Question

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Answer 1

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Answer 2

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$\huge\purple{\sf{\underline{Required \: Answer}}}$

Given:

$\quad\bullet\displaystyle\bf{ \: \dfrac{tan^2A}{tan^2A-1}+\dfrac{cosec^2A}{sec^2A-cosec^2A}=\dfrac{1}{1-2cos^2A}}$

Properties :

$\quad \bullet \bf \: {sin}^{2} x + {cos}^{2}x = 1 \:$

$\: \quad \bullet \bf \: tanx = \frac{sinx}{cosx}$

$\quad \bullet \bf \: cosecx= \frac{1}{sinx}$

$\quad \bullet \bf \: secx = \frac{1}{cosx}$

$\huge\green{\sf{\underline{Solution:}}}$

$\\ \qquad\implies \displaystyle \sf \: \frac{ {tan}^{2}x }{ {tan}^{2} x - 1} + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x}$

$\: \qquad \ \displaystyle \underline{\sf { \star \: bring \: it \: into \: sinx \: and \: cosx}}$

$\: \\ \qquad \implies \displaystyle \sf \: \frac{ \frac{ {sin}^{2}x}{ {cos}^{2}x} }{ \frac{ {sin}^{2}x }{ {cos}^{2}x } - 1 } + \frac{ \frac{1}{ {sin}^{2} x} }{ \frac{1}{ {cos}^{2} x} - \frac{1}{ {sin}^{2}x } }$

$\qquad \implies \displaystyle \sf \: \frac{ \frac{ {sin}^{2}x }{ {cos}^{2}x } }{ \frac{ {sin}^{2}x - {cos}^{2}x }{ {cos}^{2} x} } + \frac{ \frac{1}{ {sin}^{2}x } }{ \frac{ {sin}^{2}x - {cos}^{2} x }{ {sin}^{2}x {cos}^{2} x } }$

$\\ \qquad \implies \displaystyle \sf \: \frac{ {sin}^{2}x }{ {sin}^{2}x - {cos}^{2}x } + \frac{ {cos}^{2}x}{ {sin}^{2}x - {cos}^{2}x}$

$\qquad \implies \displaystyle \sf \: \frac{ {sin}^{2} x + {cos}^{2}x }{ {sin}^{2}x - {cos}^{2}x }$

$\: \qquad \implies \displaystyle \sf \: \frac{1}{ {sin}^{2}x - {cos}^{2} x }$

$\: \qquad \displaystyle \sf \star \: put \: {sin}^{2}x = 1 - {cos}^{2} x$

$\: \qquad \implies \displaystyle \sf \: \frac{1}{1 - 2 {cos}^{2}x }$

$\bigstar\ \large \underline{ \bf{hence \: proved}}$

$\\ \therefore \boxed{ \displaystyle \sf{ \underline{ \: \frac{ {tan}^{2}x }{ {tan}^{2}x - 1 } + \frac{ {cosec}^{2}x }{ {sec}^{2}x - {cosec}^{2} x} = \frac{1}{1 - 2 {cos}^{2}x } }}}$

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