Question

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Answer 1

$\\ \\{\Large{\underline{\underline{\bigstar\:\:\:\:{\pmb{\sf{Solution\::}}}}}}}$

$\sf\longmapsto \:\:\:\: \frac{1- \sin\theta\cos\theta}{\cos \theta\big(\sec \theta-\cosec\theta\big)}\cdot\frac{\sin^2\theta-\cos^2\theta}{\sin^3+\cos^3\theta}$

$\sf\longmapsto \:\:\:\: \frac{\cancel{\big(1-\sin\theta\cos\theta\big)}}{1-\cos\theta\cosec\theta}\cdot\frac{\big(\sin\theta-\cos\theta\big)\cancel{\big(\sin\theta+\cos\theta\big)}}{{\cancel{\big({\sin\theta+\cos\theta\big)}\big(\sin^2\theta+\cos^2\theta-\sin\theta\cos\theta\big)}}}\:\:\:\:\:\:\:\big[\because \: \bf {a^3+b^3=(a+b)(a^2-ab+b^2)}\big]$$\sf\longmapsto \:\:\:\: \frac{\sin\theta\cancel{\big({\sin\theta-\cos\theta}}\big)}{\cancel{\big({\sin\theta-\cos\theta}\big)}}$

$\tt\longmapsto \:\:\:\:\blue{\sin\theta}\:\:\:\:\:\:\bf proved.$

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