Show that $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16}=\pi$
(A) Let $\sin ^{-1} \frac{12}{13}=x$,$\cos ^{-1} \frac{4}{5}=y$,and $\tan ^{-1} \frac{63}{16}=z$.
Then $\sin x = \frac{12}{13}$,$\cos y = \frac{4}{5}$,and $\tan z = \frac{63}{16}$.
From these,we find $\cos x = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$,$\sin y = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$.
Thus,$\tan x = \frac{\sin x}{\cos x} = \frac{12/13}{5/13} = \frac{12}{5}$ and $\tan y = \frac{\sin y}{\cos y} = \frac{3/5}{4/5} = \frac{3}{4}$.
Now,calculate $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - (\frac{12}{5} \times \frac{3}{4})} = \frac{\frac{48+15}{20}}{1 - \frac{36}{20}} = \frac{63/20}{-16/20} = -\frac{63}{16}$.
Since $\tan(x+y) = -\frac{63}{16}$ and $\tan z = \frac{63}{16}$,we have $\tan(x+y) = -\tan z = \tan(\pi - z)$.
Since $x, y, z$ are in the first quadrant,$x+y+z = \pi$.
Then $\sin x = \frac{12}{13}$,$\cos y = \frac{4}{5}$,and $\tan z = \frac{63}{16}$.
From these,we find $\cos x = \sqrt{1 - (\frac{12}{13})^2} = \frac{5}{13}$,$\sin y = \sqrt{1 - (\frac{4}{5})^2} = \frac{3}{5}$.
Thus,$\tan x = \frac{\sin x}{\cos x} = \frac{12/13}{5/13} = \frac{12}{5}$ and $\tan y = \frac{\sin y}{\cos y} = \frac{3/5}{4/5} = \frac{3}{4}$.
Now,calculate $\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\frac{12}{5} + \frac{3}{4}}{1 - (\frac{12}{5} \times \frac{3}{4})} = \frac{\frac{48+15}{20}}{1 - \frac{36}{20}} = \frac{63/20}{-16/20} = -\frac{63}{16}$.
Since $\tan(x+y) = -\frac{63}{16}$ and $\tan z = \frac{63}{16}$,we have $\tan(x+y) = -\tan z = \tan(\pi - z)$.
Since $x, y, z$ are in the first quadrant,$x+y+z = \pi$.
Explore More
Similar Questions
If $\tan ^{-1}(x+1)+\tan ^{-1} x+\tan ^{-1}(x-1)=\tan ^{-1} 3$,then for $x < 0$ the value of $500 x^4+270 x^2+997=$
The number of real solutions of $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^2+x+1}=\frac{\pi}{2}$ is
$\sec ^2(\tan ^{-1} 2) + \operatorname{cosec}^2(\cot ^{-1} 3)$ is equal to
DifficultAP EAMCET 2001
View Solution$\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots$ up to $\infty$ is equal to
DifficultKCET 2009
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:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo