Question

prove the following identity: sec^2A/tan^2A+cosec^2A/cot^2A=sec^2A.cosec^2A​

Answer 1

Given,

$\large \boxed{\mathsf{\dfrac{sec^{2}A }{tan^{2}A} +\dfrac{cosec^{2}A }{cot^{2}A}=sec^{2}A.cosec^{2}A}}$

$\mathsf{On\: taking\: left\: hand \:side,}$

$\mathsf{\dfrac{sec^{2}A}{tan^{2}A} +\dfrac{cosec^{2}A}{cot^{2}A}}$

Since,

$\large \boxed{\mathsf{tan^{2}A =\dfrac{sin^{2}A}{cos^{2}A}}}$

$\large \boxed{\mathsf{cot^{2}A =\dfrac{cos^{2}A}{sin^{2}A}}</p><p>}$

So,

$\mathsf{\dfrac{sec^{2}A }{\dfrac{sin^{2}A}{cos^{2}A} } +\dfrac{cosec^{2}A }{\dfrac{cos^{2}A}{sin^{2}A} }}$

$\mathsf{\dfrac{sec^{2}A \times cos^{2}A }{sin^{2}A} +\dfrac{cosec^{2}A \times sin^{2}A}{cos^{2}A}}}$

Since,

$\mathsf{\large \boxed{\mathsf{sec^{2}A =\dfrac{1}{cos^{2}A}}}\\ \\\large \boxed{\mathsf{ {cosec^{2}A =\dfrac{1}{sin^{2}A}}}}}$

So,

$\mathsf{\dfrac{\dfrac{1}{cos^{2}A}\timescos^{2}A }{sin^{2}A}+\dfrac{ \dfrac{1}{ sin^{2}A} \times sin^{2}A}{cos^{2}A}}}$

On solving we get,

$\mathsf{\dfrac{1}{sin^{2}A} +\dfrac{1}{cos^{2}A} }$

$\mathsf{\dfrac{cos^{2}A + sin^{2}A}{sin^{2}A.cos^{2}A} }$

$since,\\\\\large \boxed{\mathsf{cos^{2}A+sin^{2}A =1}}$

$\mathsf{\dfrac{1}{sin^{2}A.cos^{2}A} }$

On taking seperately,

$\mathsf{\dfrac{1}{sin^{2}A}.\dfrac{1}{cos^{2}A} }$

Since,

$\mathsf{\large \boxed{\mathsf{sec^{2}A =\dfrac{1}{cos^{2}A}}}\\ \\\large \boxed{\mathsf{ {cosec^{2}A =\dfrac{1}{sin^{2}A}}}}}$

$\mathsf{=sec^{2}A.cosec^{2}A}$

L.H.S = R.H.S

HENCE PROVED