If θ does not lie in the second quadrant and θ = \; -3/4, then \; θ/2 + 2θ =

Correct Answer is : −9775-97/7575−97

Correct Answer is : −9775-97/7575−97

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TS EAMCET 19-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 22 of 160

Marks: +1, -0

\theta

\tan \theta = \; \frac{-3}{4}

\tan \; \frac{\theta}{2} + \sin 2\theta =

Solution:

Given,

\tan \theta = - \; \frac{3}{4}

\theta

does not lie in second quadrant, then

\theta

lies in fourth quadrant as

\tan \theta

is negative only in second and fourth quadrant.

\theta \in \left( \; \frac{3\pi}{2}, 2\pi \right) \; \Rightarrow \; \frac{\theta}{2} \in \left( \; \frac{3\pi}{4}, \pi \right)

\Rightarrow \; \; 2\theta \in (3\pi, 4\pi)

\Rightarrow \frac{\theta}{2}

lies in second quadrant.

2\theta

lies in third and fourth quadrant.

\tan \theta = - \; \frac{3}{4} \Rightarrow \frac{2 \tan \frac{\theta}{2}}{1 - \tan^2 \frac{\theta}{2}} = - \; \frac{3}{4}

\Rightarrow 8 \tan \frac{\theta}{2} = -3 + 3 \tan^2 \frac{\theta}{2}

\Rightarrow 3 \tan^2 \frac{\theta}{2} - 8 \tan \frac{\theta}{2} - 3 = 0

\Rightarrow 3 \tan^2 \frac{\theta}{2} - 9 \tan \frac{\theta}{2} + \tan \frac{\theta}{2} - 3 = 0

\Rightarrow \; \; \tan \frac{\theta}{2} = - \frac{1}{3} \; \; \left[ \because \tan \frac{\theta}{2} \neq 3 \right]

\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}

= \frac{2 \left( -\frac{3}{4} \right)}{1 + \frac{9}{16}} = -\frac{3}{2} \times \frac{16}{25} = -\frac{24}{25}

\therefore \tan \frac{\theta}{2} + \sin 2\theta = -\frac{1}{3} - \frac{24}{25} = \frac{-25-72}{75} = -\frac{97}{75}

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