Question

sin^-1 (√1-cosx)/ V1+cosx)

please give correct ans

don't give st.upid answer

Answer 1

Answer:

$\sin^{-1}(tan\frac{x}{2})$

Step-by-step explanation:

Given Question :

$\sin^{-1}\bigg(\dfrac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}} \bigg)$

To find :

The value of the given expression

Solution ;

$\sin^{-1}\bigg(\dfrac{\sqrt{1-\cos x}}{\sqrt{1+\cos x}} \bigg)$

Multiply and divide by $\sqrt{1-\cos x}$ inside the bracket

$: \longmapsto \sin^{-1}\bigg(\dfrac{\sqrt{1-\cos x} \times \sqrt{1-\cos x}}{\sqrt{1+\cos x }\times \sqrt{1-\cos x}} \bigg)$

$: \longmapsto \sin^{-1}\bigg(\dfrac{1-\cos x}{\sqrt{1-\cos ^2x} } \bigg)$

$: \longmapsto \sin^{-1}\bigg(\dfrac{1-\cos x}{\sqrt{\sin ^2x} } \bigg)$

$\text{We know that, } \boxed {\sin^2x + cos^2x = 1}$

$: \longmapsto \sin^{-1}\bigg(\dfrac{2\sin^2{\frac{x}{2}}}{\sin x} } \bigg)$

$\text{We know that, } \boxed {1-cosx =2sin^2\frac{x}{2} }$

$: \longmapsto \sin^{-1}\bigg(\dfrac{2\sin^2{\frac{x}{2}}}{2\sin \frac{x}{2}\cos \frac{x}{2}} } \bigg)$

$\text{We know that, } \boxed {sinx=2sin\frac{x}{2} \cos\frac{x}{2}}$

$: \longmapsto \sin^{-1}\bigg(\dfrac{\sin{\frac{x}{2}}}{\cos \frac{x}{2}} } \bigg)$

$: \longmapsto \underline {\boxed{\sin^{-1}(tan\frac{x}{2})}}$

This is the final answer.

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Hope it helps!!