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Question

(i) (sec2θ1)cot2θ=1 (ii) (sec2θ1)(cosec2θ1)=1

(iii) (1cos2θ)sec2θ=tan2θ

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Solution

(i) (sec2θ1)cot2θ=1L.H.S=(sec2θ1)cot2θ=tan2θ×cot2θ=(1cot2θ)×(cotθ)=1=R.H.S

(ii) (sec2θ1)(cosec2θ1)=1L.H.S=(sec2θ1)(cosec2θ1)=(tan2θ)(1sin2θ1)=(tan2θ)(1sin2θsin2θ)=(sin2θcos2θ)(cos2θsin2θ)=1=R.H.S

(iii) (1cos2θ)sec2θ=tan2θL.H.S=(1cos2θ)sec2θ=(sin2θ)(1cos2θ)=sin2θcos2θ=tan2θ=R.H.S


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