જો frac{{5π }}{2} < x < 3π ,હોય તો frac{{√ {1

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3.Trigonometrical Ratios, Functions and Identities

normal

જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =

$-cot \frac{x}{2}$

$cot \frac{x}{2}$

$ tan \frac{x}{2}$

$-tan \frac{x}{2}$

Solution

On rationalizing ; we get

$\frac{{1 – \sin x\, + 1 + \sin x\, + 2|\cos x|}}{{1 – \sin x – 1 + \sin x}}$ $=$ $\frac{{2\left( {1 + |\cos x|} \right)}}{{ – \,2\,(\sin x)}}$ $=$$\frac{{1 – \cos x}}{{ – (\sin x)}}$

Standard 11

Mathematics

$2 \sin(\frac{\pi}{8}) \sin (\frac{2 \pi}{8}) \sin (\frac{3 \pi}{8}) \sin (\frac{5 \pi}{8}) \sin (\frac{6 \pi}{8}) \sin (\frac{7 \pi}{8})$ ની કિમંત મેળવો.

medium

(JEE MAIN-2021)

જો $x + \frac{1}{x} = 2\,\cos \theta ,$ તો ${x^3} + \frac{1}{{{x^3}}} = $

easy

$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $

easy

સાબિત કરો કે : $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$

medium

$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $

easy

જો $\frac{{5\pi }}{2} < x < 3\pi $,હોય તો $\frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ =

Solution

Similar Questions

$2 \sin(\frac{\pi}{8}) \sin (\frac{2 \pi}{8}) \sin (\frac{3 \pi}{8}) \sin (\frac{5 \pi}{8}) \sin (\frac{6 \pi}{8}) \sin (\frac{7 \pi}{8})$ ની કિમંત મેળવો.

જો $x + \frac{1}{x} = 2\,\cos \theta ,$ તો ${x^3} + \frac{1}{{{x^3}}} = $

$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $

સાબિત કરો કે : $\cot x \cot 2 x-\cot 2 x \cot 3 x-\cot 3 x \cot x=1$

$\frac{{\sin \theta + \sin 2\theta }}{{1 + \cos \theta + \cos 2\theta }} = $

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