Inverse Circular Functions Ques 12
- If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1} \frac{4}{9}$, where the inverse trigonometric functions take only the principal values, then the correct option(s) is/are
(2015 Adv.)
(a) $\cos \beta>0$
(b) $\sin \beta<0$
(c) $\cos (\alpha+\beta)>0$
(d) $\cos \alpha<0$
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Answer:
Correct Answer: 12.(b,c,d)
Solution:
Formula:
- Here, $\alpha=3 \sin ^{-1} (\frac{6}{11})$ and $\beta=3 \cos ^{-1} (\frac{4}{9})$ as $\frac{6}{11}>\frac{1}{2}$
$ \begin{aligned} & \Rightarrow \quad \sin ^{-1} (\frac{6}{11})>\sin ^{-1} (\frac{1}{2})=\frac{\pi}{6} \\ & \therefore \quad \alpha=3 \sin ^{-1} \frac{6}{11}>\frac{\pi}{2} \\ & \Rightarrow \quad \cos \alpha<0 \end{aligned} $
Now, $\quad \beta=3 \cos ^{-1} (\frac{4}{9})$
As $\quad \frac{4}{9}<\frac{1}{2} \Rightarrow \cos ^{-1} (\frac{4}{9})>\cos ^{-1} (\frac{1}{2})=\frac{\pi}{3}$
$\therefore \quad \beta=3 \cos ^{-1} (\frac{4}{9})>\pi$
$\therefore \quad \cos \beta<0$ and $\sin \beta<0$
Now, $\alpha+\beta$ is slightly greater than $\frac{3 \pi}{2}$.
$\therefore \quad \cos (\alpha+\beta)>0$
BETA
