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JEE Mathematics Solved Question Paper 2001

Multiple Choice Questions

51.

$\cos^{2} (\frac{π}{6} + θ) - \sin^{2} (\frac{π}{6} + θ) is equal to$

$\frac{1}{2} \cos (2 θ)$
0
$- \frac{1}{2} \cos (2 θ)$
$\frac{1}{2}$

52.

$In ∆ ABC, \frac{\cos (C) + \cos (A)}{c + a} + \frac{\cos (B)}{b} is equal to$

$\frac{1}{a}$
$\frac{1}{b}$
$\frac{c + a}{b}$
1

53.

$In ∆ ABC, \frac{a}{b^{2} - c^{2}} + \frac{c}{b^{2} - a^{2}} = 0, then B is equal to$

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

54.

$In ∆ ABC, a^{2} \sin (2 C) + c^{2} \sin (2 A) is equal to$

$∆$
$2 ∆$
$3 ∆$
$4 ∆$

55.

$The shadow of a tower standing on a level ground is found to be 60 m longer when the sun' s altitude is 30 ° than when it is 45 ° . The height of the tower is$

$30 m$
$90 m$
$60 \sqrt{3} m$
$30 (\sqrt{3} + 1) m$

56.
$If cosecθ = \frac{p + q}{p - q}, then \cot (\frac{π}{4} + \frac{θ}{4}) is equal to$

$\frac{\sqrt{p}}{q}$

$\sqrt{\frac{q}{p}}$

$\sqrt{pq}$

$pq$

$\sqrt{\frac{q}{p}}$

$\frac{1}{sinθ} = \frac{p + q}{p - q} \frac{1 + \tan^{2} \frac{θ}{2}}{2 \tan \frac{θ}{2}} = \frac{p + q}{p - q} Applying componendo and dividendo, \frac{1 + \tan^{2} \frac{θ}{2} + 2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2} - 2 \tan \frac{θ}{2}} = \frac{p + q + p - q}{p + q - p - q} \Rightarrow \frac{{(1 + \tan \frac{θ}{2})}^{2}}{{(1 - \tan \frac{θ}{2})}^{2}} = \frac{2 p}{2 q} = \frac{p}{q} \Rightarrow \frac{\tan \frac{π}{4} + \tan \frac{θ}{2}}{1 - t an \frac{π}{4} \tan \frac{θ}{2}} = \sqrt{\frac{p}{q}} \Rightarrow \tan (\frac{θ}{2} + \frac{π}{4}) = \sqrt{\frac{p}{q}} \Rightarrow \cot (\frac{p}{4} + \frac{θ}{2}) = \sqrt{\frac{q}{p}}$