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TS EAMCET 4 May 2018 Shift 1 Solved Paper

Section: Mathematics

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Question : 24 of 160

Marks: +1, -0

2

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

\sec^{-1} \frac{5\sqrt{2}}{7}

\tan^{-1} \frac{1}{8}

=

$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{8}$

Solution:

Consider the trigonometric function.

2 \tan^{-1} \frac{1}{5}+\sec^{-1} \frac{5 \sqrt{2}}{7}+2 \tan^{-1} \frac{1}{8}

Let

I=2 \tan^{-1} \frac{1}{5}+\sec^{-1} \frac{5\sqrt{2}}{7}+2 \tan^{-1} \frac{1}{8}

Its value is calculated as,

I=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}{40}} \right]+\sec^{-1} \frac{5\sqrt{2}}{7}

=2 \tan^{-1} \frac{13}{39}+\sec^{-1} \frac{5\sqrt{2}}{7}

Solve further,

I=2 \tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{7}

=\tan^{-1} \frac{\frac{2}{3}}{1-\frac{1}{9}}+\tan^{-1} \frac{1}{7}

=\tan^{-1} \frac{6}{8}+\tan^{-1} \frac{1}{7}

=\tan^{-1} \left( \frac{\frac{3}{4}+\frac{1}{7}}{1-\left( \frac{3}{4} \times \frac{1}{7} \right)} \right)

Solve further,

I=\tan^{-1} \left( \frac{25}{25} \right)

=\tan^{-1}(1)

=\frac{\pi}{4}

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