Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Trigonometric Functions

Multiple Choice Questions

41.

$What is \frac{1}{\sin (10 °)} - \frac{\sqrt{3}}{\cos (10 °)} = ?$

42.

$If k = \sin (\frac{π}{18}) \sin (\frac{5 π}{18}) \sin (\frac{7 π}{18}), then what is the value of k ?$

$\frac{1}{2}$
$\frac{1}{4}$
$\frac{1}{8}$
$\frac{1}{16}$

43.

$The expression \frac{\sin (α) + \sin (β)}{\cos (α) + \cos (β)} = ?$

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

44.

What is the value of $\tan (18 °)$ ?

$\frac{\sqrt{5} - 1}{\sqrt{10 + 2 \sqrt{5}}}$
$\frac{\sqrt{5} - 1}{\sqrt{10 + \sqrt{5}}}$
$\frac{\sqrt{10 + 2 \sqrt{5}}}{\sqrt{5} - 1}$
$\frac{\sqrt{10 + \sqrt{5}}}{\sqrt{5} - 1}$

45.

$If \tan (α + β) = 2 and \tan (α + β) = 1, then \tan (2 α) = ?$

- 3
- 2
$- \frac{1}{3}$
1

46.

Consider the following for triangle ABC :

$I . \sin (\frac{B + C}{2}) = \cos (\frac{A}{2}) II . \tan (\frac{B + C}{2}) = \cot (\frac{A}{2}) III . \sin (B + C) = \sin (A) IV . \tan (B + C) = - \tan (A)$

Which of the above are correct ?

1 and 3
1 and 2
1 and 4
2 and 3

47.

$If \sin (θ) = 3 \sin (θ + 2 α), then the value of \tan (θ + α) + 2 \tan (α) is equal to$

48.
$If \sec (θ) - \csc (θ) = \frac{4}{3}, then what is (\sin (θ) - \cos (θ)) = ?$

- 2 only

$\frac{1}{2} only$

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

Neither $\frac{1}{2}$ nor - 2

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

$∵ \sec (θ) - \csc (θ) = \frac{4}{3} \Rightarrow \frac{1}{\cos (θ)} - \frac{1}{\sin (θ)} = \frac{4}{3} \Rightarrow \frac{\sin (θ) - \cos (θ)}{\cos (θ) \times \sin (θ)} = \frac{4}{3} \Rightarrow \sin (θ) - \cos (θ) = \frac{4}{3} \cos (θ) . \sin (θ) Now, {(\sec (θ) - \csc (θ))}^{2} = \frac{16}{9} \Rightarrow \sec^{2} (θ) + \csc^{2} (θ) = 2 \sec (θ) \csc (θ) = \frac{16}{9} \Rightarrow \frac{1}{\cos^{2} (θ)} + \frac{1}{\sin^{2} (θ)} - \frac{2}{\sin (θ) \cos (θ)} = \frac{16}{9} \Rightarrow \frac{\sin^{2} (θ) + \cos^{2} (θ) - 2 \sin (θ) \cos (θ)}{\cos^{2} (θ) \sin^{2} (θ)} = \frac{16}{9} \Rightarrow \frac{1 - 2 \sin (θ) \cos (θ)}{\cos^{2} (θ) \sin^{2} (θ)} = \frac{16}{9} \Rightarrow 16 \cos^{2} (θ) \sin^{2} (θ) + 18 \cos (θ) \sin (θ) - 9 = 0 \Rightarrow 16 \cos^{2} (θ) \sin^{2} (θ) + 24 \cos (θ) \sin (θ) - 6 \cos (θ) \sin (θ) - 9 = 0 \Rightarrow 8 \cos (θ) \sin (θ) (2 \cos (θ) \sin (θ) + 3) - 3 (2 \cos (θ) \sin (θ) + 3) = 0 \Rightarrow (2 \cos (θ) \sin (θ) + 3) (8 \cos (θ) \sin (θ) - 3) = 0 \Rightarrow \cos (θ) \sin (θ) = - \frac{3}{2} or \cos (θ) \sin (θ) = \frac{3}{8} ∴ \sin (θ) - \cos (θ) = \frac{4}{3} \times \frac{3}{8} = \frac{1}{2}$