Exercise 16 A
Question 1
Ans: 1−cos2θsin2θ=1
L⋅H⋅S=1−cos2θsin2θ [∵1−cos2θ=sin2θ]
=sin2θsin2θ=1=R⋅H⋅S
Question 2
Ans: 1−sin2θcos2θ=1
L.H.S =1−sin2θcos2θ[1−sin2θ=cos2θ]
=cos2θcos2θ=1=R⋅H⋅S
Question 3
Ans: sinA⋅cotA=cosA
L.H.S=sinA⋅cotA
=sinA⋅cosAsinA [cotA=cosAsinA]
=cosA=R.H.S
Question 4
Ans: 1cos2θ−tan2θ=1
L.H.S=1cos2θ−tan2θ[1cos2θ=sec2θ]
=sec2θ−tan2θ
=1=R⋅H⋅S
Question 5
Ans: tan2cos2A=1−cos2A
L⋅H⋅S=tan2Acos2A [tan2A=sin2Acos2A]
=sin2Acos2Acos2A
=sin2A=1−cos2A=R⋅H⋅S
Question 6
Ans: tanθ=sinθ√1−sin2θ
⇒R.H.S=sinθ√1−sin2θ [1−sin2θ=cos2θ]
=sinθ√cos2θ
=sinθcosθ=tanθ= L.H.S
Question 7
Ans: 1+cosθsin2θ=11−cosθ
L.H.S =1+cosθsin2θ=1+cosθ1−cos2θ[sin2θ=1−cos2θ]
=1+cosθ(1−cosθ)(1−cosθ)[a2−b2=(a−b)(a+b)
=11−cosθ= R.H.S
Question 8
Ans: cot2θ(1−cos2θ)=cos2θ[cotθ=cosθsinθ]
L⋅H⋅S=cot2θ(1−cos2θ) [1−cos2θ=sin2θ]
=cos2θsin2θ×sin2θ
=cos2θ = R.H.S
Question 9
Ans: tan2θ(1−sin2θ)=sin2θ
L.H.S =tan2θ(1−sin2θ)
=sin2θcos2θ×cos2θ[tanθ=sinθcosθ][1−sin2θ=cos2θ]
=sin2θ=R⋅H⋅S
Question 10
Ans: (1−sin2θ)sic2θ=1
L.H.S =(1−sin2θ)sec2θ[1−sin2θ=cos2θ]
=cos2θ⋅sec2θ
=1= R.H.S
Question 11
Ans:
(1−cos2θ)cosec2θ=1L⋅H⋅S=(1−cos2θ)cosec2θ1−cos2θ=sin2θcosecθ=1sinθ=sin2θ×1sin2θ1−=R⋅H⋅S
Question 12
Ans: sin2θ+11+tan2θ=1
L⋅H⋅S=sin2θ+11+tan2θ
=sin2θ+1sec2θ[1+tan2θ=sec2θ]
≐sin2θ+cos2θ [1sec2θ=cos2θ]
R⋅H.S=1
Question 13
Ans: cos2θ+11+cot2θ=1
L⋅HS=cos2θ+11+cot2θ
[1+cot2θ=cossc2θ]
=cos2θ+1cosec2θ
{1cosec=sinθ
=cos2θ+sin2θ
R⋅H⋅S=1
Question 14
Ans: sin2θ+cos2θsec2θ−tan2θ=1
L⋅H⋅S=sin2θ+cos2θsec2θ−tan2θ=1
sin2θ+cos2θ=1
sec2θ−tan2θ=1
R⋅H⋅S=1
Question 15
Ans: [cos2Asin2A+1]tan2A=1cos2A
L⋅H⋅S=[cos2⋅Asin2A+1]tan2θ
cosAsinA=cotA
=(cot2A+1)tan2θ
cos2A+1=cosec2A
=cosec2A×tan2A
cosecA=1sinA
=1sin2A×sin2Acos2A
R⋅H⋅S=1cos2A
Question 16
Ans: sin2θ+sin2θcos2θ=sin2θ.
L⋅H⋅S=sin4θ+sin2θcos2θ=sin2θ(sin2θ+cos2θsin2θ+cos2θ=1=sin2θ×1R⋅H⋅S=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
L⋅H⋅S=sin2Acot2A+cos2Atan2A=sin2A×cos2Asin2A+cos2A×sin2Acos2AcosAsinA=cotθ,sinθcosθ=tanθ=cos2A+sin2A⋅=1 R.H.S =1
Question 20
Ans: tanθ+cotθ=secθ⋅cosecθ⋅.
