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Question

# Prove the following trigonometric identities.(i) 1+cosθ+sinθ1+cosθ−sinθ=1+sinθcosθ(ii) sinθ−cosθ+1sinθ+cosθ−1=1secθ−tanθ(iii) cosθ−sinθ+1cosθ+sinθ−1=cosecθ+cotθ(iv) (sinθ+cosθ)(tanθ+cotθ)=secθ+cosecθ

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Solution

## (i) We have to prove the following identity-1+cosθ+sinθ1+cosθ−sinθ=1+sinθcosθConsider the LHS.1+cosθ+sinθ1+cosθ−sinθ=[1+cosθ+sinθ1+cosθ−sinθ][1+cosθ+sinθ1+cosθ+sinθ]=(1+cosθ+sinθ)2(1+cosθ)2−sin2θ=(1+cosθ)2+sin2θ+2(1+cosθ)sinθ1+cos2θ+2cosθ−sin2θ=1+cos2θ+2cosθ+sin2θ+2sinθ+2sinθcosθ1+cos2θ+2cosθ−sin2θ=2+2(cosθ+sinθ+sinθcosθ)2cos2θ+2cosθ [∵cos2θ+sin2θ=1]=2(1+cosθ+sinθ+sinθcosθ)2cosθ(1+cosθ)=2[1+cosθ+sinθ(1+cosθ)]2cosθ(1+cosθ)=2(1+cosθ)(1+sinθ)]2cosθ(1+cosθ)=(1+sinθ)cosθ= RHSHence proved.(ii) We have to prove the following identity-sinθ−cosθ+1sinθ+cosθ−1=1secθ−tanθConsider the LHS.sinθ−cosθ+1sinθ+cosθ−1=[sinθ−cosθ+1sinθ+cosθ−1][sinθ−cosθ+1sinθ+cosθ+1]=(sinθ+1)2−cos2θ(sinθ+cosθ)2−1=sin2θ+1+2sinθ−cos2θsin2θ+cos2θ+2sinθcosθ−1=2sin2θ+2sinθ2sinθcosθ=2sinθ(1+sinθ)2sinθcosθ=(1+sinθ)cosθ=[(1+sinθ)cosθ][(1−sinθ)(1−sinθ)]=[(1−sin2θ)cosθ(1−sinθ)]=[(cosθ)(1−sinθ)]=[1(secθ−tanθ)]=RHS [Divide numerator and denominator by cosθ] Hence proved.(iii) We have to prove the following identity-cosθ−sinθ+1cosθ+sinθ−1=cosecθ+cotθConsider the LHS.cosθ−sinθ+1cosθ+sinθ−1=cosθ−sinθ+1cosθ+sinθ−1×cosθ+sinθ+1cosθ+sinθ+1=(cosθ+1)2−sin2θ(cosθ+sinθ)2−12=cos2θ+1+2cosθ−sin2θcos2θ+sin2θ+2cosθsinθ−1=cos2θ+1+2cosθ−(1−cos2θ)1+2cosθsinθ−1 =2cos2θ+2cosθ2cosθsinθ=2cosθ(cosθ+1)2 cosθsinθ=cosθ+1sinθ=cosθsinθ+1sinθ=cotθ+cosecθ= RHSHence proved.(iv) (sinθ+cosθ)(tanθ+cotθ)=secθ+cosecθConsider the LHS.(sinθ+cosθ)(tanθ+cotθ)=(sinθ+cosθ)(sinθcosθ+cosθsinθ)=(sinθ+cosθ)(sin2θ+cos2θsinθcosθ) [∵sin2θ+cos2θ=1]=(sinθ+cosθ)1sinθcosθ=(sinθ+cosθ)(sinθcosθ)=1cosθ+1sinθ=secθ+cosecθ= RHSHence proved.

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