Question

$\sf \huge \orange{ \: Prove \: that : - }$

$\displaystyle \: \sf \green{ \bigg(1 + \frac{1}{ { \tan(x) }^{2} } \bigg) + \bigg(1 + \frac{1}{ { \cot(x) }^{2} } \bigg) = \frac{1}{ { \sin(x) }^{4} - \sin(x) {}^{2} } }$
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Answer 1

(1-tanx)^ ^ 2 + (1-cotx)^ ^ 2

=(1-2tan x +tan ^2x)+(1 1-2cotx+ ot^ ^ 2x)

=sec^2x+csc^2x-2(tanx+cotx)

=sec ^2x+csc^2x-2(sinx/cosx+cosx/sinx)

=sec ^2x+csc^2x-2[( sin^ ^ 2x+cos^ ^ 2x)/(cos x^ * ! sinx)]

=sec^ ^ 2x+c ^2x -2[1/(cos x *sinx)]

=sec^2x+csc^2x -2[(1/cos x )*(1/sinx)]

=sec ^2x+csc^2x-2secx ^ * cscx

=( x-csc x)^ ^

I hope this is helpful

Answer 2

Appropriate Question :- Prove that

$\displaystyle \: \rm { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) = \dfrac{1}{ { \sin}^{2}x - \sin{}^{4} x} }$

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

Consider LHS

$\displaystyle \: \sf { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) }$

We know,

$\boxed{ \rm{ \:tanx = \frac{1}{cotx} \: }}$

$\boxed{ \rm{ \:cotx = \frac{1}{tanx} \: }}$

So, using these results, we get

$\rm \: = \: (1 + {cot}^{2}x) + (1 + {tan}^{2}x)$

$\rm \: = \: {cosec}^{2}x + {sec}^{2}x$

can be further rewritten as

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

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

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

$\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: (1 - {sin}^{2}x)}$

$\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: - \: {sin}^{4}x}$

Hence,

$\boxed{ \rm{ \:\rm { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) = \dfrac{1}{ { \sin}^{2}x - \sin{}^{4} x} }}}$

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

Formulae Used :-

$\boxed{ \rm{ \:1 + {tan}^{2}x = {sec}^{2}x \: }}$

$\boxed{ \rm{ \:1 + {cot}^{2}x = {cosec}^{2}x \: }}$

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

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

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

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

Additional information :-

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