Question

prove that sina+cosa/sina-cosa+sina-cosa/sina+cosa=2/sin2a-cos2a =2 sec^2a/tan^2a-1​

Answer 1

To Prove :-

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

Solution :-

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

$\longrightarrow \sf \dfrac{(sinA+cosA)(sinA+cosA) +(sinA-cosA)(sinA-cosA)}{(sinA-cosA)(sinA+cosA)} \\\\\longrightarrow \dfrac{(sinA+cosA)^2+ (sinA-cosA)^2}{sinA^2-cosA^2} \\\\\longrightarrow \dfrac{sin^2A+cos^2A+2sinAcosA+sin^2A+cos^2A-2sinAcosA}{sinA^2A-cos^2A} \\\\\bullet \bf \ sin^2A+cos^2A = 1 \\\\\longrightarrow\sf \dfrac{2}{sin^2A-cos^2A}$

Divide numerator and denominator by $\sf cos^2A$ .

$\longrightarrow \sf \dfrac{\dfrac{2}{cos^2A}}{\dfrac{sin^2A}{cos^2A}- \dfrac{cos^2A}{cos^2A}} \\\\\\\bullet \bf \ \dfrac{1}{cosA } = secA \\\\\bullet \ \dfrac{sinA}{cosA} = tanA \\\\\longrightarrow\sf \dfrac{2sec^2A}{tan^2A-1}$

$\bf \therefore \dfrac{sinA+cosA}{sinA-cosA} +\dfrac{sinA-cosA}{sinA+cosA}= \dfrac{2}{sin^2A-cos^2A} = \dfrac{2sec^2A}{tan^2A-1}$

Hence proved.