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Trigonometric Functions

Multiple Choice Questions

911.

If sin(A) + sin(B) + sin(C) = 0 and cos(A) + cos(B) + cos(C) = 0, then cos (A + B) + cos (B + C) + cos(C + A) is equal to

cos(A + B + C)
2
1
0

912.

If $\tan (θ) . \tan (120 ° - θ) \tan (120 ° + θ) = \frac{1}{\sqrt{3}}$ , then $θ$ is equal to

$\frac{nπ}{3} + \frac{π}{18}, n \in Z$
$\frac{nπ}{3} + \frac{nπ}{12}, n \in Z$
$\frac{nπ}{12} + \frac{π}{12}, n \in Z$
$\frac{nπ}{3} + \frac{π}{6}, n \in Z$

913.

${(\frac{1 + \cos (\frac{π}{8}) - isin (\frac{π}{8})}{1 + \cos (\frac{π}{8}) + isin (\frac{π}{8})})}^{8} = ?$

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

914.

$If (1 + \tan (α)) (1 + \tan (4 α)) = 2, α \in (0, \frac{π}{16}), then α = ?$

$\frac{π}{20}$
$\frac{π}{30}$
$\frac{π}{40}$
$\frac{π}{50}$

915.

$If \cos (θ) = \frac{\cos (α) - \cos (β)}{1 - \cos (α) \cos (β)}, then one of the values of \tan (\frac{θ}{2}) is$

$\cot (\frac{β}{2}) \tan (\frac{α}{2})$
$\tan (\frac{α}{2}) \tan (\frac{β}{2})$
$\tan (\frac{β}{2}) \cot (\frac{α}{2})$
$\tan^{2} (\frac{α}{2}) \tan^{2} (\frac{β}{2})$

916.

The value of the expression $\frac{1 + \sin (2 α)}{\cos (2 α - 2 π) \tan (α - \frac{3 π}{4})} - \frac{1}{4} \sin (2 α) [\cot (\frac{α}{2}) + \cot (\frac{3 π}{2} + \frac{α}{2})] is$

0
1
$\sin^{2} (\frac{α}{2})$
$\sin^{2} (α)$

917.

$If \frac{1}{6} \sin (θ), \cos (θ) and \tan (θ) are in geometric progression, then the solution set of θ is$

$2 nπ \pm (\frac{π}{6})$
$2 nπ \pm (\frac{π}{3})$
$nπ + {(- 1)}^{n} (\frac{π}{3})$
$nπ + (\frac{π}{3})$

918.

$In ∆ ABC if x = \tan (\frac{B - C}{2}) \tan (\frac{A}{2}), y = \tan (\frac{C - A}{2}) \tan (\frac{B}{2}) and z = \tan (\frac{A - B}{2}) \tan (\frac{C}{2}), then (x + y + z) = ?$

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

919.
$If A > 0, B > 0 and A + B = \frac{π}{3}, then the maximum value of Atan (B) is$

$\frac{1}{\sqrt{3}}$

$\frac{1}{3}$

$\frac{1}{2}$

$\sqrt{3}$

$\frac{1}{3}$

$Let y = \tan (A) \tan (B) y = \tan (A) \tan (\frac{π}{3} - A) [∵ A + B = \frac{π}{3}] \Rightarrow \frac{dy}{dA} = \sec^{2} (A) \tan (\frac{π}{3} - A) - \sec^{2} (\frac{π}{3} - A) \tan (A) For maxima or minima, put \frac{dy}{dA} = 0 ∴ \sec^{2} (A) \tan (\frac{π}{3} - A) = \tan (A) \sec^{2} (\frac{π}{3} - A) \Rightarrow (1 + \tan^{2} (A)) \tan (\frac{π}{3} - A) = \tan (A) [1 + \tan (\frac{π}{3} - A)] \Rightarrow \tan (\frac{π}{3} - A) + \tan^{2} (A) \tan (\frac{π}{3} - A) = \tan (A) + \tan (A) \tan^{2} (\frac{π}{3} - A) \Rightarrow [\tan (\frac{π}{3} - A) - \tan (A)] [1 - \tan (A) \tan (\frac{π}{3} - A)] = 0 \Rightarrow \tan (\frac{π}{3} - A) = \tan (A) \Rightarrow \frac{π}{3} - A = A \Rightarrow 2 A = \frac{π}{3} \Rightarrow A = \frac{π}{6} Maximum value of \tan (A) \tan (B) = \tan (\frac{π}{6}) \tan (\frac{π}{6}) = \frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}} = \frac{1}{3}$