Question

The value of $\frac{2\left(\sin 2x+2{\cos }^{2}x-1\right)}{\cos x-\sin x-\cos 3x+\sin 3x}$ is
(a) cos x
(b) sec x
(c) cosec x
(d) sin x

Answer 1

(c) cosec x

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \frac{2\left(\sin 2x+2{\cos }^{2}x-1\right)}{\cos x-\sin x-\cos 3x+\sin 3x} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{\cos x-\sin x-4{\cos }^{3}x+3\cos x+3\sin x-4{\sin }^{3}x} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{4\cos x-4{\cos }^{3}x+2\sin x-4{\sin }^{3}x} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{4\cos x\left(1-{\cos }^{2}x\right)+2\sin x\left(1-2{\sin }^{2}x\right)} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{4\cos x{\sin }^{2}x+2\sin x\cos 2x} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{2\times 2\sin x\cos x\sin x+2\sin x\cos 2x} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{2\sin 2x\sin x+2\sin x\cos 2x} \\ =\frac{2\left(\sin 2x+\cos 2x\right)}{2\sin x\left(\sin 2x+\cos 2x\right)} \\ =\frac{1}{\sin x} \\ =\mathrm{cosec}x \end{gathered}$$