Question

The value of the expression $\frac{{\sin }^{3}x}{1+\cos x}+\frac{{\cos }^{3}x}{1-\sin x}$ is/are

• A. $\sqrt{2}\cos \left(\frac{\pi }{4}-x\right)$
• B. $\sqrt{2}\cos \left(\frac{\pi }{4}+x\right)$
• C. $\sqrt{2}\sin \left(\frac{\pi }{4}-x\right)$
• D. None of these

Answer 1

Answer: (A)

Let $A=\frac{{\sin }^{3}x}{1+\cos x}+\frac{{\cos }^{3}x}{1-\sin x}$ $=\frac{({\sin }^{3}x+{\cos }^{3}x)+({\cos }^{4}x-{\sin }^{4}x)}{(1+\cos x)(1-\sin x)}$ $=\frac{\begin{array}{rl}& [(\sin x+\cos x)(1-\sin x.\cos x)]\\ & +[(\cos x+\sin x)(\cos x-\sin x)]\end{array}}{(1+\cos x)(1-\sin x)}$ $=\frac{(\sin x+\cos x)[(1-\sin x.\cos x)(\cos x-\sin x)]}{1+\cos x-\sin x-\sin x.\cos x}$ $=\sin x+\cos x$ $=\sqrt{2}\left\{\frac{1}{\sqrt{2}}.\sin x+\frac{1}{\sqrt{2}}.\cos x\right\}$ $=\sqrt{2}\left\{\cos \frac{\pi }{4}.\sin x+\sin \frac{\pi }{4}.\cos x\right\}$ $=\sqrt{2}\sin \left(\frac{\pi }{4}+x\right)$ $=\sqrt{2}\cos \left(\frac{\pi }{4}-x\right)$