Question

(sinA ÷ 1 - cotA) + (cosA ÷ 1 - tanA) = sinA + cosA



It is a proving question.​

Answer 1

$LHS = \frac{sinA}{1-cotA} + \frac{cosA}{1-tanA}$

⇒$\frac{sinA}{1-\frac{cosA}{sinA} } + \frac{cosA}{1-\frac{sinA}{cosA}}$

⇒$\frac{sin^2A}{sinA-cosA}+ \frac{cos^2A}{cosA-sinA}$

⇒$\frac{sin^2a}{sinA-cosA}-\frac{cos^2A}{sinA-cosA}$  

⇒$\frac{sin^2A-cos^2A}{sinA-cosA}$

Now, since

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

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

∴$LHS = sinA + cosA = RHS$

Hence proved :)