જો alpha = cos^{-1}left(frac{3}{5}right) અને | vedclass

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(A) આપેલ છે કે $\cos \alpha = \frac{3}{5}$. $\cos^2 \alpha + \sin^2 \alpha = 1$ હોવાથી,$\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$ મળે.તે

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{9}{5\sqrt{10}}\right)$

Vedclass · Answer

(A) આપેલ છે કે $\cos \alpha = \frac{3}{5}$. $\cos^2 \alpha + \sin^2 \alpha = 1$ હોવાથી,$\sin \alpha = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$ મળે.તેથી,$	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{4/5}{3/5} = \frac{4}{3}$ થાય.આપણને $	an \beta = \frac{1}{3}$ આપેલ છે.સૂત્ર $	an(\alpha - \beta) = \frac{	an \alpha - 	an \beta}{1 + 	an \alpha 	an \beta}$ નો ઉપયોગ કરતા:$	an(\alpha - \beta) = \frac{\frac{4}{3} - \frac{1}{3}}{1 + (\frac{4}{3})(\frac{1}{3})} = \frac{1}{1 + \frac{4}{9}} = \frac{1}{\frac{13}{9}} = \frac{9}{13}$.હવે,જો $	an(\alpha - \beta) = \frac{9}{13}$ હોય,તો $\sin(\alpha - \beta) = \frac{9}{\sqrt{9^2 + 13^2}} = \frac{9}{\sqrt{81 + 169}} = \frac{9}{\sqrt{250}} = \frac{9}{5\sqrt{10}}$ થાય.તેથી,$\alpha - \beta = \sin^{-1}\left(\frac{9}{5\sqrt{10}}ight)$ મળે.

જો $\alpha = \cos^{-1}\left(\frac{3}{5}\right)$ અને $\beta = \tan^{-1}\left(\frac{1}{3}\right)$,જ્યાં $0 < \alpha, \beta < \frac{\pi}{2}$,તો $\alpha - \beta$ ની કિંમત શોધો.

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કિંમત શોધો: $\operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$

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