Question

cotx cos^x - tanx sin^x =2 cot2x

Answer 1

$\textbf{To prove:}$

$cotx\,cos^2x-tanx\,sin^2x=2\,cot2x$

$\textbf{Solution:}$

$\text{Consider,}$

$cotx\,cos^2x-tanx\,sin^2x$

$=(\dfrac{cosx}{sinx})cos^2x-(\dfrac{sinx}{cosx})sin^2x$

$=\dfrac{cos^3x}{sinx}-\dfrac{sin^3x}{cosx}$

$=\dfrac{cos^4x-sin^4x}{sinx\,cosx}$

$=\dfrac{(cos^2x)^2-(sin^2x)^2}{sinx\,cosx}$

$\text{Using the algebraic identitty}$

$\boxed{\bf\,a^2-b^2=(a+b)(a-b)}$

$=\dfrac{(cos^2x+sin^2x)(cos^2x-sin^2x)}{sinx\,cosx}$

$=\dfrac{(1)(cos^2x-sin^2x)}{sinx\,cosx}$

$=\dfrac{2(cos^2x-sin^2x)}{2\,sinx\,cosx}$

$\text{Using the following identity}$

$\bf\,cos2A=cos^2A-sin^2A$

$\bf\;sinA=2\:sinA\:cosA$

$=2\,\dfrac{cos2x}{sin2x}$

$=2\,cot2x$

$\therefore\bf\,cotx\,cos^2x-tanx\,sin^2x=2\,cot2x$

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Answer 2

Answer:

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