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RRB Technician Grade 1 Model Paper 3

Question : 51 of 100

Marks: +1, -0

\frac{sec\theta +4cot\theta }{4tan\theta -5cos\theta }

is:

Solution:

Concept:

Use trigonometric ratios and the Pythagorean identity to find all required values.

Explanation:

Given

5\sin\theta = 4

, so

\sin\theta = \frac{4}{5}

In a right triangle, let perpendicular

P=4

and hypotenuse

H=5

Using

H^2 = P^2 + B^2

5^2 = 4^2 + B^2

B^2 = 25 - 16 = 9

, giving base

B = 3

Thus

\cos\theta = \frac{B}{H} = \frac{3}{5}

Also,

\sec\theta = \frac{1}{\cos\theta} = \frac{5}{3}

\tan\theta = \frac{P}{B} = \frac{4}{3}

and

\cot\theta = \frac{1}{\tan\theta} = \frac{3}{4}

Now substitute into the expression:

\frac{\sec\theta + 4\cot\theta}{4\tan\theta - 5\cos\theta} = \frac{\frac{5}{3} + 4\cdot\frac{3}{4}}{4\cdot\frac{4}{3} - 5\cdot\frac{3}{5}}

Simplify numerator:

\frac{5}{3} + 3 = \frac{14}{3}

Simplify denominator:

\frac{16}{3} - 3 = \frac{7}{3}

Therefore, the value is

\frac{\frac{14}{3}}{\frac{7}{3}} = 2

Answer:

2

(Option D).

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