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Question: If \(\sin^{- 1}(A + iB) = x + iy,\) then \(\frac{A}{B}\) equals....

If sin⁡−1(A+iB)=x+iy,\sin^{- 1}(A + iB) = x + iy, then AB\frac{A}{B} equals.

A

tan⁡xtanh⁡y\frac{\tan x}{\tanh y}

B

tanh⁡xtan⁡y\frac{\tanh x}{\tan y}

C

tanh⁡xtanh⁡y\frac{\tanh x}{\tanh y}

D

cos⁡xcos⁡hy\frac{\cos x}{{\cos h}y}

Answer

tan⁡xtanh⁡y\frac{\tan x}{\tanh y}

Explanation

Solution

sin⁡−1(A+iB)=x+iy\sin^{- 1}(A + iB) = x + iy ⇒ A+iB=sin⁡(x+iy)A + iB = \sin(x + iy)

⇒A+iB=sin⁡xcos⁡hy+icos⁡xsin⁡hyA + iB = \sin x{\cos h}y + i\cos x{\sin h}y

∴AB=sin⁡xcos⁡hycos⁡xsin⁡hy=tan⁡xtan⁡hy\frac{A}{B} = \frac{\sin x{\cos h}y}{\cos x{\sin h}y} = \frac{\tan x}{{\tan h}y}.