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Trigonometric Functions

Multiple Choice Questions

91.

The value of

$\tan (α) + 2 \tan (2 α) + 4 \tan (4 α) + . . . + 2^{n - 1} \tan (2^{n - 1} α) + 2^{n} \cot (2^{n} α)$ is

$\cot (2^{n} α)$
$2^{n} \tan (2^{n} α)$
0
$\cot (α)$

92.

If $\tan (\frac{απ}{4}) = \cot (\frac{βπ}{4})$ , then

$α + β = 0$
$α + β = 2 n$
$α + β = 2 n + 1$
$α + β = 2 (2 n + 1), \forall n is an integer$

93.

The principal value of $\sin^{- 1} (\tan (- \frac{5 π}{4}))$ is

$\frac{π}{4}$
- $\frac{π}{4}$
$\frac{π}{2}$
$- \frac{π}{2}$

94.

The value of $\cos (\frac{π}{15}) \cos (\frac{2 π}{15}) \cos (\frac{4 π}{15}) \cos (\frac{8 π}{15})$

$\frac{1}{16}$
$- \frac{1}{16}$
1
0

95.

The principal amplitude of ${(\sin (40 °) + icos (40 °))}^{5}$

70 $°$
- 110 $°$
110°
- 70°

Short Answer Type

96.
Find the general solution of $\sec (θ) + 1 = (2 + \sqrt{3}) \tan (θ)$

$Given, \sec (θ) + 1 = (2 + \sqrt{3}) \tan (θ) On squaring both sides, we get {(\sec (θ) + 1)}^{2} = {(2 + \sqrt{3})}^{2} \tan^{2} (θ) \Rightarrow {(\sec (θ) + 1)}^{2} = (4 + 3 + 4 \sqrt{3}) (\sec^{2} (θ) - 1) \Rightarrow (\sec (θ) + 1) \{(\sec (θ) + 1) - (7 + 4 \sqrt{3}) (\sec (θ) - 1)\} = 0 \Rightarrow (\sec (θ) + 1) \{(8 + 4 \sqrt{3}) - (6 + 4 \sqrt{3}) \sec (θ)\} = 0 \Rightarrow \sec (θ) + 1 = 0 or (8 + 4 \sqrt{3}) - (6 + 4 \sqrt{3}) = 0 \Rightarrow \cos (θ) = \frac{\sqrt{3}}{2} = \cos (\frac{π}{6}) \Rightarrow θ = 2 nπ \pm \frac{π}{6} But θ = 2 nπ - \frac{π}{6} does not satisfying the given equation .$