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If \( \Large A+B+C= \pi \ and\ \cos A = B \cos C,\ then\ \tan B \tan C \) is equal to:

A) \( \Large \frac{1}{2} \)
B) 2
C) 1
D) \( \Large -\frac{1}{2} \)

Correct Answer:


B) 2

Description for Correct answer:
Since, \( \Large A+B+C = \pi \)

or \( \Large A = \pi - \left(B+C\right) \)

We have, \( \Large \cos A = \cos B \cos C \)

=> \( \Large \cos \left[ \pi - \left(B+C\right) \right] = \cos B \cos C \)

=> \( \Large -\cos \left(B+C\right) = \cos B \cos C \)

=> \( \Large -\left[ \cos B \cos C - \sin B \sin C \right] = \cos B \cos C \)

=> \( \Large \sin B \sin C = 2 \cos B \cos C \)

=> \( \Large \tan B \tan C = 2 \)

Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio

Comments

Similar Questions

1). The period of \( \Large \sin^{2} \theta \) is

A). \( \Large \pi ^{2} \)

B). \( \Large \pi \)

C). \( \Large 2 \pi \)

D). \( \Large \frac{ \pi }{2} \)

2). If \( \Large y = \sin^{2} \theta + cosec^{2} \theta \), \( \Large \theta \ne 0 \), then

A). \( \Large y = 0 \)

B). \( \Large y \le 2 \)

C). \( \Large y \ge -2 \)

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3). If \( \alpha \) is a root of \( \Large 25 \cos^{2} \theta + 5 \cos \theta - 12 = 0,\) \( \Large \frac{\pi}{2} < \alpha < \pi \), then \( \Large sin 2\alpha \) is equal to

A). \( \Large \frac{24}{25} \)

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D). \( \Large -\frac{13}{18} \)

4). The value of \( \Large \cos \frac{ \pi }{65} \cos \frac{2 \pi }{65} \cos \frac{4 \pi }{65}.....\cos \frac{32 \pi }{65} \) is

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5). If \( \Large \tan x + \cot x = 2,\ then\ \sin^{2n}x + \cos^{2n}x \) is equal to:

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6). If \( \Large \sin A + \cos B = a\ and\ \sin B + \cos A = b,\) then \( \Large \sin \left(A+B\right) \) is equal to:

A). \( \Large \frac{a^{2}+b^{2}}{2} \)

B). \( \Large \frac{a^{2}-b^{2}+2}{2} \)

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7). \( \Large \frac{2 \sin \alpha }{1+ \cos\ \alpha + \sin \alpha } = x \), then \( \Large \frac{1- \cos \alpha - \sin \alpha }{\cos \alpha } \) equal to

A). \( \Large \frac{1}{x} \)

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8). If \( \Large \sec \theta + \tan \theta = 1 \), then root of the equation \( \Large \left(a-2b+c\right)x^{2} + \left(b-2c+a\right)x + \left(c-2a+b\right) = 0 \) is:

A). \( \Large \sec \theta \)

B). \( \Large \tan \theta \)

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A). \( \Large -\frac{3}{16} \)

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