1.

(i) sin6θ+cos6θ=1−3sin2θcos2θ (ii) sin4θ−cos4θ=sin2θ−cos2θ (iii) cosec4θ−cosec2θ=cot4θ+cot2θ

Answer»

(i) sin6θ+cos6θ=13sin2θcos2θ

(ii) sin4θcos4θ=sin2θcos2θ

(iii) cosec4θcosec2θ=cot4θ+cot2θ



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