જો cot^{-1}[(cos alpha)^{1/2}] - tan^{-1}[(co | vedclass

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(A) આપેલ છે: $\cot^{-1}[(\cos \alpha)^{1/2}] - 	an^{-1}[(\cos \alpha)^{1/2}] = x$નિત્યસમ $\cot^{-1}(y) = 	an^{-1}(\frac{1}{y})$ નો ઉપયોગ કરતા:$	an^

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

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(A) આપેલ છે: $\cot^{-1}[(\cos \alpha)^{1/2}] - 	an^{-1}[(\cos \alpha)^{1/2}] = x$નિત્યસમ $\cot^{-1}(y) = 	an^{-1}(\frac{1}{y})$ નો ઉપયોગ કરતા:$	an^{-1}[\frac{1}{\sqrt{\cos \alpha}}] - 	an^{-1}[\sqrt{\cos \alpha}] = x$સૂત્ર $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}(\frac{A-B}{1+AB})$ નો ઉપયોગ કરતા:$	an^{-1}[\frac{\frac{1}{\sqrt{\cos \alpha}} - \sqrt{\cos \alpha}}{1 + (\frac{1}{\sqrt{\cos \alpha}})(\sqrt{\cos \alpha})}] = x$$	an^{-1}[\frac{\frac{1-\cos \alpha}{\sqrt{\cos \alpha}}}{1+1}] = x$$	an x = \frac{1-\cos \alpha}{2\sqrt{\cos \alpha}}$હવે,આપણે $\sin x$ શોધવાની જરૂર છે. કાટકોણ ત્રિકોણની મદદથી જ્યાં સામેની બાજુ $1-\cos \alpha$ અને પાસેની બાજુ $2\sqrt{\cos \alpha}$ છે:કર્ણ = $\sqrt{(1-\cos \alpha)^2 + (2\sqrt{\cos \alpha})^2} = \sqrt{1 - 2\cos \alpha + \cos^2 \alpha + 4\cos \alpha} = \sqrt{1 + 2\cos \alpha + \cos^2 \alpha} = \sqrt{(1+\cos \alpha)^2} = 1+\cos \alpha$તેથી,$\sin x = \frac{	ext{સામેની બાજુ}}{	ext{કર્ણ}} = \frac{1-\cos \alpha}{1+\cos \alpha} = \frac{2\sin^2(\alpha/2)}{2\cos^2(\alpha/2)} = 	an^2(\frac{\alpha}{2})$.

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

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