Question

sin a - cos a ÷ sin a + cos a + sin a + cos a ÷ sin a - cos a = 2 ÷ 2sin^2 a-1​

Answer 1

Answer:

Step-by-step explanation:

Given :

To prove :

$\frac{sinA-cosA}{sinA+cosA} +\frac{sinA+cosA}{sinA-cosA} = \frac{2}{2sin^2A-1}$

Solution :

We know that,

$(a+b)^2=a^2+2ab+b^2$

$(a-b)^2=a^2-2ab+b^2$

$a^2-b^2=(a+b)(a-b)$

$sin^2A+cos^2A=1$

LHS = $\frac{sinA-cosA}{sinA+cosA} +\frac{sinA+cosA}{sinA-cosA}$

⇒ $\frac{(sinA-cosA)^2+(sinA+cosA)^2}{sinA+cosA)(sinA-cosA)}$

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

⇒ $\frac{2sin^2A+2cos^2A}{sin^2A-(1-sin^2A)}$

⇒ $\frac{2(sin^2A + cos^2A)}{sin^2A-1+sin^2A}$

⇒ $\frac{2(1)}{2sin^2A-1}$

⇒ $\frac{2}{2sin^2A-1}$ = RHS

Hence, proved,.