L⋅H⋅S=tanθ+cotθ
[tanθ=sinθcosθ,cotθ=cosθsinθ]
=sinθcosθ+cosθsinθ
=sin2θ+cos2θcosθsinθ
=[sin2θ+cos2θsinθ]
=1cosθsinθ
[1cosθ=secθ,1sinθ=cosecθ]
R⋅H⋅S=Secθ⋅cosecθ
Question 21
Ans: (tanA+cotA)sinAcosA=1
L⋅H⋅S=(tanA+cotA)(sinAcosA)
[tanθ=sinθcosθ,cotθ≐cosθsinθ]=[sinAcosA+cosAsinA]sinAcosA.=sin2A+cos2AcosAsinA×sinAcosA[sin2θ+Cos2θ=1]
=1cosAsinA×sinAcosA
R.H.S = 1
Question 22
Ans: 1+cosθ−sin2θsinθ(1+cosθ)=cotθL⋅H⋅S=1+cosθ−sin2θsinθ(1+cosθ)=1−sin2θ+cosθsinθ(1+cosθ)[1−sin2θ=cos2θ]=cosθ+cos2θsinθ(1+cosθ
=cosθ(1+cosθ)sinθ(1+cosθ)
=cosθsinθ
[cosθsinθ=cotθ]
R⋅H⋅S=cotθ
Question 23
Ans:
11−cosθ+11+cosθ=2cosec2θ
L.H.S =11−cosθ+11+cosθ=(a+b)(a−b)=a2−b2=1+cosθ+1−cosθ(1−cosθ)(1+cosθ)=21−cos2θ
[1−cos2θ=sin2θ]=2sin2θ.[1sinθ=cosecθ]
R⋅H⋅S=2cosec2θ
Question 24
Ans: 1−tan2θcot2θ−1=tan2θ.
L⋅H⋅S=1−tan2θcot2θ−1=1−tan2θ1tan2θ−1
=1−tan2θ1−tan2θtan2θ
=tan2θ=R.H.S
Question 25
Ans: 1secθ+tanθ=1−sinθcosθ
L⋅H⋅S=1Secθ+tanθ
=secθ−tanθ(secθ+tanθ)(secθ−tanθ)
=secθ−tanθsec2θ−tan2θ[sec2θ−tan2θ=1]
=secθ−tanθ1
=1cosθ−sinθcosθ
R.H.S=1−sinθcosθ
Question 26
Ans: (cosecA−sinA)⋅(secA−cosA)(tanA+cotA)=1
L.H.S =(cosecA−sinA)⋅(secA−cosA)(tanA+cotA)
[cosecA=1sinA,secA=1cosA,tanA=sinAcosA
cotA=cosAsinA]
=[1sinA−sinA][1cosA−cosA][sinAcosA+cosAsinA]
=1−sin2AsinA×1−cos2AcosA×sin2A+cos2AsinAcosA
=[sin2A+cos2A=1]sinA×sin2AcosA×1sinAcosA
=cos2Asin2Asin2Acos2A
R.H.S =1
Question 27
Ans: 1+sinθ1−sinθ=(secθ+tanθ)2
L.H.S. =1+sinθ1−sinθ=(1+sinθ)(1+sinθ)(1−sinθ)(1+sinθ)=(1+sinθ)21−sin2θ
[1−sin2θ=cos2θ]=(1+sinθ)2cos2θ=[1+sinθcosθ]2=[1cosθ+sinθcosθ]2
1cosθ=secθ,sinθconθ=tanθ
R⋅H⋅S=(secθ+tanθ)2
Question 28
Ans: 1+cosθ1−cosθ=(cosecθ+cotθ)2
L⋅H⋅S=1+cosθ1−cosθ=(1+cosθ)(1+cosθ)(1−cosθ)(1+cosθ)⋅[1−cos2θ=sin2θ]
=(1+cos(θ)21−cos2θ
=(1+cosθ)2sin2θ
=[1+cosθsinθ]2
=[1sinθ+cosθsinθ]2
(1sinθ=cosecθ,cosθsinθ=cotθ)
R⋅H⋅S=(cosθ+θ+cotθ)2
Question 29
Ans: cotθ+cosecθ−1cotθ−cosecθ+1=cosecθ+cotθ−1+cosθsinθ
L⋅H⋅S=cotθ+cosecθ−1cotθ−cosecθ+1
=cotA+cosecθ−(cosec2θ−cot2θ)cotθ−cosecθ+1
=cotθ+cosecθ+(cot2θ−cosec2θ)cotθ−cosecθ+1
=(cotθ+cosecθ)+cotθ+cosecθ)(cotθ−cosecθ)cotθ−cosecθ+1
=(cotθ+cosecθ)(1+cotθ−cosecθ)cotθ−cosecθ+1
=(cotc+cosecθ)(cotθ−cosecθ+1)(cotθ−cosecθ+1
=cotθ+cosecθ
=cosecθ+cotθ
=1sinθ+cosθsinθ.
R⋅H⋅S=1+cosθsinθ
Question 30
Ans: tanθ+cotθ=2
(tanθ+cotθ)2=4 [Square both side]
tan2θ+cot2θ+2tanθcotθ=4
[tanθ⋅cotθ=1]
tan2θ+cot2θ+2×1=4tan2θ+cot2θ+2=4tan2θ+cot2θ=4−2=2tan2θ+cot2θ=2
Question 31
Ans: a2+b2=(sinθ+cosθ)2+(sinθ−cosθ)2
=−sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ−2sinϵcosθ
=2sin2θ+2cos2θ
=2(sin2θ+(cos2θ)
=2×1
=2
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