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SSC CHSL 19 Mar 2020 Shift 3 Paper

Question : 57 of 100

Marks: +1, -0

The value of

\frac{\cos 29^{\circ} \csc 61^{\circ} \tan 45^{\circ} + 2 \sin 35^{\circ} \sec 55^{\circ}}{3 \sin^{2} 42^{\circ} + 3 \sin^{2} 48^{\circ}}

Solution:

Given,

\frac{\cos 29^{\circ} \csc 61^{\circ} \tan 45^{\circ} + 2 \sin 35^{\circ} \sec 55^{\circ}}{3 \sin^{2} 42^{\circ} + 3 \sin^{2} 48^{\circ}}

= \frac{\cos 29^{\circ} \times \frac{1}{\sin 61^{\circ}} \times 1 + 2 \sin 35^{\circ} \times \frac{1}{\cos 55^{\circ}}}{3 \sin^{2} 42^{\circ} + 3 [\sin (90^{\circ} - 42^{\circ})]^{2}}

= \frac{\cos 29^{\circ} \times \frac{1}{\cos 29^{\circ}} \times 1 + 2 \sin 35^{\circ} \times \frac{1}{\sin 35^{\circ}}}{3 \sin^{2} 42^{\circ} + 3 \cos^{2} 42^{\circ}}

[\because \sin(90-\theta) = \cos \theta

and

\cos(90-\theta) = \sin \theta]

= \frac{1+2}{3 \times (\sin^{2} 42^{\circ} + \cos^{2} 42^{\circ})}

= \frac{3}{3 \times 1}

[\because \sin^{2} \theta + \cos^{2} \theta = 1]

= 1

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