Question

Prove that
$(1 - {tan}^{4} x )\div (1 + {tan}^{4}x) = (cosx + sinx) \div (cosx - sinx)$
​

Answer 1

$\huge{\underline{\bf\red{Questi {\mathbb{O}} n} :}}$

To prove :-

$\boxed{ \frac{1 - { \tan }^{4}x }{1 + { \tan }^{4}x } = \frac{ \cos x + \sin x }{ \cos x - \sin x } }$

$\huge{\underline{\bf\blue{Soluti {\mathbb{O}} n}\ :}}$

LHS

$\begin{aligned}& = \ \frac{1 - { \tan }^{4} x}{1 + { \tan }^{4} x} \\ \\ & = \frac{1 - \dfrac{ { \sin }^{4}x }{ { \cos}^{4}x } }{1 + \dfrac{ { \sin }^{4}x }{ { \cos}^{4}x }} \\ \\ & = \frac{ \quad \dfrac{ { \cos }^{4}x - { \sin }^{4} x }{ \cancel{ { \cos}^{4}x }} \quad }{ \dfrac{ { \cos }^{4}x + { \sin }^{4} x }{ \cancel{{ \cos}^{4}x }} } \\ \\ & = \frac{ { \cos}^{4} x - { \sin}^{4} x}{{ \cos}^{4} x + { \sin}^{4} x} \\ \\ & = \frac{ ({ \cos }^{2}x - { \sin }^{2}x )( { \cos }^{2} x + { \sin}^{2} x)}{ {( { \cos }^{2}x + { \sin}^{2}x) }^{2} - 2 \cos x\sin x} \\ \\ & = \frac{ ({ \cos }^{2}x - { \sin }^{2}x )(1)}{ 1 - 2 \cos x\sin x} \\ \\ & = \frac{ ({ \cos }^{2}x - { \sin }^{2}x )}{ { \cos }^{2}x + { \sin}^{2}x - 2 \cos x\sin x} \\ \\ & = \frac{ ({ \cos }^{2}x - { \sin }^{2}x )}{ {( \cos x - \sin x) }^{2} } \\ \\ & = \frac{ \cancel{( \cos x - \sin x)}( \cos x + \sin x) }{ \cancel{( \cos x - \sin x)}(\cos x - \sin x) } \\ \\ & = \frac{ \cos x + \sin x }{ \cos x - \sin x} & \end{aligned}$

= RHS

${\underline{\boxed{\dots \sf\green{ HENCE}}}}$

$\overline{ \qquad \underline{ \sf{ \red{V}E \orange{R}I \green{F}I} \purple{E}D \qquad }}\!\!\! \large{\mid}\!\! :$

$\red{\rule{5.5cm}{0.02cm}}$

Identities Used :-

$\boxed{ {x}^{2} - {y}^{2} = (x - y)(x + y)}$

$\boxed{ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy }$

$\boxed{ \quad { \cos }^{2} \theta + { \sin}^{2} \theta = 1 \quad }$

$\boxed{ \qquad \tan x = \dfrac{ \sin x }{ \cos x } \qquad }$

$\purple{\rule{7.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$