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Trigonometric Functions

Multiple Choice Questions

901.

$The set of solutions of the system of equations x + y = \frac{2 π}{3} and \cos (x) + \cos (y) = \frac{3}{2}, where x, y are real, is$

$\{(x, y) \cos (\frac{x - y}{2}) = \frac{1}{2}\}$
$\{(x, y) \sin (\frac{x - y}{2}) = \frac{1}{2}\}$
$\{(x, y) \cos (x - y) = \frac{1}{2}\}$
Empty set

902.

If in the angles of a triangle are in the ratio1 : 1 : 4, then the ratio of the perimeter of the triangle to its largest side is

$\sqrt{2} + 2 : \sqrt{3}$
3 : 2
$\sqrt{3} + 2 . \sqrt{2}$
$2 + \sqrt{3} : \sqrt{3}$

903.

$If \cos (x) = \tan (y), \cot (y) = \tan (z) and \cot (z) = \tan (x) then \sin (x) = ?$

$\frac{\sqrt{5} + 1}{4}$
$\frac{\sqrt{5} - 1}{4}$
$\frac{\sqrt{5} + 1}{2}$
$\frac{\sqrt{5} - 1}{2}$

904.

$\tan (81 °) - \tan (63 °) - \tan (27 °) + \tan (9 °) = ?$

905.

If x and y are acute angles such that cos(x) + cos(y) = $\frac{3}{2}$ sin(x) + sin(y) = $\frac{3}{4}$ , then sin(x + y) equals to

$\frac{2}{5}$
$\frac{3}{4}$
$\frac{3}{5}$
$\frac{4}{5}$

906.

$The sum of the solutions in (0, 2 π) the equation \cos (x) \cos (\frac{π}{3} - x) \cos (\frac{π}{3} + x) = \frac{1}{4} is$

$4 π$
$π$
$2 π$
$3 π$

907.

$In any ∆ ABC, \frac{(a + b + c) (b + c - a) (c + a - b) (a + b - c)}{4 b^{2} c^{2}} = ?$

$\sin^{2} (B)$
$\cos^{2} (A)$
$\cos^{2} (B)$
$\sin^{2} (A)$

908.

$\sum_{k = 1}^{6} [\sin (\frac{2 kπ}{7} - icos (\frac{2 kπ}{7}))] = ?$

909.
If s $\sin (θ) + \cos (θ) = p$ and $\tan (θ) + \cot (θ) = q$ , then q(p² - 1) is equal to

$\frac{1}{2}$

2

1

3

$Given, \sin (θ) + \cos (θ) = p . . . (i) and \tan (θ) + \cot (θ) = q . . . (ii) From Eq (i) {(\sin (θ) + \cos (θ))}^{2} = p^{2} \sin^{2} (θ) + \cos^{2} (θ) + 2 \sin (θ) \cos (θ) = p^{2} 1 + 2 \sin (θ) \cos (θ) = p^{2} [∵ \sin^{2} (θ) + \cos^{2} (θ) = 1] \Rightarrow 1 + \sin (2 θ) = p^{2} [∵ \sin (2 θ) = 2 \sin (θ) \cos (θ)] . . . (iii) From Eq (ii), \tan (θ) + \cot (θ) = q \Rightarrow \tan (θ) + \frac{1}{\tan (θ)} = q$

$\Rightarrow \frac{\tan^{2} (θ) + 1}{\tan (θ)} = q \Rightarrow \frac{\tan^{2} (θ) + 1}{2 \tan (θ)} = \frac{q}{2} [∵ \csc (2 θ) = \frac{q}{2}] ∴ \csc (2 θ) = \frac{q}{2} \Rightarrow \frac{1}{\sin (2 θ)} = \frac{q}{2} \Rightarrow \frac{1}{p^{2} - 1} = \frac{q}{2} [∵ From Eq (iii)] \Rightarrow 2 = q (p^{2} - 1) \Rightarrow q (p^{2} - 1) = 2$