Question

Prove that, $tan[\frac{1}{2}sin^{-1}(\frac{2x}{1-x^2})+\frac{1}{2}cos^{-1}(\frac{1-y^2}{1+y^2})]=\frac{x+y}{1-xy}$

Answer 1

Step-by-step explanation:

LHS = tan[1/2 sin⁻¹(2x/1-x²)+1/2 cos⁻¹(1-y²/1+y²)].....................................1

WKT,

sin⁻¹(2x/1-x²) = 2tan⁻¹x

and

cos⁻¹(1-y²/1+y²) = 2tan⁻¹y

Substituting above two in 1,

= tan[(1/2) × 2tan⁻¹x + (1/2) × 2tan⁻¹y]

= tan[tan⁻¹x + tan⁻¹y]

Now, we also know that,

tan⁻¹x + tan⁻¹y = tan⁻¹((x+y)/(1-xy))

Therefore,

= tan[tan⁻¹((x+y)/(1-xy))]

= (x+y)/(1-xy)

Hence Proved