Inverse Circular Functions 1 Question 2
2. The greater of the two angles $A=2 \tan ^{-1}(2 \sqrt{2}-1)$ and $B=3 \sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}$ is …… .
$(1989,2 M)$
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Answer:
Correct Answer: 2. $(A)$
Solution:
- Given, $A=2 \tan ^{-1}(2 \sqrt{2}-1)$
and $B=3 \sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{3}{5}$
Here, $\quad A=2 \tan ^{-1}(2 \sqrt{2}-1)$
$ =2 \tan ^{-1}(2 \times 1.414-1) $
$ =2 \tan ^{-1}(1.828) $
$\therefore \quad A>2 \tan ^{-1}(\sqrt{3})=2 \cdot \frac{\pi}{3}=\frac{2 \pi}{3}$
To find the value of $B$, we first say
$ \sin ^{-1} \frac{1}{3}<\sin ^{-1} \frac{1}{2}=\frac{\pi}{6} $
so that $\quad 0<3 \sin ^{-1} \frac{1}{3}<\frac{\pi}{2}$
Now, $\quad 3 \sin ^{-1} \frac{1}{3}=\sin ^{-1} 3 \cdot \frac{1}{3}-4 \cdot \frac{1}{27}$
$ \begin{aligned} & =\sin ^{-1} \frac{23}{27} \\ & =\sin ^{-1}(0.851)<\sin ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{3} \\ & \sin ^{-1} \frac{3}{5}=\sin ^{-1}(0.6)<\sin ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{3} \\ & \therefore \quad B<\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3} \end{aligned} $
Thus, $A>\frac{2 \pi}{3}$ and $B<\frac{2 \pi}{3}$
Hence, greater angle is $A$.
