Inverse Circular Functions 3 Question 1
1. The value of $\sin ^{-1} \frac{12}{13}-\sin ^{-1} \frac{3}{5}$ is equal to
(a) $\pi-\sin ^{-1} \frac{63}{65}$
(b) $\frac{\pi}{2}-\sin ^{-1} \frac{56}{65}$
(c) $\frac{\pi}{2}-\cos ^{-1} \frac{9}{65}$
(d) $\pi-\cos ^{-1} \frac{33}{65}$
(2019 Main, 12 April I)
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Answer:
Correct Answer: 1. (b)
Solution:
$ \begin{aligned} & \text { Key Idea Use formulae } \\ & \text { (i) } \sin ^{-1} x-\sin ^{-1} y \\ & =\sin ^{-1}\left(x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}\right) \text { if } x^{2}+y^{2} \leq 1 \\ & \text { or if } x y>0 \text { and } x^{2}+y^{2}>1 \forall x, y \in[-1,1] \\ & \text { (ii) } \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}} \text { and } \\ & \text { (iii) } \sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2} \end{aligned} $
We have,
$ \begin{aligned} \sin ^{-1} \frac{12}{13} & -\sin ^{-1} \frac{3}{5} \\ & =\sin ^{-1} \frac{12}{13} \sqrt{1-\frac{3}{5}^{2}}-\frac{3}{5} \sqrt{1-\frac{12}{13}^{2}} \end{aligned} $
$\left[\because \sin ^{-1} x-\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^{2}}-y \sqrt{1-x^{2}}\right)\right.$,
if $x^{2}+y^{2} \leq 1$ or if $x y>0$ and $\left.x^{2}+y^{2}>1 \forall x, y \in[-1,1]\right]$
$=\sin ^{-1} \frac{12}{13} \times \frac{4}{5}-\frac{3}{5} \times \frac{5}{13} $
$=\sin ^{-1} \frac{48-15}{65} $
$=\sin ^{-1} \frac{33}{65} $
$=\cos ^{-1} \sqrt{1-\frac{33}{65}} $
$=\cos ^{-1} \sqrt{\frac{3136}{4225}} $ $\because \sin ^{-1} x=\cos ^{-1} \sqrt{1-x^{2}}$
$=\cos ^{-1} \frac{56}{65}=\frac{\pi}{2}-\sin ^{-1} \frac{56}{65} $ $\because \sin ^{-1} \theta + \cot ^{-1} \theta = \frac {\pi}{2} $
