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TS EAMCET 2017 Code A Solved Paper

Section: Mathematics

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Question : 12 of 160

Marks: +1, -0

If

\csc\theta - \cot\theta = 2017

, then quadrant which θ lies is

I
IV
III
II

Solution:

Consider the given trigonometric equation

\csc\theta - \cot\theta = 2017 \text{..... (1)}

We know that,

\csc^{2}\theta - \cot^{2}\theta = 1

(\csc\theta - \cot\theta)(\csc\theta + \cot\theta) = 1

\csc\theta + \cot\theta = \frac{1}{\csc\theta - \cot\theta}

\csc\theta + \cot\theta = \frac{1}{2017} \text{..... (2)}

Add equation (1) and (2).

2\csc\theta = 2017 + \frac{1}{2017}

\csc\theta = \frac{1}{2}\left[2017 + \frac{1}{2017}\right] > 0

Therefore,

\theta

lies in I or II quadrant.

Subtracting equation (1) from (2),

2\cot\theta = \frac{1}{2017} - 2017

\cot\theta = \frac{1}{2}\left(\frac{1}{2017} - 2017\right) < 0

Therefore,

\theta

lies in II and III quadrant.

Hence,

\theta

lies in II quadrant

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