Inverse Trigonometric Functions Question 4
Question 4 - 31 January - Shift 1
If $\sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0,0<\alpha<13$,
then $\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha)$ is equal to
(1) $\pi$
(2) 16
(3) 0
(4) $16-5 \pi$
Show Answer
Answer: (1)
Solution:
Formula: Identity of subtraction of inverse tangent function and formula for sin inverse of sin of x.
$ \cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4} $
$\therefore \sin ^{-1} \frac{\alpha}{17}=\tan ^{-1} \frac{77}{36}-\tan ^{-1} \frac{3}{4}=\tan ^{-1}(\frac{\frac{77}{36}-\frac{3}{4}}{1+\frac{77}{36} \cdot \frac{3}{4}})$
$\sin ^{-1} \frac{\alpha}{17}=\tan ^{-1} \frac{8}{15}=\sin ^{-1} \frac{8}{17}$
$\Rightarrow \frac{\alpha}{17}=\frac{8}{17} \Rightarrow \alpha=8$
$\therefore \sin ^{-1}(\sin 8)+\cos ^{-1}(\cos 8)$
$=3 \pi-8+8-2 \pi$
$=\pi$
