3 Marks Question

Question 13 Marks

Solve : $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Answer

$y=\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$
Now put the value of principal :$
\begin{aligned}
\tan ^{-1}(1) & =\frac{\pi}{4}, \cos ^{-1}\left(-\frac{1}{2}\right)=\pi-\cos ^{-1}\left(\frac{1}{2}\right) \\
& =\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \\
\sin ^{-1}\left(-\frac{1}{2}\right) & =-\sin ^{-}\left(\frac{1}{2}\right)=-\frac{\pi}{6}
\end{aligned}
$
Put these values in the equation (i) :
$\begin{aligned} y & =\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}=\frac{3 \pi+8 \pi-2 \pi}{12} \\ & =\frac{9 \pi}{12}=\frac{3 \pi}{4} \text {}\end{aligned}$

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Question 23 Marks

If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$ then find the value of $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$

Answer

We know that $0 \leq \cos ^{-1} x \leq \pi \forall x \in[-1,1]$
$\therefore \cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$
$\Rightarrow \cos ^{-1} \alpha=\cos ^{-1} \beta=\cos ^{-1} \gamma=\pi$
$\Rightarrow \alpha=\beta=\gamma=-1$
$\therefore \alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$
$\quad=(-1)(-1-1)+(-1)(-1-1)+(-1)(-1-1)$
$\quad=2+2+2=6 \text {}$

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