Question

Prove that,
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$\sf{ \frac{ {tan}^{2} \: A}{ {tan}^{2} \: A - 1 } + \frac{ {cosec}^{2} \: A}{ {sec}^{2} \: A - {cosec}^{2} \: A } = \frac{1}{1 - 2 {cos}^{2} \: \: A } }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \dfrac{ {tan}^{2} \: A}{ {tan}^{2} \: A - 1 } + \dfrac{ {cosec}^{2} \: A}{ {sec}^{2} \: A - {cosec}^{2} \: A}$

can be rewritten as

$\rm \: =  \:\dfrac{\dfrac{ {sin}^{2} A}{ {cos}^{2} A} }{ \: \: \dfrac{ {sin}^{2} A}{ {cos}^{2} A} - 1 \: \: } + \dfrac{\dfrac{1}{ {sin}^{2} A} }{\dfrac{1}{ {cos}^{2} A} - \dfrac{1}{ {sin}^{2} A} }$

can be further rewritten as

$\rm \: =  \:\dfrac{\dfrac{ {sin}^{2} A}{ {cos}^{2} A} }{ \: \: \dfrac{ {sin}^{2} A - {cos}^{2} A}{ {cos}^{2} A} \: \: } + \dfrac{\dfrac{1}{ {sin}^{2} A} }{\dfrac{ {sin}^{2} A - {cos}^{2} A}{ {cos}^{2} A \: {sin}^{2} A} }$

$\rm \: = \: \dfrac{ {sin}^{2} A}{ {sin}^{2} A - {cos}^{2} A} + \dfrac{ {cos}^{2}A}{{sin}^{2} A - {cos}^{2} A}$

$\rm \: = \: \dfrac{ {sin}^{2} A + {cos}^{2}A }{ {sin}^{2} A - {cos}^{2} A}$

$\rm \: = \: \dfrac{1}{ {sin}^{2} A - {cos}^{2} A}$

can be further rewritten as

$\rm \: = \: \dfrac{1}{1 - {cos}^{2} A - {cos}^{2} A}$

$\rm \: = \: \dfrac{1}{1 - 2{cos}^{2} A}$

Hence,

$\color{green}\rm\implies \:\dfrac{ {tan}^{2} \: A}{ {tan}^{2} \: A - 1 } + \dfrac{ {cosec}^{2} \: A}{ {sec}^{2} \: A - {cosec}^{2} \: A } = \: \dfrac{1}{1 - 2{cos}^{2} A}$

$\rule{190pt}{2pt}$

Formulae Used :-

$\boxed{ \rm{ \:tanx = \frac{sinx}{cosx} \: }}$

$\boxed{ \rm{ \:secx = \frac{1}{cosx} \: }}$

$\boxed{ \rm{ \:cosecx = \frac{1}{sinx} \: }}$

$\boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

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