Question

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

• A. $\frac{\tan x}{\tan hy}$
• B. $\frac{\tan hx}{\tan y}$
• C. $\frac{\tan hx}{\tan hy}$
• D. $\frac{\cos x}{\cos hy}$

Answer 1

The correct option is A

$\frac{\tan x}{\tan hy}$

Explanation for the correct option.

Compute the required value:

Given that, ${\sin }^{-1}(A+iB)=x+iy$.

Taking $\sin$ both sides we have:

$$\begin{gathered} (A+iB)=\sin (x+iy) \\ (A+iB)=\sin \left(x\right)\cos \left(hy\right)+i\cos \left(x\right)\sin \left(hy\right) \end{gathered}$$

Comparing real & imaginary parts we have:

$$\begin{gathered} A=\sin \left(x\right)\cos \left(hy\right) \\ B=\cos \left(x\right)\sin \left(hy\right) \end{gathered}$$

Therefore the value is:

$$\begin{gathered} \frac{A}{B}=\frac{[\sin \left(x\right)\cos \left(hy\right)]}{[\cos \left(x\right)\sin \left(hy\right)]} \\ =\frac{\tan \left(x\right)}{\text{tan}\left(\text{hy}\right)} \end{gathered}$$

Hence, option A is correct.