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SSC Selection Post 2019 PYQ — Question 51

SSC Selection Post - Graduation Level 2019 - 16 Oct 2019 • Quantitative Aptitude • Q51

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If x = cosec 490 , then tan2 490 + 1sin2490sec2410\frac{1}{sin^2 49^0 sec^2 41^0}+ sin2 410 + 1cosec2490\frac{1}{cosec^2 49^0}is equal to:

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Correct answer: C

Explanation

The correct answer is Option 3 i.e. 1+x21+x^2.

x=cosec49∘=1sin⁡49∘x = cosec 49^\circ = \frac{1}{\sin 49^\circ}

tan⁡249∘+1sin⁡249∘sec⁡241∘+sin⁡241∘+1csc⁡249∘\tan^2 49^\circ + \frac{1}{\sin^2 49^\circ \sec^2 41^\circ} + \sin^2 41^\circ + \frac{1}{\csc^2 49^\circ}

tan⁡249∘=sin⁡249∘cos⁡249∘=1/x2cos⁡249∘=1x2cos⁡249∘1sin⁡249∘sec⁡241∘=x2cos⁡241∘sin⁡241∘=cos⁡249∘1csc⁡249∘=sin⁡249∘=1x2\tan^2 49^\circ = \frac{\sin^2 49^\circ}{\cos^2 49^\circ} = \frac{1/x^2}{\cos^2 49^\circ} = \frac{1}{x^2 \cos^2 49^\circ} \\[5pt] \frac{1}{\sin^2 49^\circ \sec^2 41^\circ} = x^2 \cos^2 41^\circ \\[5pt] \sin^2 41^\circ = \cos^2 49^\circ \\[5pt] \frac{1}{\csc^2 49^\circ} = \sin^2 49^\circ = \frac{1}{x^2}

1x2cos⁡249∘+x2cos⁡241∘+cos⁡249∘+1x2Since cos⁡241∘=sin⁡249∘=1x2, we get:1x2cos⁡249∘+x2⋅1x2+cos⁡249∘+1x2=1x2cos⁡249∘+1+cos⁡249∘+1x2=1+x2\frac{1}{x^2 \cos^2 49^\circ} + x^2 \cos^2 41^\circ + \cos^2 49^\circ + \frac{1}{x^2} \\[5pt] \text{Since } \cos^2 41^\circ = \sin^2 49^\circ = \frac{1}{x^2}, \text{ we get:} \\ \frac{1}{x^2 \cos^2 49^\circ} + x^2 \cdot \frac{1}{x^2} + \cos^2 49^\circ + \frac{1}{x^2} \\[5pt] = \frac{1}{x^2 \cos^2 49^\circ} + 1 + \cos^2 49^\circ + \frac{1}{x^2} \\[5pt] = 1 + x^2