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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}=$
  • A
    $\frac{\pi}{6}$
  • ✓
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{3}$
  • D
    $\frac{\pi}{8}$
✓

Answer

Correct option: B.
$\frac{\pi}{4}$
(B) $2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}$
$=2\left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
$=2 \tan ^{-1}\left[\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5}\left(\frac{1}{8}\right)}\right]+\tan ^{-1} \sqrt{\left(\frac{5 \sqrt{2}}{7}\right)^2-1}$
$\ldots\left[\because \sec ^{-1} x=\tan ^{-1} \sqrt{x^2-1}\right]$
$=2 \tan ^{-1} \frac{13}{39}+\tan ^{-1} \frac{1}{7}$
$=2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$
$=\tan ^{-1}\left[\frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{3}\right)^2}\right]+\tan ^{-1} \frac{1}{7}$
$\ldots\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right.$, if $\left.-1<x<1\right]$
$=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}$
$=\tan ^{-1}\left[\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}\left(\frac{1}{7}\right)}\right]=\tan ^{-1}(1)=\frac{\pi}{4}$

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