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BITSAT 2012

Exam: BITSAT
Year: 2012
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

BITSAT 2012 · Mathematics · Inverse Trigonometric Functions

^ -1 ((1)/(4))+ ^ -1 ((2)/(9)) equal to-

\tan ^{-1}

\left(\frac{1}{4}\right)+\tan ^{-1}

\left(\frac{2}{9}\right)

equal to-

Options

A $\frac{1}{2} \cos ^{-1}\left(\frac{3}{5}\right)$
B $\frac{1}{2} \sin ^{-1}\left(\frac{3}{5}\right)$
C $\frac{1}{2} \tan ^{-1}\left(\frac{3}{5}\right)$
D $\tan ^{-1}\left(\frac{1}{2}\right)$

Correct answer

D. $\tan ^{-1}\left(\frac{1}{2}\right)$

Step-by-step solution

\tan ^{-1}

\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)

\quad=\tan ^{-1}\left[\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right]=\tan ^{-1}\left[\frac{1}{2}\right]

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