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Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions1 Mark
Question
In a $\triangle\text{ABC},$ if C is a right angle, then $\tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{b}}\Big)=$
  1. $\frac{\pi}{3}$
  2. $\frac{\pi}{4}$
  3. $\frac{5\pi}{2}$
  4. $\frac{\pi}{6}$
✓

Answer

  1. $\frac{\pi}{4}$
Solution:
We know,
$\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\therefore\ \tan^{-1}\Big(\frac{\text{a}}{\text{b}+\text{c}}\Big)+\tan^{-1}\Big(\frac{\text{b}}{\text{c}+\text{a}}\Big)=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{a}}{\text{b}+\text{c}}+\frac{\text{b}}{\text{c}+\text{a}}}{1-\frac{\text{a}}{\text{b}+\text{c}}\times\frac{\text{b}}{\text{c}+\text{a}}}\end{pmatrix}$
$=\tan^{-1}=\begin{pmatrix}\frac{\frac{\text{ac}+\text{a}^2+\text{b}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}{\frac{\text{ac}+\text{c}^2+\text{bc}}{(\text{b}+\text{c})(\text{c}+\text{a})}}\end{pmatrix}$
$=\tan^{-1}\Big(\frac{\text{ac}+\text{c}^2+\text{bc}}{\text{ac}+\text{c}^2+\text{bc}}\Big)$ $\big[\because\text{a}^2+\text{b}^2=\text{c}^2\big]$
$=\tan^{-1}(1)$
$=\tan^{-1}\Big(\tan\frac{\pi}{4}\Big)$
$=\frac{\pi}{4}$

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