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Question Number 217205 by peter frank last updated on 05/Mar/25

Answered by som(math1967) last updated on 06/Mar/25

$$\cos {\tan }^{-1}\left(\sin {\cot }^{-1}x\right)$$$$=\cos {\tan }^{-1}\left(\sin {\sin }^{-1}\frac{1}{\sqrt{1+x^2}}\right)$$$$=\cos {\tan }^{-1}\frac{1}{\sqrt{1+x^2}}$$$$cos{\cos }^{-1}\frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}=\sqrt{\frac{1+x^2}{2+x^2}}={\left(\frac{1+x^2}{2+x^2}\right)}^{\frac{1}{2}}$$

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