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TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

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Question : 22 of 160

Marks: +1, -0

If

\frac{\cos(\theta_1+\theta_2)}{\cos(\theta_1-\theta_2)}

\frac{\cos(\theta_3-\theta_4)}{\cos(\theta_3+\theta_4)}=0

\cot\theta_1 \cdot \cot\theta_2 \cdot \cot\theta_3 \cdot \cot\theta_4 =

1
-1
2
$\frac{1}{2}$

Solution:

We have,

\; \frac{\cos(\theta_1+\theta_2)}{\cos(\theta_1-\theta_2)} + \; \frac{\cos(\theta_3-\theta_4)}{\cos(\theta_3+\theta_4)} = 0

\Rightarrow \; \frac{\cos(\theta_1+\theta_2)}{\cos(\theta_1-\theta_2)} = - \; \frac{\cos(\theta_3-\theta_4)}{\cos(\theta_3+\theta_4)}

\Rightarrow \; \frac{\cos(\theta_1+\theta_2)-\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)+\cos(\theta_1-\theta_2)}

= \; \frac{-\cos(\theta_3-\theta_4)-\cos(\theta_3+\theta_4)}{-\cos(\theta_3-\theta_4)+\cos(\theta_3+\theta_4)}

[using componendo and dividendo rule]

\Rightarrow \; \frac{-2 \sin\theta_1 \sin\theta_2}{2 \cos\theta_1 \cos\theta_2} = \; \frac{-2 \cos\theta_3 \cos\theta_4}{-2 \sin\theta_3 \sin\theta_4}

\Rightarrow -\tan\theta_1 \tan\theta_2 = \cot\theta_3 \cot\theta_4

\Rightarrow \cot\theta_1 \cot\theta_2 \cot\theta_3 \cot\theta_4 = -1

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