Question

If $\theta =co{s}^{-1}\left(4/5\right)+ta{n}^{-1}\left(2/3\right)$ then

• A. $sin\theta =\frac{17}{5\sqrt{13}}$
• B. $cos\theta =\frac{6}{5\sqrt{13}}$
• C. $tan\theta =\frac{17}{6}$
• D. $cot\theta =\frac{17}{6}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $tan\theta =\frac{17}{6}$
B $cos\theta =\frac{6}{5\sqrt{13}}$
C $sin\theta =\frac{17}{5\sqrt{13}}$

Given, $\theta ={\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)$
Let ${\cos }^{-1}\left(\frac{4}{5}\right)=\alpha \Rightarrow \cos \alpha =\frac{4}{5}\Rightarrow \sin \alpha =\frac{3}{5}\Rightarrow \tan \alpha =\frac{3}{4}$
And
${\tan }^{-1}\left(\frac{2}{3}\right)=\beta \Rightarrow \tan \beta =\frac{2}{3}\Rightarrow \sin \beta =\frac{2}{\sqrt{13}}\Rightarrow \cos \beta =\frac{3}{\sqrt{13}}$
Therefore, $\theta =\alpha +\beta$
Gives
$\sin \theta =\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta =\frac{17}{5\sqrt{13}}$
$\cos \theta =\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta =\frac{6}{5\sqrt{13}}$
$\tan \theta =\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\frac{17}{6}$
$\cot \theta =\frac{6}{17}$