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Grade 12Differential Calculus

differentiate tan^-1(cos x/1+sin x) with respect to sec^-1(x)

taniska

12 Years agoGrade 12

2 Answers

Arun Kumar

12 Years ago

Hi

$\\a=tan^{-1}(\frac{cosx}{1+sinx})=tan^{-1}(\frac{cos^2{\frac{x}{2}}-sin^2{\frac{x}{2}}}{(sin{\frac{x}{2}}+cos\frac{x}{2})^2})=tan^{-1}(\frac{cos{\frac{x}{2}}-sin{\frac{x}{2}}}{sin{\frac{x}{2}}+cos\frac{x}{2}}) \\=>a=tan^{-1}(\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}})=tan^{-1}tan(\frac{\pi}{4}-\frac{x}{2})=\frac{\pi}{4}-\frac{x}{2} \\b=sec^{-1}x \\\\ \frac{da}{dx}=\frac{-1}{2} \\\frac{db}{dx}=\frac{-1}{x*\sqrt{x^2-1}} \\=>\frac{da}{db}=\frac{\frac{-1}{2}}{\frac{-1}{x*\sqrt{x^2-1}}}$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

Rishi Sharma

5 Years ago

Dear Student,
Please find below in the attached image the solution to your problem.

Thanks and Regards

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