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BITSAT 2009

Exam: BITSAT
Year: 2009
Subject: Mathematics
Chapter: Matrices

BITSAT 2009 · Mathematics · Matrices

Find the value of (2 ^ -1 (1)/(5)+ ^ -1 (1)/(2)+ 2 ^ -1 (1)/(8) )

Find the value of (2 \tan ^{-1}

\frac{1}{5}+\tan ^{-1} \frac{1}{2}+

2 \tan ^{-1}

\frac{1}{8}

)

Options

A(\pi / 4)
B(\pi / 2)
C(3 \pi / 4)
DNone of these

Correct answer

A. (\pi / 4)

Step-by-step solution

(

\begin{aligned} & 2 \tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{2}+2 \tan ^{-1} \frac{1}{8}\\ & =2 \tan ^{-1} \frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}}+\tan ^{-1} \frac{1}{7}\\ & =2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{2 \cdot \frac{1}{3}}{1-\frac{1}{9}}+\tan ^{-1} \frac{1}{7}\\ & =\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}}\\ & =\tan ^{-1} \frac{25}{25}=\tan ^{-1}1=45^{\circ}=\frac{\pi}{4} \end{aligned}

)

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