[tex] \sf \huge \orange{ \: Prove \: that : - }[/tex]
[tex] \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} } } [/tex]
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(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
Verified answer
Appropriate Question :- Prove that
[tex]\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} } \\ [/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
Consider LHS
[tex]\displaystyle \: \sf { \bigg(1 + \frac{1}{ { \tan}^{2}x} \bigg) + \bigg(1 + \frac{1}{ { \cot}^{2} x} \bigg) } \\ [/tex]
We know,
[tex]\boxed{ \rm{ \:tanx = \frac{1}{cotx} \: }} \\ [/tex]
[tex]\boxed{ \rm{ \:cotx = \frac{1}{tanx} \: }} \\ [/tex]
So, using these results, we get
[tex]\rm \: = \: (1 + {cot}^{2}x) + (1 + {tan}^{2}x) \\ [/tex]
[tex]\rm \: = \: {cosec}^{2}x + {sec}^{2}x \\ [/tex]
can be further rewritten as
[tex]\rm \: = \: \dfrac{1}{ {sin}^{2}x } + \dfrac{1}{ {cos}^{2} x} \\ [/tex]
[tex]\rm \: = \: \dfrac{ {cos}^{2}x + {sin}^{2}x}{ {sin}^{2}x \: {cos}^{2}x} \\ [/tex]
[tex]\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: {cos}^{2}x} \\ [/tex]
[tex]\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: (1 - {sin}^{2}x)} \\ [/tex]
[tex]\rm \: = \: \dfrac{ 1}{ {sin}^{2}x \: - \: {sin}^{4}x} \\ [/tex]
Hence,
[tex]\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} }}} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Formulae Used :-
[tex]\boxed{ \rm{ \:1 + {tan}^{2}x = {sec}^{2}x \: }} \\ [/tex]
[tex]\boxed{ \rm{ \:1 + {cot}^{2}x = {cosec}^{2}x \: }} \\ [/tex]
[tex]\boxed{ \rm{ \:secx = \frac{1}{cosx} \: }} \\ [/tex]
[tex]\boxed{ \rm{ \:cosecx = \frac{1}{sinx} \: }} \\ [/tex]
[tex]\boxed{ \rm{ \: {sin}^{2}x + {cos}^{2}x = 1 \: }} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional information :-
[tex]\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}[/tex]