cot ^{ - 1}left[ frac{sqrt {1 - sin x} + sqrt | vedclass

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Question

(D) ધારો કે $y = \cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right]$.છેદનું સંમેયીકરણ

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) ધારો કે $y = \cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right]$.છેદનું સંમેયીકરણ કરતા:$y = \cot ^{ - 1}\left[ \frac{(\sqrt {1 - \sin x} + \sqrt {1 + \sin x})^2}{(\sqrt {1 - \sin x})^2 - (\sqrt {1 + \sin x})^2} \right]$અંશ અને છેદનું વિસ્તરણ કરતા:$y = \cot ^{ - 1}\left[ \frac{(1 - \sin x) + (1 + \sin x) + 2\sqrt {(1 - \sin x)(1 + \sin x)}}{(1 - \sin x) - (1 + \sin x)} \right]$$y = \cot ^{ - 1}\left[ \frac{2 + 2\sqrt {1 - \sin ^2 x}}{-2\sin x} \right] = \cot ^{ - 1}\left[ \frac{2(1 + \cos x)}{-2\sin x} \right]$અડધા ખૂણાના સૂત્રો $1 + \cos x = 2\cos ^2(x/2)$ અને $\sin x = 2\sin(x/2)\cos(x/2)$ નો ઉપયોગ કરતા:$y = \cot ^{ - 1}\left[ \frac{2(2\cos ^2(x/2))}{-2(2\sin(x/2)\cos(x/2))} \right] = \cot ^{ - 1}\left[ -\frac{\cos(x/2)}{\sin(x/2)} \right]$$y = \cot ^{ - 1}\left( -\cot(x/2) \right) = \cot ^{ - 1}\left( \cot(\pi - x/2) \right) = \pi - \frac{x}{2}$.

$\cot ^{ - 1}\left[ \frac{\sqrt {1 - \sin x} + \sqrt {1 + \sin x}}{\sqrt {1 - \sin x} - \sqrt {1 + \sin x}} \right] = $

Similar Questions

જો $\cot^{-1}[(\cos \alpha)^{1/2}] - \tan^{-1}[(\cos \alpha)^{1/2}] = x$ હોય,તો $\sin x = $

જો $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$ હોય,તો $x + y + z$ ની કિંમત શું થાય?

$\begin{aligned} & \text{જો } \cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\} \\ & b>a, \text{ હોય તો } x= \end{aligned}$

સાબિત કરો કે $2 \sin ^{-1} \frac{3}{5} = \tan ^{-1} \frac{24}{7}$.

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