Question

Value of cot^2x(secx-1)/1+sinx + sec^2x(sinx-1)/1+secx=?
Best answer will be awarded brainliest by me

Answer 1

Answer:

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Answer 2

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$\purple\longmapsto\footnotesize\:cot^2x(\dfrac{secx-1}{1+sinx})+sec^2x(\dfrac{sinx-1}{1+secx}) \\ \\ \purple\longmapsto\footnotesize\:\frac{cot^2x(secx-1)(secx+1)+sec^2x(sinx-1)(sinx+1)}{(1+sinx)(1+secx)} \\ \\\purple\longmapsto\footnotesize \:\frac{cot^2x(sec^2x-1)+sec^2x(sin^2x-1)}{(1+sinx)(1+secx)} \\ \\\purple\longmapsto\footnotesize\frac{cot^2x(tan^2x)+sec^2x(-cos^2x)}{(1+sinx)(1+secx)} \\ \\ \purple\longmapsto\footnotesize\frac{cot^2x(tan^2x)-sec^2x(cos^2x)}{(1+sinx)(1+secx)} \\ \\\purple\longmapsto\footnotesize\frac{\frac{1}{tan^2x}(tan^2x)-\frac{1}{cos^2x}(cos^2x)}{(1+sinx)(1+secx)} \\ \\\purple\longmapsto\footnotesize\frac{1-1}{(1+sinx)(1+secx)} \\ \\ \purple\longmapsto\footnotesize \:\frac{0}{(1+sinx)(1+secx)} \\ \\ \purple\longmapsto \: \pink{\large\boxed{\boxed{\bold{0}}}}$

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