Answer the question [ 4 mark question ]

Question 14 Marks

Prove that $\tan ^{-1}\left(\frac{63}{16}\right)=\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)$

Answer

Suppose, $\sin ^{-1} \frac{5}{13}=\alpha \therefore \sin \alpha=\frac{5}{13}$
$\tan \alpha=\frac{\sin \alpha}{\sqrt{1-\sin ^2 \alpha}}=\frac{\frac{5}{13}}{\sqrt{1-\frac{25}{169}}}$
$=\frac{5}{13} \times \frac{13}{12}=\frac{5}{12}$
$\text { and } \beta=\cos ^{-1} \frac{3}{5} \therefore \cos \beta=\frac{3}{5}$
$\sin \beta=\sqrt{1-\cos ^2 \beta}$
$=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
$\tan \beta=\frac{\sin \beta}{\cos \beta}=\frac{4}{5} \times \frac{5}{3}=\frac{4}{3}$
$\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}$
$=\frac{15+48}{36-20}=\frac{63}{16}$
$\Rightarrow \alpha+\beta=\tan ^{-1}\left(\frac{63}{16}\right)$
$\Rightarrow \sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{63}{16}\right)$
Hence proved.

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Question 24 Marks

Prove that $ \begin{array}{l} \tan ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right]=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x ,0 < x < 1 \end{array} $

Answer

To Prove
$\tan ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right]=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x, 0 < x < 1$
$\text{L.H.S.} = \tan ^{-1}\left[\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right]$
Suppose $ x=\cos 2 \theta$
$\therefore 1+x =1+\cos 2 \theta=2 \cos ^2 \theta$
$1-x =1-\cos 2 \theta=2 \sin ^2 \theta$
$\therefore \text { L.H.S. } =\tan ^{-1}\left[\frac{\sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta}}{\sqrt{2 \cos ^2 \theta}-\sqrt{2 \sin ^2 \theta}}\right]$
$ =\tan ^{-1}\left[\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}\right]$
$ =\tan ^{-1}\left[\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}\right]$
$ =\tan ^{-1}\left[\frac{1+\tan \theta}{1-\tan \theta}\right]$
Here divide the numerator and denominator by $\cos \theta$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\theta\right)\right]$
$=\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x (\because \cos 2 \theta=x)$
$=\text { RHS }$ Hence proved.

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Question 34 Marks

Prove that $\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} \frac{x}{y}\right)+\tan \left(\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} \frac{x}{y}\right)=\frac{2 y}{x}$

Answer

Suppose that $\frac{1}{2} \cos ^{-1} \frac{x}{y}=\theta$ then
$\cos ^{-1} \frac{x}{y}=2 \theta \therefore \frac{x}{y}=\cos 2 \theta$
$\text { L.H.S }=\tan \left(\frac{\pi}{4}+\theta\right)+\tan \left(\frac{\pi}{4}-\theta\right)$
$=\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}+\frac{\tan \frac{\pi}{4}-\tan \theta}{1+\tan \frac{\pi}{4} \tan \theta}$
$=\frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta}$
$=\frac{(1+\tan \theta)^2+(1-\tan \theta)^2}{(1-\tan \theta)(1+\tan \theta)}$
$=\frac{1+2 \tan \theta+\tan ^2 \theta+1-2 \tan \theta+\tan ^2 \theta}{\left(1-\tan ^2 \theta\right)}$
$=\frac{2+2 \tan ^2 \theta}{\left(1-\tan ^2 \theta\right)}=\frac{2\left(1+\tan ^2 \theta\right)}{1-\tan ^2 \theta}$
$=\frac{2}{\frac{\left(1-\tan ^2 \theta\right)}{1+\tan ^2 \theta}}=\frac{2}{\cos 2 \theta}$
Putting the value
$=\frac{2}{x / y}=\frac{2 y}{x}=\text{R.H.S.}$
Hence proved.

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Question 44 Marks

$\cos ^{-1} \frac{1-a^2}{1+a^2}+\cos ^{-1} \frac{1-b^2}{1+b^2}=2 \tan ^{-1} x$

Answer

Suppose that then
$a =\tan \theta_1, b=\tan \theta_2$
$\theta_1 =\tan ^{-1} a, \theta_2=\tan ^{-1} b$
$\therefore \frac{1-a^2}{1+a^2}=\frac{1-\tan ^2 \theta_1}{1+\tan ^2 \theta_1}=\cos 2 \theta_1$
and $ \frac{1-b^2}{1+b^2}=\frac{1-\tan ^2 \theta_2}{1+\tan ^2 \theta_2}=\cos 2 \theta_2$
hence from given equation :$\cos ^{-1}\left(\cos 2 \theta_1\right)+\cos ^{-1}\left(\cos 2 \theta_2\right)=2 \tan ^{-1} x$
$2 \theta_1+2 \theta_2=2 \tan ^{-1} x$
$\theta_1+\theta_2=\tan ^{-1} x$
$\Rightarrow \tan \left(\theta_1+\theta_2\right)=x$
$\Rightarrow \frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}=x$
$\Rightarrow \frac{a+b}{1-a b}=x$
$\therefore x=\frac{a+b}{1-a b} \text {}$

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Mathematics Answer the question [ 4 mark question ]

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