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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions4 Marks
Question
Show that $2 \cot ^{-1} \frac{3}{2}+\sec ^{-1} \frac{13}{12}=\frac{\pi}{2}$
✓

Answer

$2 \cot ^{-1} \frac{3}{2}=2 \tan ^{-1} \frac{2}{3} \quad \ldots\left[\because \cot ^{-1} x=\tan ^{-1}\left(\frac{1}{x}\right)\right]$
$=\tan ^{-1}\left[\frac{2 \times \frac{2}{3}}{1-\left(\frac{2}{3}\right)^2}\right]$
$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$
$=\tan ^{-1}\left[\frac{\left(\frac{4}{3}\right)}{1-\frac{4}{9}}\right]$
$=\tan ^{-1}\left(\frac{4}{3} \times \frac{9}{5}\right)=\tan ^{-1} \frac{12}{5}$
$\ldots$ (1)
Let $\sec ^{-1} \frac{13}{12}=\alpha$
Then, $\sec \alpha=\frac{13}{12}$, where $0<\alpha<\frac{\pi}{2}$
$\therefore \tan \alpha>0$
Now, $\tan \alpha=\sqrt{\sec ^2 \alpha-1}$
$=\sqrt{\frac{169}{144}-1}=\sqrt{\frac{25}{144}}=\frac{5}{12}$
$\therefore \alpha=\tan ^{-1} \frac{5}{12}=\cot ^{-1} \frac{12}{5} \ldots\left[\because \tan ^{-1} x=\cot ^{-1}\left(\frac{1}{x}\right)\right]$
$\therefore \sec ^{-1} \frac{13}{12}=\cot ^{-1} \frac{12}{5}$
$\ldots(2)$
Now, LHS $=2 \cot ^{-1} \frac{3}{2}+\sec ^{-1} \frac{13}{12}$
$=\tan ^{-1} \frac{12}{5}+\cot ^{-1} \frac{12}{5} \quad \ldots[$ By (1) and (2)]
$\begin{array}{ll}=\frac{\pi}{2} \quad \cdots\left[\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right] \\ =\text { RHS. }\end{array}$

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