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Question

The expression $$\sin A . \cos A \left (\dfrac{\tan^{2}\ A}{\tan A - 1} + \dfrac{\cot^2 A}{\cot A - 1} \right ) $$ can be written as: 


A
cosAcosecA+1
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B
tanA+secA
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C
secA+cosecA
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D
sinAcosA+1
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Solution

The correct option is D $$\sin A \cos A + 1 $$
 $$\dfrac{\tan^{2}\ A}{\tan A - 1} + \dfrac{\cot^2 A}{\cot A - 1} $$

$$ = \dfrac{\tan A}{\dfrac{(\tan A - 1)}{\tan A}} + \dfrac{\cot A}{\dfrac{(\cot A - 1)}{\cot A}} $$

$$ = \dfrac{\tan A}{1 - \cot A} + \dfrac{\cot A}{1 - \tan A} $$

$$ = \dfrac{\tan^2 A}{\tan A - 1} - \dfrac{\cot A}{\tan A - 1} $$

$$ = \dfrac{\tan^2 A - \cot A}{\tan A - 1} $$ 

$$ = \dfrac{\tan^3 A - 1}{\tan^2 A - \tan A} $$

$$ = \dfrac{(\tan A - 1) (\tan^2 A + \tan A + 1)}{\tan A (\tan A - 1)} $$ 

$$ = \dfrac{\sec^2 A + \tan A}{\tan A } $$ 

$$ = \dfrac{1}{\sin A \cos A} + 1 $$ 

 $$\therefore \sin A \cdot  \cos A \left (\dfrac{\tan^{2}\ A}{\tan A - 1} + \dfrac{\cot^2 A}{\cot A - 1} \right ) = 1 + \sin A \cos A $$

Mathematics

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