Question

sin^2 A tan A+ cos^2 A cot A+ 2 sin A cosA = tan A+ cot A​

Answer 1

Answer:

$\Large\underline{\underline{\sf{ \color{magenta}{\qquad Given \qquad} }}}$

$LHS = \sin² \: A \cot² \: A + \cos² \: A \tan² \: A$

$= \sin² \: A \: \times \frac{ \ \cos² \: A }{ \sin² A } + \cos² \: A \: \times \frac{ \sin² A }{ \cos² A }$

$\cos²A + \sin²A$

$= 1$

$= RHS$

$\boxed{\sf\red{Hence Proved}}$

Answer 2

To prove :

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sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A

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Solution :

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Consider sin² A tan A + cos² A cot A + 2 sin A cos A

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$\sf{ = {sin}^{2}A \dfrac{sin \: A}{cos \: A} + {cos}^{2}A \dfrac{cos \: A}{sin \:A } + 2 \: sin \:A \: cos \: A }$

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$\sf{ =\dfrac{ {sin}^{3}A }{cos \: A} + \dfrac{ {cos}^{3}A }{sin \: A} + 2 \: sin \: A \: cos \: A}$

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$\sf{ = \dfrac{ {sin}^{4} A + {cos}^{4}A + 2 \: {sin}^{2} A \: {cos}^{2} A }{sin \:A \: cos \: A} }$

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$\sf{= \dfrac{( {sin}^{2} A + {cos}^{2}A) {}^{2} }{sin \:A \: cos \:A } }$

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$\sf{ = \dfrac{1}{sin \: A \: cos \: A} \: \: \: ...(1) }$

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Now, consider tan A + cot A

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$\sf{ = \dfrac{sin \: A}{cos \: A} + \dfrac{cos \:A }{sin \:A } }$

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$\sf{ = \dfrac{ {sin}^{2}A + {cos}^{2} A }{sin \: A \: cos \: A} }$

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$\sf{ = \dfrac{1 }{sin \: A \: cos \: A} \: \: \: ...(2) }$

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From (1) and (2), we have,

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sin² A tan A + cos² A cot A + 2 sin A cos A = tan A + cot A

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Hence proved..