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Trigonometric Functions

Multiple Choice Questions

251.

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

252.

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

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

253.

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

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

254.

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

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

255.

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

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

256.

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

257.

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

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

258.

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

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

259.

$In ∆ ABC, if bcos (θ) = c - a, (where θ is an acute angle), then (c - a) \tan (θ) = ?$

$2 \sqrt{ca} \cos (\frac{B}{2})$
$2 \sqrt{ac} \sin (\frac{B}{2})$
$2 cacos (\frac{B}{2})$
$2casin (\frac{B}{2})$

260.
$If \tan (α) and \tan (β) are the roots of the equation x^{2} + px + q = 0, then the value of \sin^{2} (α + β) + pcos (α + β) \sin (α + β) + {qcos}^{2} (α + β)$

p + q

p

q

$\frac{p}{p + q}$

$(c) Since, \tan (α) and \tan (β) are the roots of the equation x^{2} + px + q = 0 \tan (α) + \tan (β) = - p and \tan (α) . \tan (β) = q \Rightarrow \tan (α + β) = \frac{\tan (α) + \tan (β)}{1 - \tan (α) \tan (β)} = \frac{- p}{1 - q} = \frac{p}{q - 1} Now consider \sin^{2} (α + β) + pcos (α + β) \sin (α + β) + {qcos}^{2} (α + β) = \cos^{2} (α + β) (\tan^{2} (α + β) + ptan (α + β) + q) = \frac{1}{\sec^{2} (α + β)} (\tan^{2} (α + β) + ptan (α + β) + q) = \frac{1}{1 + \tan^{2} (α + β)} (\tan^{2} (α + β) + ptan (α + β) + q) = \frac{1}{1 + \frac{p^{2}}{{(q - 1)}^{2}}} [\frac{p^{2}}{{(q - 1)}^{2}} + \frac{p^{2}}{q - 1} + q] = \frac{{(q - 1)}^{2}}{{(q - 1)}^{2} + p^{2}} [\frac{p^{2} + p^{2} (q - 1) + q {(q - 1)}^{2}}{{(q - 1)}^{2}}] = \frac{p^{2} + p^{2} q - p^{2} + q {(q - 1)}^{2}}{p^{2} + {(q - 1)}^{2}} = \frac{q (p^{2} + {(q - 1)}^{2})}{p^{2} + {(q - 1)}^{2}}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

260. If tanα and tanβ are the roots of the equation x2 + px + q = 0, then the value ofsin2α + β + pcosα + βsinα + β + qcos2α + β p + q p q pp + q

260.
$If \tan (α) and \tan (β) are the roots of the equation x^{2} + px + q = 0, then the value of \sin^{2} (α + β) + pcos (α + β) \sin (α + β) + {qcos}^{2} (α + β)$

p + q

p

q

$\frac{p}{p + q}$