Math, asked by yash842004, 1 year ago

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Answered by lakshay9122

Hope it will help you...

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Answered by Anonymous

Taking L.H.S

$\dfrac{sin \: A \: + \: cos \: A}{sin \: A\: - \: cos \: A}$ + $\dfrac{sin \: A \: - \: cos \: A}{sin \: A \: + \: cos \: A}$

Taking LCM we get;

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

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

= $\dfrac{1 \: + \: 1}{ {sin}^{2} A \: - \: {cos}^{2} A}$

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

= $\dfrac{2}{(1 \: - \: {sin}^{2} A)\:-{cos}^{2}\:A}$

= $\dfrac{2}{1 \: - \: {cos}^{2} A \:-\:{cos}^{2}A}$

= $\dfrac{2}{1\:-\: 2{cos}^{2} A}$

Now..

L.H.S = R.H.S

Hence proved.

yash842004: thanks..

Anonymous: Welcome :)

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