Question

Question:-
$\sf \: To \: \:prove \\ \\ \sf \: : \implies \sf \dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA} = \dfrac{2}{sin^2A - cos^2A} = \dfrac{2}{2sin^2A - 1}$
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Answer 1

complete Question:

Prove:

$\sf \: : \implies \sf \dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA} = \dfrac{2}{sin^2A - cos^2A} = \dfrac{2}{2sin^2A - 1} = \frac{2}{1 - 2 \: cos {}^{2}A }$

Given:

$\sf \: : \implies \sf \dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA} = \dfrac{2}{sin^2A - cos^2A} = \dfrac{2}{2sin^2A - 1} = \frac{2}{1 - 2 \: cos {}^{2}A }$

Solution:

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

using trigonometric identities :

$\\ { \underline{ \boxed{ \bf{ \pink{\sin^2\theta+\cos^2\theta=1}}}}}$

${ \underline{ \boxed{ \bf{ \red{\sin^2\theta=1-\cos^2\theta}}}}}$

${ \underline{ \boxed{ \bf{ \green{cos^2\theta=1-\sin^2\theta.}}}}}$

We have,

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

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

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

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

$\sf:\implies\dfrac{2\times1}{\sin^2A-\cos^2A}$

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

$\sf:\implies\dfrac{2}{\sin^2A-(1-\sin^2A)}$

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

Hence proved.

Answer 2

To prove :-

$\sf\dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA} = \dfrac{2}{sin^2A - cos^2A} = \dfrac{2}{2sin^2A - 1}$

Here as a first step we'll divide the whole equation into 3 parts which we have to prove .

On taking Last part first i.e ,

$\dfrac{2}{2sin^2A - 1}$

Now , we know the identity i.e ,

$\leadsto \sf \cos {}^{2}A + \sin {}^{2}A = 1 \\ \leadsto \sf \: sin {}^{2} A = 1 - \cos {}^{2} A \: \: (1)$

Now , in denominator we have 2sin²A - 1 which can be written from the eq (1) as

$\bf \frac{2}{2(1 - \cos {}^{2}A) - 1 } = \frac{2}{2 - 2 \cos {}^{2}A - 1 } = \frac{2}{1 - 2 \cos {}^{2} A }$

So , In Third part we got ,

$</strong><strong>\sf \</strong><strong>frac{2}{1 - 2 \cos {}^{2} A }$

Now , Take the second part which is as follow :

$\frac{2}{sin^2A - cos^2A} = \sf\frac{2}{1 - \cos {}^{2} A - \cos {}^{2} A } \\ = \frac{2}{1 - 2 \cos {}^{2} A}$

So , we have made our second part equal to the third part .

Now , take the first part I.e ,

$\dfrac{sinA + cosA}{sinA - cosA} + \dfrac{sinA - cosA}{sinA + cosA}$

After taking decimal and solving we get

$\frac{2( \sin {}^{2} A + \cos {}^{2}A )}{ \sin {}^{2}A - \cos {}^{2}A }$

By the square identity, we get

$\frac{2}{ \sin {}^{2} A - \cos {}^{2} A }$

Now , solve this in the same manner as we did in second part and there we get our answer

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

So , here we proved all parts or sides equal to the above equation.

Thankyou