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Question-217205




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) (sin cot^(−1) x)  =cos tan^(−1) (sin sin^(−1) (1/( (√(1+x^2 )))))  =cos tan^(−1) (1/( (√(1+x^2 ))))  coscos^(−1) ((√(1+x^2 ))/( (√(2+x^2 ))))=(√((1+x^2 )/(2+x^2 )))=(((1+x^2 )/(2+x^2 )))^(1/2)
$$\mathrm{cos}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{cot}^{−\mathrm{1}} {x}\right) \\ $$$$=\mathrm{cos}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{sin}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right) \\ $$$$=\mathrm{cos}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${cos}\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}=\sqrt{\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{2}} }}=\left(\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{2}} }\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

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