Question

(SinA - cosA ) /(SinA+ cosA)
+
(SinA +cosA ) (SinA - cos A) =(2 /Sin^2 A-1)​

Answer 1

ANSWER:

To Prove:

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

Proof:

$:\longrightarrow\dfrac{\sin A-\cos A}{\sin A+\cos A}+\dfrac{\sin A+\cos A}{\sin A-\cos A}=\dfrac{2}{2sin^2A-1}\\\\\text{Solving LHS,}\\\\:\implies\dfrac{\sin A-\cos A}{\sin A+\cos A}+\dfrac{\sin A+\cos A}{\sin A-\cos A}\\\\\text{Taking LCM,}\\\\:\implies \dfrac{(\sin A-\cos A)(\sin A-\cos A)+(\sin A+\cos A)(\sin A+\cos A)}{(\sin A+\cos A)(\sin A-\cos A)} \\\\:\implies \dfrac{(\sin A-\cos A)^2+(\sin A+\cos A)^2}{(\sin A+\cos A)(\sin A-\cos A)} \\\\\text{We know that, a^2-b^2=(a+b)(a-b)\:\:\:\&\:\:\:(a\pm b)^2=a^2+b^2\pm2ab So,}$

$:\implies \dfrac{(\sin^2A+\cos^2A-2\sin A\cos A)+(\sin^2A+\cos^2A+2\sin A\cos A)}{(\sin^2A-\cos^2A)}\\\\:\implies \dfrac{\sin^2A+\cos^2A-2\sin A\cos A+\sin^2A+\cos^2A-2\sin A\cos A}{sin^2A-\cos^2A}\\\\:\implies \dfrac{(\sin^2A+\cos^2A)+(\sin^2A+\cos^2A)}{sin^2A-\cos^2A} \\\\\text{We know that, \sin^2\theta+\cos^2\theta=1 . So,}\\\\:\implies\dfrac{1+1}{\sin^2A-\cos^2A} \\\\\text{We know that, \cos^2\theta=1-\sin^2\theta . So,}\\\\:\implies\dfrac{2}{\sin^2A-(1-\sin^2A)}\\\\:\implies \dfrac{2}{\sin^2A-1+\sin^2A}\\\\\bf{:\implies \dfrac{2}{2\sin^2A-1} = RHS}\\\\\text{\underline{HENCE PROVED!!}}$

Formulae Used:

$\:\:\:\:\bullet\:\:\:\:a^2-b^2=(a+b)(a-b)\\\\\:\:\:\:\bullet\:\:\:\: (a\pm b)^2=a^2+b^2\pm2ab\\\\\:\:\:\:\bullet\:\:\:\:\sin^2\theta+\cos^2\theta=1(or \cos^2\theta=1-\sin^2\theta)$