Question

prove LHS=RHS pls help me ​

Answer 1

AnswEr:-

To Prove:-

$\\ \tt\longrightarrow \dfrac{1}{ \sec A + \tan A } = \sec A - \tan A.$

ID Used:-

$\\ \tt\longrightarrow \sec \theta = \frac{1}{ \cos \theta } ... \: \: id(1) .\\ \\ \tt\longrightarrow \tan \theta = \frac{ \sin \theta}{ \cos \theta}... \: \: id(2). \\ \\ \tt\longrightarrow \sin {}^{2} \: theta + \cos {}^{2} \theta= 1. \\ \\ \tt\longrightarrow \cos {}^{2} \theta = 1 - \sin {}^{2} \theta... \: \: id(3). \\ \\ \tt\longrightarrow (a+b)(a-b) = a^2-b^2</p><p>..\ \ id(4).$

Proof:-

Lets simplify LHS:-

$\\ \tt\implies \dfrac{ 1}{ \sec A + \tan A} . \\ \\ \tt = \frac{1}{ \frac{1}{ \cos A} + \frac{ \sin A}{ \cos A} }. \\ \\ \rm[by \: \: using \: \: id(1) \: \: and \: \: id(2)]</p><p>. \\ \\ \tt = \frac{1}{ \frac{1 + \sin A }{ \cos A} } . \\ \\ \tt = \frac{ \cos A}{1 + \sin A} . \\ \\ \tt = \frac{ \cos A}{1 + \sin A} \times \frac{1 - \sin A}{1 - \sin A}. \\ \\ \tt = \frac{ \cos A(1 - \sin A}{(1 + \sin A)(1 - \sin A)} . \\ \\ \tt = \frac{ \cos A(1 - \sin A)}{ {1}^{2} - { \sin}^{2}A}. \\ \\ \rm[by \: \: using \: \: id(3)].</p><p> \\ \\ \tt = \frac{ \cos A( 1 - \sin A)}{ {\cos}^{2} A } . \\ \\ \rm[by \: \: using \: \: id(4)]</p><p>. \\ \\ \tt = \frac{ \cancel{ \cos A}(1 - \sin A)}{ \cancel{\cos A }\times \cos A}. \\ \\ \tt = \frac{ 1 - \sin A}{ \cos A} . \\ \\ \tt = \frac{1}{ \cos A} - \frac{ \sin A}{ \cos A} . \\ \\ \tt = \sec A - \tan A = rhs. \\ \\ \rm[by \: \: using \: \: id(1) \: \: and \: \: id(2)]</p><p>. \\ \\ \rm{ \underline {\underline{ \: \: \: \: HENCE \: \: \: \: PROVED.</p><p> \: \: \: \: }}}$

$\rule{200}2$

More Trigonometric IDs:-

$\\ \tt\leadsto 1 + \cot^2\theta = cosec^2\theta.\\ \\ \tt\leadsto 1 + \tan^2\theta = \sec^2\theta. \\ \\ \tt\leadsto \sin\theta = \dfrac{1}{cosec\theta}.\\ \\ \tt\leadsto cosec\theta = \dfrac{1}{\sin\theta}.\\ \\ \tt\leadsto \sec\theta = \dfrac{1}{\cos\theta}.\\ \\ \tt\leadsto\cot\theta = \dfrac{1}{\tan\theta}. \\ \\ \tt\leadsto\tan\theta = \dfrac{1}{\cot\theta}.$

$\rule{200}2$