S Chand Class 10 CHAPTER 16 TRIGONOMETRY Exercise 16 A

 Exercise 16 A

Question 1

Ans: 1cos2θsin2θ=1
LHS=1cos2θsin2θ [1cos2θ=sin2θ]
=sin2θsin2θ=1=RHS

Question 2

Ans: 1sin2θcos2θ=1
L.H.S =1sin2θcos2θ[1sin2θ=cos2θ]
=cos2θcos2θ=1=RHS

Question 3

Ans: sinAcotA=cosA
L.H.S=sinAcotA
=sinAcosAsinA  [cotA=cosAsinA]
=cosA=R.H.S

Question 4

Ans: 1cos2θtan2θ=1
L.H.S=1cos2θtan2θ[1cos2θ=sec2θ]
=sec2θtan2θ
=1=RHS

Question 5

Ans: tan2cos2A=1cos2A

LHS=tan2Acos2A [tan2A=sin2Acos2A]
=sin2Acos2Acos2A
=sin2A=1cos2A=RHS

Question 6

Ans: tanθ=sinθ1sin2θ
 R.H.S=sinθ1sin2θ [1sin2θ=cos2θ]
=sinθcos2θ
=sinθcosθ=tanθ= L.H.S

Question 7

Ans: 1+cosθsin2θ=11cosθ
L.H.S =1+cosθsin2θ=1+cosθ1cos2θ[sin2θ=1cos2θ]
=1+cosθ(1cosθ)(1cosθ)[a2b2=(ab)(a+b)
=11cosθ= R.H.S

Question 8

Ans: cot2θ(1cos2θ)=cos2θ[cotθ=cosθsinθ]
LHS=cot2θ(1cos2θ) [1cos2θ=sin2θ]
=cos2θsin2θ×sin2θ
=cos2θ = R.H.S

Question 9

Ans: tan2θ(1sin2θ)=sin2θ
L.H.S =tan2θ(1sin2θ)
=sin2θcos2θ×cos2θ[tanθ=sinθcosθ][1sin2θ=cos2θ]
=sin2θ=RHS

Question 10

Ans: (1sin2θ)sic2θ=1
L.H.S =(1sin2θ)sec2θ[1sin2θ=cos2θ]
=cos2θsec2θ
=1= R.H.S

Question 11

Ans: 
(1cos2θ)cosec2θ=1LHS=(1cos2θ)cosec2θ1cos2θ=sin2θcosecθ=1sinθ=sin2θ×1sin2θ1=RHS

Question 12

Ans: sin2θ+11+tan2θ=1 
LHS=sin2θ+11+tan2θ
=sin2θ+1sec2θ[1+tan2θ=sec2θ]
sin2θ+cos2θ [1sec2θ=cos2θ]
RH.S=1

Question 13

Ans:  cos2θ+11+cot2θ=1
LHS=cos2θ+11+cot2θ
[1+cot2θ=cossc2θ]
=cos2θ+1cosec2θ
{1cosec=sinθ
=cos2θ+sin2θ
RHS=1

Question 14

Ans: sin2θ+cos2θsec2θtan2θ=1
LHS=sin2θ+cos2θsec2θtan2θ=1
sin2θ+cos2θ=1
sec2θtan2θ=1
RHS=1

Question 15

Ans: [cos2Asin2A+1]tan2A=1cos2A
LHS=[cos2Asin2A+1]tan2θ
cosAsinA=cotA
=(cot2A+1)tan2θ
cos2A+1=cosec2A
=cosec2A×tan2A
cosecA=1sinA
=1sin2A×sin2Acos2A
RHS=1cos2A

Question 16

Ans: sin2θ+sin2θcos2θ=sin2θ.
LHS=sin4θ+sin2θcos2θ=sin2θ(sin2θ+cos2θsin2θ+cos2θ=1=sin2θ×1RHS=sin2θ

Question 17

Ans: sin4θ+2sin2θcos2θ+cos4θ=1
 L.H.S =sin2θ+2sin2θ(a)2θ+cos4θa2+2ab+b2=(a+b)2=(sin2θ)2+2sin2θcos2θ+(cos2θ)2=(sin2θ+cos2θ)2sin2θ+cos2θ=1=(1)2
R.H.S= 1

Question 18

Ans: sin4Acosec2A+cos4Asec2A=1
L.H.S =sin4Acosec2A+cos4Asec2A
=sin4A×1sin2A+cos4A×1cos2A  [cosecθ=1sinθ]
sin2θ+cos2θ=1 [secθ=1cosθ]
R.H.S =1

Question 19

Ans: sin2Acot2A+cos2Atan2A=1
LHS=sin2Acot2A+cos2Atan2A=sin2A×cos2Asin2A+cos2A×sin2Acos2AcosAsinA=cotθ,sinθcosθ=tanθ=cos2A+sin2A=1 R.H.S =1

