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Question

Prove the following questions : 
$$ \dfrac{tan \,A}{ 1 + sec\, A} - \dfrac{tan \,A}{1 - sec\, A} = 2 \, cosec A $$ 


Solution

We will start from LHS . 
[We can take $$ tan \, A $$ as common and $$ (1 - sec \, A0(1 + sec\,A) $$ makes $$ (1 - sec^2A = tan^2 A) $$ so formulae will be used.] 

LHS $$ = \dfrac{tan \,A}{ 1 + sec\, A} - \dfrac{tan \,A}{1 - sec\, A} $$ 
 
     $$ = tan \, A  \left [ \dfrac{1}{(1 + sec\,A)} - \dfrac{1}{(1 - sec\,A)}  \right ] $$  [Taking $$ tan \, A $$ common] 

    $$ = tan \, A \left [ \dfrac{1 - sec\,A - (1 + sec\,A)}{(1 + sec\,A )(1 - sec\,A)} \right ] $$  {Taking LCM) 

    $$ = tan \,A \left [ \dfrac{-2 \, sec\,A}{(1 - sec^2A)} \right ] = \dfrac{tan \, A(-2\, sec\,A)}{-(sec^2A - 1)} $$ 

    $$ = \dfrac{-tan \, A \, 2\, sec\,A}{-tan^2A} = \dfrac{2 \, sec\,A}{tan\,A} $$ 

    $$ = \dfrac{2 \times \dfrac{1}{cos\,A}}{\dfrac{sin\,A}{cos\,A}} $$ 

    $$ = \dfrac{2}{sin\,A} = 2 \, cosec \,A = $$ RHS 

Hence , proved . 

Mathematics

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