Question

$\frac{(3+i\sin \left(\theta \right))}{(4-i\cos \left(\theta \right))},\theta \in (0,2\mathrm{\pi }),$is a real number. Then, $arg(\cos \left(\theta \right)+i\sin \left(\theta \right))=$

• A. $\mathrm{\pi }-{\tan }^{-1}\left(\frac{4}{3}\right)$
• B. $-{\tan }^{-1}\left(\frac{3}{4}\right)$
• C. $\mathrm{\pi }-{\tan }^{-1}\left(\frac{3}{4}\right)$
• D. ${\tan }^{-1}\left(\frac{4}{3}\right)$

Answer 1

The correct option is A

$\mathrm{\pi }-{\tan }^{-1}\left(\frac{4}{3}\right)$

Modify the denominator so that it doesn't contain imaginary part.

Let,

$$\begin{gathered} z=\frac{(3+i\sin \left(\theta \right))}{(4-i\cos \left(\theta \right))} \\ =\frac{(3+i\sin \left(\theta \right))}{(4-i\cos \left(\theta \right))}\times \frac{(4+i\cos \left(\theta \right))}{(4+i\cos \left(\theta \right))} \\ =\frac{(12-\sin \left(\theta \right)\cos \left(\theta \right)+i(4\sin \left(\theta \right)+3\cos \left(\theta \right)\left)\right)}{(16+{\cos }^2\left(\theta \right))} \end{gathered}$$

As $Z$ is real, therefore,

$$\begin{gathered} (4\sin \left(\theta \right)+3\cos \left(\theta \right))=0 \\ \Rightarrow \tan \left(\theta \right)=-\frac{3}{4}[90°<\theta <180°] \\ \therefore arg[\sin \left(\theta \right)+i\cos \left(\theta \right)]=\mathrm{\pi }+{\tan }^{-1}\left(\frac{\cos \left(\mathrm{\theta }\right)}{\sin \left(\mathrm{\theta }\right)}\right) \\ =\mathrm{\pi }-{\tan }^{-1}\left(\frac{4}{3}\right)\mathbf{}\left(\because \cot \left(\theta \right)=-\frac{4}{3}\right) \end{gathered}$$

Therefore, Option(A) is the correct answer..