Question

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Answer 1

Question :–

• If $\: \: { \bold{ \tan (\theta) + \sin( \theta) = m}} \: \:$ and $\: \: { \bold{ \tan (\theta) - \sin( \theta) = n}} \: \:$ , then ${ \bold{ \dfrac{ {( {m}^{2} - {n}^{2} )}^{2} }{mn} = ? }}$

ANSWER :–

$\\ \to { \boxed{ \bold{ \dfrac{ {( {m}^{2} - {n}^{2} )}^{2} }{mn} = 16 }}}$

EXPLANATION :–

GIVEN :–

• $\: \: { \bold{ \tan (\theta) + \sin( \theta) = m}} \: \:$

• $\: \: { \bold{ \tan (\theta) - \sin( \theta) = n}} \: \:$

TO FIND :–

$\\ \implies { \bold{ \dfrac{ {( {m}^{2} - {n}^{2} )}^{2} }{mn} = ? }}$

SOLUTION :–

${ \bold{ = \dfrac{ {( {m}^{2} - {n}^{2} )}^{2} }{mn} }}$

• We should write this as –

$\\ { \bold{ = \dfrac{ {[( m - n)(m + n)]}^{2} }{mn} \: \: [ \: \because (a + b)(a - b) = {a}^{2} - {b}^{2} }} ]$

• Now put the values –

$\\ { \bold{ = \dfrac{ {[( \tan (\theta) + \sin( \theta) - (\tan (\theta) - \sin( \theta)))(\tan (\theta) + \sin( \theta) + \tan (\theta) - \sin( \theta))]}^{2} }{ [ \tan (\theta) + \sin( \theta) ][ \tan (\theta) - \sin( \theta)]} }}$

$\\ { \bold{ = \dfrac{ {[( 2 \sin( \theta) )(2\tan (\theta) ]}^{2} }{ [ \tan ^{2} (\theta) - \sin^{2} ( \theta)]} }}$

$\\ { \bold{ = \dfrac{ {[( 4 \sin( \theta) )( \frac{ \sin( \theta) }{ \cos( \theta) } )]}^{2} }{ [ \frac{ {sin}^{2} { \theta}}{ {cos}^{2} \theta } - \sin^{2} ( \theta)]} }}$

$\\ { \bold{ = \frac{16 \times {sin}^{2} \theta \times \dfrac{ {sin}^{2} \theta }{ cos^{2} \theta } }{ { \sin}^{2} \theta( \dfrac{1}{ {cos}^{2} \theta } - 1)} }}$

$\\ { \bold{ = \frac{16 \times \dfrac{ {sin}^{2} \theta }{ cos^{2} \theta } }{ ( \dfrac{1 -{cos}^{2} \theta}{ {cos}^{2} \theta } )} }}$

$\\ { \bold{ = \dfrac{16 \times \dfrac{ {sin}^{2} \theta }{ cos^{2} \theta } }{ ( \dfrac{{sin}^{2} \theta}{ {cos}^{2} \theta } )} }}$

$\\ { \bold{ = 16}}$

▪︎ Hence , ${ \boxed { \bold{ \dfrac{ {( {m}^{2} - {n}^{2} )}^{2} }{mn} = 16 }}}$