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UPSC CDS 1 2022 Math Paper

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Question : 83 of 100

Marks: +1, -0

If

\tan \theta + \sec \theta = 3

, then what is the value of

3 \tan \theta + 9 \sec \theta

?

15
17
19
21

Solution: 👈: Video Solution

Given:

\tan \theta + \sec \theta = 3

Formula Used:

\sec^2 \theta - \tan^2 \theta = 1

(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1

Calculation:

We have

\tan \theta + \sec \theta = 3 \qquad \cdots

(i)

According to the formula used

(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1

\Rightarrow 3(\sec \theta - \tan \theta) = 1

\Rightarrow (\sec \theta - \tan \theta) = \frac{1}{3} \qquad \text{ - (ii) }

On adding (i) and (ii), we get

2 \sec \theta = 3 + \frac{1}{3} = \frac{10}{3}

\Rightarrow \sec \theta = \frac{5}{3} \qquad \cdots \text{-(iii) }

On putting the value of

\sec \theta

in (i), we get

\tan \theta + \frac{5}{3} = 3

\Rightarrow \tan \theta = 3 - \frac{5}{3} = \frac{4}{3} \qquad \text{ - -(iv) }

Now, we have to find the value of

3 \tan \theta + 9 \sec \theta

So, from (iii) and (iv), we get

3 \times \frac{4}{3} + 9 \times \frac{5}{3}

\Rightarrow 4 + 15 = 19

\therefore

The required value of

3 \tan \theta + 9 \sec \theta

is 19 .

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