Question 20

Ans: tanθ+cotθ=secθcosecθ.
LHS=tanθ+cotθ
[tanθ=sinθcosθ,cotθ=cosθsinθ]
=sinθcosθ+cosθsinθ
=sin2θ+cos2θcosθsinθ
=[sin2θ+cos2θsinθ]
=1cosθsinθ
[1cosθ=secθ,1sinθ=cosecθ]
RHS=Secθcosecθ

Question 21

Ans:  (tanA+cotA)sinAcosA=1
LHS=(tanA+cotA)(sinAcosA)
[tanθ=sinθcosθ,cotθcosθsinθ]=[sinAcosA+cosAsinA]sinAcosA.=sin2A+cos2AcosAsinA×sinAcosA[sin2θ+Cos2θ=1]
=1cosAsinA×sinAcosA
R.H.S = 1

Question 22

Ans:   1+cosθsin2θsinθ(1+cosθ)=cotθLHS=1+cosθsin2θsinθ(1+cosθ)=1sin2θ+cosθsinθ(1+cosθ)[1sin2θ=cos2θ]=cosθ+cos2θsinθ(1+cosθ
=cosθ(1+cosθ)sinθ(1+cosθ)
=cosθsinθ
[cosθsinθ=cotθ]
RHS=cotθ

Question 23

Ans: 
 11cosθ+11+cosθ=2cosec2θ
 L.H.S =11cosθ+11+cosθ=(a+b)(ab)=a2b2=1+cosθ+1cosθ(1cosθ)(1+cosθ)=21cos2θ
[1cos2θ=sin2θ]=2sin2θ.[1sinθ=cosecθ]
RHS=2cosec2θ

Question 24

Ans:  1tan2θcot2θ1=tan2θ.
LHS=1tan2θcot2θ1=1tan2θ1tan2θ1 
=1tan2θ1tan2θtan2θ
=tan2θ=R.H.S

Question 25

Ans:  1secθ+tanθ=1sinθcosθ
LHS=1Secθ+tanθ
=secθtanθ(secθ+tanθ)(secθtanθ)
=secθtanθsec2θtan2θ[sec2θtan2θ=1]
=secθtanθ1
=1cosθsinθcosθ
R.H.S=1sinθcosθ

Question 26

Ans: (cosecAsinA)(secAcosA)(tanA+cotA)=1
L.H.S =(cosecAsinA)(secAcosA)(tanA+cotA)
[cosecA=1sinA,secA=1cosA,tanA=sinAcosA
cotA=cosAsinA]
=[1sinAsinA][1cosAcosA][sinAcosA+cosAsinA]
=1sin2AsinA×1cos2AcosA×sin2A+cos2AsinAcosA
=[sin2A+cos2A=1]sinA×sin2AcosA×1sinAcosA
=cos2Asin2Asin2Acos2A
R.H.S =1

Question 27

Ans: 1+sinθ1sinθ=(secθ+tanθ)2
 L.H.S. =1+sinθ1sinθ=(1+sinθ)(1+sinθ)(1sinθ)(1+sinθ)=(1+sinθ)21sin2θ
[1sin2θ=cos2θ]=(1+sinθ)2cos2θ=[1+sinθcosθ]2=[1cosθ+sinθcosθ]2
1cosθ=secθ,sinθconθ=tanθ
RHS=(secθ+tanθ)2

Question 28

Ans: 1+cosθ1cosθ=(cosecθ+cotθ)2
LHS=1+cosθ1cosθ=(1+cosθ)(1+cosθ)(1cosθ)(1+cosθ)[1cos2θ=sin2θ]
=(1+cos(θ)21cos2θ
=(1+cosθ)2sin2θ
=[1+cosθsinθ]2
=[1sinθ+cosθsinθ]2
(1sinθ=cosecθ,cosθsinθ=cotθ)

RHS=(cosθ+θ+cotθ)2

Question 29

Ans: cotθ+cosecθ1cotθcosecθ+1=cosecθ+cotθ1+cosθsinθ
LHS=cotθ+cosecθ1cotθcosecθ+1
=cotA+cosecθ(cosec2θcot2θ)cotθcosecθ+1
=cotθ+cosecθ+(cot2θcosec2θ)cotθcosecθ+1
=(cotθ+cosecθ)+cotθ+cosecθ)(cotθcosecθ)cotθcosecθ+1
=(cotθ+cosecθ)(1+cotθcosecθ)cotθcosecθ+1
=(cotc+cosecθ)(cotθcosecθ+1)(cotθcosecθ+1
=cotθ+cosecθ
=cosecθ+cotθ
=1sinθ+cosθsinθ.
RHS=1+cosθsinθ

Question 30

Ans: tanθ+cotθ=2
(tanθ+cotθ)2=4 [Square both side]
tan2θ+cot2θ+2tanθcotθ=4
[tanθcotθ=1]
tan2θ+cot2θ+2×1=4tan2θ+cot2θ+2=4tan2θ+cot2θ=42=2tan2θ+cot2θ=2

Question 31

Ans:  a2+b2=(sinθ+cosθ)2+(sinθcosθ)2
=sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ2sinϵcosθ
=2sin2θ+2cos2θ
=2(sin2θ+(cos2θ)
=2×1
=2



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