Prove $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$
(A) Let $\sin ^{-1} \frac{3}{5}=x$. Then,$\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.
$\therefore \tan x=\frac{3}{4} \Rightarrow x=\tan ^{-1} \frac{3}{4}$.
$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \dots(1)$.
Now,let $\cos ^{-1} \frac{12}{13}=y$. Then,$\cos y=\frac{12}{13} \Rightarrow \sin y=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$.
$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$.
$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12} \dots(2)$.
Now,consider the $L$.$H$.$S$: $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$.
Using equations $(1)$ and $(2)$,we get $\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4}$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$:
$= \tan ^{-1} \left( \frac{\frac{5}{12}+\frac{3}{4}}{1-\left(\frac{5}{12} \cdot \frac{3}{4}\right)} \right) = \tan ^{-1} \left( \frac{\frac{20+36}{48}}{\frac{48-15}{48}} \right) = \tan ^{-1} \left( \frac{56}{33} \right)$.
To convert $\tan ^{-1} \frac{56}{33}$ to $\sin ^{-1}$,let $\tan ^{-1} \frac{56}{33} = z$. Then $\tan z = \frac{56}{33}$.
Using the identity $\sin z = \frac{\tan z}{\sqrt{1+\tan^2 z}} = \frac{56/33}{\sqrt{1+(56/33)^2}} = \frac{56/33}{\sqrt{(1089+3136)/1089}} = \frac{56/33}{65/33} = \frac{56}{65}$.
Thus,$\tan ^{-1} \frac{56}{33} = \sin ^{-1} \frac{56}{65} = R.H.S$.
$\therefore \tan x=\frac{3}{4} \Rightarrow x=\tan ^{-1} \frac{3}{4}$.
$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \dots(1)$.
Now,let $\cos ^{-1} \frac{12}{13}=y$. Then,$\cos y=\frac{12}{13} \Rightarrow \sin y=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$.
$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$.
$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12} \dots(2)$.
Now,consider the $L$.$H$.$S$: $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$.
Using equations $(1)$ and $(2)$,we get $\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4}$.
Using the formula $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$:
$= \tan ^{-1} \left( \frac{\frac{5}{12}+\frac{3}{4}}{1-\left(\frac{5}{12} \cdot \frac{3}{4}\right)} \right) = \tan ^{-1} \left( \frac{\frac{20+36}{48}}{\frac{48-15}{48}} \right) = \tan ^{-1} \left( \frac{56}{33} \right)$.
To convert $\tan ^{-1} \frac{56}{33}$ to $\sin ^{-1}$,let $\tan ^{-1} \frac{56}{33} = z$. Then $\tan z = \frac{56}{33}$.
Using the identity $\sin z = \frac{\tan z}{\sqrt{1+\tan^2 z}} = \frac{56/33}{\sqrt{1+(56/33)^2}} = \frac{56/33}{\sqrt{(1089+3136)/1089}} = \frac{56/33}{65/33} = \frac{56}{65}$.
Thus,$\tan ^{-1} \frac{56}{33} = \sin ^{-1} \frac{56}{65} = R.H.S$.
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View SolutionLet $Z$ denote the set of integers. Then match the items in List-$I$ with those of the items in List-$II$.
List-$I$ List-$II$ $A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$ $I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$ $B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$ $II$. $k \pi \pm 1, k \in Z$ $C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$ $III$. $\frac{3}{2}$ $D$. $\sin ^{-1}|\sin x|=\sqrt{\sin ^{-1}|\sin x|} \Rightarrow x \in$ $IV$. $\frac{3 \pi}{8}$ $V$. $\frac{\pi}{2}$
The correct match is:
| List-$I$ | List-$II$ |
| $A$. $\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1} \frac{1}{3}$ | $I$. $k \pi \pm(-1)^k \frac{\pi}{6}, k \in Z$ |
| $B$. $\sin ^{-1}\left(\frac{(-1)^n}{2}\right), n \in Z$ | $II$. $k \pi \pm 1, k \in Z$ |
| $C$. $\tan ^{-1}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right)$ | $III$. $\frac{3}{2}$ |
| $D$. $\sin ^{-1}|\sin x|=\sqrt{\sin ^{-1}|\sin x|} \Rightarrow x \in$ | $IV$. $\frac{3 \pi}{8}$ |
| $V$. $\frac{\pi}{2}$ |
The correct match is:
If $\alpha$ and $\beta$ are the roots of the equation $6x^2 - 5x + 1 = 0$,then the value of $\tan^{-1}\alpha + \tan^{-1}\beta$ is:
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