Question

If $$\frac{\pi }{2}<x<\pi ,\text{ then }\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$$ is equal to

पर्याय

• 2 sec x

• −2 sec x

• sec x

• −sec x

Answer 1

−2 sec x

$$\sqrt{\frac{1-\sin x}{1+\sin x}}+\sqrt{\frac{1+\sin x}{1-\sin x}}$$
$$=\sqrt{\frac{(1-\sin x)(1-\sin x)}{(1+\sin x)(1-\sin x)}}+\sqrt{\frac{(1+\sin x)(1+\sin x)}{(1-\sin x)(1+\sin x)}}$$
$$=\sqrt{\frac{{(1-\sin x)}^2}{1-{\sin }^{2}x}}+\sqrt{\frac{{(1+\sin x)}^2}{1-{\sin }^{2}x}}$$
$$=\sqrt{\frac{{(1-\sin x)}^2}{{\cos }^{2}x}}+\sqrt{\frac{{(1+\sin x)}^2}{{\cos }^{2}x}}$$
$$=\frac{(1-\sin x)}{-\cos x}+\frac{(1+\sin x)}{-\cos x}[\frac{\pi }{2}<x<\pi ,\text{so }\cos x\text{ will be negative . }]$$
$$=-(\sec x-\tan x)-(\sec x+\tan x)$$
$$=-2\sec x$$