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Question

(i) sin6θ+cos6θ=13sin2θcos2θ

(ii) sin4θcos4θ=sin2θcos2θ

(iii) cosec4θcosec2θ=cot4θ+cot2θ

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Solution

i right parenthesis w e space k n o w space t h a t sin squared theta plus cos squared theta equals 1 open parentheses bold sin to the power of bold 2 bold italic theta bold plus bold cos to the power of bold 2 bold italic theta close parentheses cubed equals 1 cubed sin to the power of 6 theta plus cos to the power of 6 theta plus 3 sin to the power of 4 theta cos squared theta plus 3 sin squared theta cos to the power of 4 theta equals 1 sin to the power of 6 theta plus cos to the power of 6 theta plus 3 sin squared theta cos squared theta open parentheses sin squared theta plus cos squared theta close parentheses equals 1 sin to the power of 6 theta plus cos to the power of 6 theta plus 3 sin squared theta cos squared theta equals 1 sin to the power of 6 theta plus cos to the power of 6 theta equals 1 minus 3 sin squared theta cos squared theta

(ii) sin4θcos4θ=sin2θcos2θLHS=sin4θcos4θ=(sin2θ)2(cos2θ)2=(sin2θ+cos2θ)(sin2θcos2θ)=sin2θcos2θ=RHS

(iii)cosec4θcot4θ=cosec2θ+cot2θLHS=cosec4θcot4θ=(cosec2θ)2(cot2θ)2=(cosec2θ+cot2θ)(cosec2θcot2θ)=cosec2θ+cot2θ=RHS


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