Question

If $\alpha ={\cot }^{-1}\left(\frac{-3}{4}\right),$ then the value of $\sin \left(\frac{\alpha }{2}\right)+\cos \left(\frac{\alpha }{2}\right)$ is equal to

• A. $1$
• B. $\frac{3}{\sqrt{5}}$
• C. $\sqrt{2}\sin (\frac{1}{2}{\cot }^{-1}(-\frac{3}{4})+{\cot }^{-1}1)$
• D. $\sqrt{2}\sin (\pi -{\tan }^{-1}(1)-\frac{1}{2}{\tan }^{-1}\frac{4}{3})$

Answer 1

Answer: (B), (C), (D)

$\sqrt{2}\sin (\pi -{\tan }^{-1}(1)-\frac{1}{2}{\tan }^{-1}\frac{4}{3})$

$\alpha ={\cot }^{-1}\left(\frac{-3}{4}\right)$
$\therefore \alpha \in (\frac{\pi }{2},\pi )$
$\alpha =\pi -{\cot }^{-1}\left(\frac{3}{4}\right)$
$\Rightarrow {\tan }^{-1}\left(\frac{4}{3}\right)=\pi -\alpha$

Let $y^2={(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})}^2$
$\Rightarrow y^2=1+\sin \alpha$
$\Rightarrow y^2=1+\frac{4}{5}=\frac{9}{5}$
$\Rightarrow y=\frac{3}{\sqrt{5}}$

$\sqrt{2}\sin (\frac{1}{2}{\cot }^{-1}(-\frac{3}{4})+{\cot }^{-1}1)$
$=\sqrt{2}\sin (\frac{\pi }{4}+\frac{\alpha }{2})$
$=\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}$

$\sqrt{2}\sin (\pi -{\tan }^{-1}(1)-\frac{1}{2}{\tan }^{-1}\frac{4}{3})$
$=\sqrt{2}\sin (\pi -\frac{\pi }{4}-\frac{1}{2}(\pi -\alpha ))$
$=\sqrt{2}\sin (\frac{\pi }{4}+\frac{\alpha }{2})$
$=\cos \frac{\alpha }{2}+\sin \frac{\alpha }{2}$