Question

$\huge \rm \blue{ \: Prove \: that : - } \\ \displaystyle \: \rm \green{ \bigg(1 + \frac{1}{ { \tan(x) }^{2} } \bigg) + \bigg(1 + \frac{1}{ { \cot(x) }^{2} } \bigg) = \frac{1}{ { \sin(x) }^{4} - \sin(x) {}^{2} } }$
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Answer 1

Prove that:-

$</p><p>\begin{gathered}\displaystyle \: \rm { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) = \dfrac{1}{ { \sin}^{2}x - \sin{}^{4} x} } \\ \end{gathered} </p><p>$

Solution:-

consider LHS

$\large\underline{\sf{Solution-}} </p><p></p><p>\begin{gathered}\displaystyle \: \sf { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) } \\ \end{gathered} </p><p>$

we know that,,

$</p><p>\begin{gathered}\boxed{ \rm{ \:tanx = \frac{1}{cotx} \: }} \\ \end{gathered} </p><p> \: \: </p><p>$

$\begin{gathered}\boxed{ \rm{ \:cotx = \frac{1}{tanx} \: }} \\ \end{gathered} </p><p>$

so,using these results ,we get,,

$=(1+cot {}^{2} x)+(1+tan {}^{2} x)</p><p></p><p>$

$\begin{gathered}\rm \: = \: {cosec}^{2}x + {sec}^{2}x \\ \end{gathered}</p><p></p><p>$

can be further rewritten as

$</p><p>\begin{gathered}\rm \: = \: \dfrac{1}{ {sin}^{2}x } + \dfrac{1}{ {cos}^{2} x} \\ \end{gathered}</p><p></p><p>$

$\begin{gathered}\rm \: = \: \dfrac{ {cos}^{2}x + {sin}^{2}x}{ {sin}^{2}x \: {cos}^{2}x} \\ \end{gathered}</p><p></p><p>$

$\begin{gathered}\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: {cos}^{2}x} \\ \end{gathered}</p><p></p><p>\begin{gathered}\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: (1 - {sin}^{2}x)} \\ \end{gathered}</p><p></p><p>$

$\begin{gathered}\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: - \: {sin}^{4}x} \\ \end{gathered}</p><p></p><p>$

$</p><p></p><p>\begin{gathered}\boxed{ \rm{ \:\rm { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) = \dfrac{1}{ { \sin}^{2}x - \sin{}^{4} x} }}} \\ \end{gathered}(1+tan2x1)+(1+cot2x1)$

$</p><p></p><p>\begin{gathered}\boxed{ \rm{ \:1 + {tan}^{2}x = {sec}^{2}x \: }} \\ \end{gathered} </p><p></p><p></p><p></p><p>$

$\begin{gathered}\boxed{ \rm{ \:1 + {cot}^{2}x = {cosec}^{2}x \: }} \\ \end{gathered} </p><p></p><p></p><p>$

$\begin{gathered}\boxed{ \rm{ \:secx = \frac{1}{cosx} \: }} \\ \end{gathered} </p><p></p><p>$

$\begin{gathered}\boxed{ \rm{ \:cosecx = \frac{1}{sinx} \: }} \\ \end{gathered} </p><p></p><p>$

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

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