જો cot ^{-1}(sqrt{cos alpha})-tan ^{-1}(sqrt{ | vedclass

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(B) આપેલ સમીકરણ: $\cot ^{-1}(\sqrt{\cos \alpha})-	an ^{-1}(\sqrt{\cos \alpha})=x$ નિત્યસમ $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ નો ઉપયોગ કરતા: $\

Vedclass · Accepted Answer

$	an ^2\left(\frac{\alpha}{2}ight)$

Vedclass · Answer

(B) આપેલ સમીકરણ: $\cot ^{-1}(\sqrt{\cos \alpha})-	an ^{-1}(\sqrt{\cos \alpha})=x$ નિત્યસમ $\cot ^{-1}(y) = 	an ^{-1}(\frac{1}{y})$ નો ઉપયોગ કરતા: $	an ^{-1}\left(\frac{1}{\sqrt{\cos \alpha}}ight)-	an ^{-1}(\sqrt{\cos \alpha})=x$ સૂત્ર $	an ^{-1}(A) - 	an ^{-1}(B) = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$ નો ઉપયોગ કરતા: $	an ^{-1}\left[\frac{\frac{1}{\sqrt{\cos \alpha}}-\sqrt{\cos \alpha}}{1+\frac{1}{\sqrt{\cos \alpha}} \cdot \sqrt{\cos \alpha}}ight]=x$ $	an ^{-1}\left[\frac{\frac{1-\cos \alpha}{\sqrt{\cos \alpha}}}{2}ight]=x$ $	an x = \frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}$ હવે,$\sin x = \frac{1-\cos \alpha}{1+\cos \alpha}$ $\sin x = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = 	an^2\left(\frac{\alpha}{2}ight)$.

જો $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$ હોય,તો $\sin x$ નું મૂલ્ય શું થાય?

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