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AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 22 of 160

Marks: +1, -0

The value of

\tan \left(\;\frac{7\pi}{8}\right)

is

$2\sqrt{2}-1$
$1-\sqrt{2}$
$1+\sqrt{2}$
$\;\frac{1}{1+\sqrt{2}}$

Solution: 👈: Video Solution

Observe that

\tan \left(\;\frac{7\pi}{8}\right) = \tan \left(\pi - \;\frac{\pi}{8}\right)

= -\tan \left(\;\frac{\pi}{8}\right)

Now, using

\tan (2\theta) = \; \frac{2\tan\theta}{1-\tan^2\theta}

Let

\theta = \; \frac{\pi}{8}

So,

\tan \left(2 \cdot \;\frac{\pi}{8}\right) = \; \frac{2\tan \frac{\pi}{8}}{1-\tan^2 \;\frac{\pi}{8}}

\Rightarrow \tan \;\frac{\pi}{4} = \; \frac{2\tan \frac{\pi}{8}}{1-\tan^2 \;\frac{\pi}{8}}

On putting

\tan \;\frac{\pi}{8} = x

,

\;1 = \; \frac{2x}{1-x^2}

\; \Rightarrow \; \; 1-x^2-2x=0 \Rightarrow x^2+2x-1=0

\; \Rightarrow \; \; x = -1-\sqrt{2},\; \sqrt{2}-1

\; \; \text{But,}\; \tan \;\frac{\pi}{8} > 0 \; \text{so, only take}\; \tan \;\frac{\pi}{8} = \sqrt{2}-1

\; \; \text{Thus,}\; \tan \left(\;\frac{7\pi}{8}\right) = -\tan \left(\;\frac{\pi}{8}\right) = -(\sqrt{2}-1) = 1-\sqrt{2}

\;

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