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Question 11 Mark

Find the values of the following :(i) $\sin \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{3}{5}\right)\right]$
(ii) $\cos \left[\cos ^{-1}\left(-\frac{1}{2}\right)+\tan ^{-1} \sqrt{3}\right]$

Answer

(i) We have if $-1 \leq x \leq 1$ then
$ \text { Here }-1<\frac{3}{5}<1$
$\therefore \quad \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$
$\quad \sin \left[\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\frac{\pi}{2}\right.$
$\text { (ii) } \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3} \text { and } \tan ^{-1} \sqrt{3}=\frac{\pi}{3} \text {. }$
$\therefore \cos \left[\cos ^{-1}\left(-\frac{1}{2}\right)+\tan ^{-1} \sqrt{3}\right]$
$=\cos \left(\frac{2 \pi}{3}+\frac{\pi}{3}\right)$
$=\cos \pi$
$=-1 .$

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Question 21 Mark

Find the values of the following(i) $\sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)$
(ii) $\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
(iii) $\sin \left(\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right)$
(iv) $\sin \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{5}{12}\right)$

Answer

(i) $\sin ^{-1}\left(\sin \left(\frac{5 \pi}{3}\right)\right)=\sin ^{-1}\left(\sin \left(2 \pi-\frac{\pi}{3}\right)\right)$
$ =\sin ^{-1}\left(\sin \left(-\frac{\pi}{3}\right)\right)=-\frac{\pi}{3}, \text { as }-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \quad \sin ^{-1}\left(\sin \frac{5 \pi}{3}\right)=-\frac{\pi}{3} $
(ii) $\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}$, as $\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$(iii) We have $\cos ^{-1}(-x)=\pi-\cos ^{-1} x$
$ \therefore \sin \left(\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right)=\sin \left(\pi-\cos ^1 \frac{1}{\sqrt{2}}\right)$
$=\sin \left(\pi-\frac{\pi}{4}\right)=\sin \left(\frac{3 \pi}{4}\right)=\frac{1}{\sqrt{2}}$
$\therefore \quad \sin \left(\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right)=\frac{1}{\sqrt{2}} $
(iv) Let $\cos ^{-1} \frac{4}{5}=\theta$ and $\tan ^{-1} \frac{5}{12}=\phi$
$ \therefore \quad \cos \theta=\frac{4}{5} \text { and } \tan \phi=\frac{5}{12}$
$\therefore \quad \sin \theta=\frac{3}{5} \text { and } \sin \phi=\frac{5}{13}, \cos \phi=\frac{12}{13}$
$\quad \sin \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{5}{12}\right)=\sin (\theta+\phi)$
$=\quad \sin \theta \cos \phi+\cos \theta \sin \phi$
$=\quad \frac{3}{5} \frac{12}{13}+\frac{4}{5} \frac{5}{13}=\frac{56}{65}$
$\therefore \quad \sin \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{5}{12}\right)=\frac{56}{65} $

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Question 31 Mark

Find the principal values of the following :
(i) $\sin ^{-1}\left(-\frac{1}{2}\right)$
(ii) $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
(iii) $\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$

Answer

(i) $\sin ^{-1}\left(-\frac{1}{2}\right)$

We have, $\sin \left(-\frac{1}{2}\right)=-\frac{1}{2}$, where $-\frac{1}{2} \in[-1,1]$ and $-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}

The principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$ is $-\frac{\pi}{6}$(ii) $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
We have, $\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$, where $\frac{\sqrt{3}}{2} \in[-1,1]$ and $\frac{\pi}{6} \in[0, \pi]$
$
\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}
$
The principal value of $\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is $\frac{\pi}{6}$.
(iii) $\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$
We have, $\cot \frac{2 \pi}{3}=\frac{-1}{\sqrt{3}} \quad$, where $\frac{-1}{\sqrt{3}} \in R$ and $\frac{\pi}{6} \in(0, \pi)$
$\therefore \quad$ The principal value of $\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ is $\frac{2 \pi}{3}$

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Question 41 Mark

In $\triangle A B C$ if $a=13, b=14, c=15$ then find the values of(i) $\cos A$
(ii) $\sin \frac{A}{2}$
(iii) $\cos \frac{A}{2}$
(iv) $\tan \frac{A}{2}$
(v) $A (\triangle ABC )$
(vi) $\sin A$

Answer

$s =\frac{a+b+c}{2}=\frac{13+14+15}{2}=21$
$ (s-a)=21-13=8$
$(s-b)=21-14=7$
$(s-c)=21-15=6 $
(i)$cos A =\frac{b^2+c^2-a^2}{2 b c}$
$ =\frac{13^2+15^2-14^2}{2(13)(15)}=\frac{198}{390}=\frac{33}{65}$
(ii)$ \sin \frac{A}{2} =\sqrt{\frac{(s-b(s-c)}{b c}}$
$=\sqrt{\frac{7 \times 6}{14 \times 15}}=\frac{1}{\sqrt{5}}$
{(iii)} $ \cos \frac{A}{2} =\sqrt{\frac{s(s-a)}{b c}}$
$ =\sqrt{\frac{21 \times 8}{14 \times 15}}=\frac{2}{\sqrt{5}} $
(iv) $\tan \frac{A}{2}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}=\frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}}=\frac{1}{2}$
(v)$A (\Delta ABC ) =\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{21 \times 8 \times 7 \times 6}=84 \text { sq. unit } $
(vi) $\quad \sin A =2 \sin \frac{A}{2} \cos \frac{A}{2}$
$=2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}}=\frac{4}{5}$

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Question 51 Mark

Prove the following
(i) $2 \tan ^{-1}\left(-\frac{1}{3}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\frac{\pi}{2}$
(ii) $2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)=\frac{\pi}{4}$

Answer

$\text { (i) } 2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{3}\right) \text {, as } x y=\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}<1$
$=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\frac{1}{3} \frac{1}{3}}\right)=\tan ^{-1} \frac{3}{4}=\theta, \text { (say) }$
$ \therefore \quad \tan \theta=\frac{3}{4} \quad \therefore 0<\theta<\frac{\pi}{2}$
$\therefore \quad \sin \theta=\frac{3}{5} \quad \therefore \theta=\sin ^{-1} \frac{3}{5}$
$\therefore \quad 2 \tan ^{-1} \frac{1}{3}=\tan ^{-1} \frac{3}{4}=\sin ^{-1} \frac{3}{5}$
$\therefore \quad 2 \tan ^{-1} \frac{1}{3}+\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{3}{5}+\cos ^{-1} \frac{3}{5}=\frac{\pi}{2} $
(ii) $2 \tan ^{-1} \frac{1}{3}=\tan ^{-1} \frac{3}{4}$ as seen in (i)
$ \therefore 2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7} \text { and } x y=\frac{3}{4} \times \frac{1}{7}=\frac{3}{28}<1$
$=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \frac{1}{7}}\right)=\tan ^{-1} 1=\frac{\pi}{4} $

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