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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 41 of 120

Marks: +1, -0

\csc\theta = \frac{29}{21}

0 < \theta < 90^{\circ}

Solution: 👈: Video Solution

Given that

\csc\theta = \frac{29}{21}

\therefore \sin\theta = \frac{21}{29}

\therefore \cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{21}{29}\right)^2}

\sqrt{\left(1 - \frac{21}{29}\right)\left(1 + \frac{21}{29}\right)}

\sqrt{\frac{8}{29} \times \frac{50}{29}} = \sqrt{\frac{8 \times 2 \times 25}{29 \times 29}}

= \frac{4 \times 5}{29} = \frac{20}{29}

\therefore \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{21}{29}}{\frac{20}{29}} = \frac{21}{20}

\therefore 4\sec\theta + 4\tan\theta = \frac{4}{\cos\theta} + \frac{4\sin\theta}{\cos\theta}

= \frac{4}{\frac{20}{29}} + 4 \times \frac{21}{20}

= 4 \times \frac{29}{20} + 4 \times \frac{21}{20}

= 4\left(\frac{29}{20} + \frac{21}{20}\right) = 4\left(\frac{29+21}{20}\right)

= 4 \times \frac{50}{20} = 4 \times \frac{5}{2} = 10

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