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UPSC CDS 1 2021 Math Paper

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Question : 48 of 100

Marks: +1, -0

If

6 + 8 \tan \theta = \sec \theta

8 - 6 \tan \theta = k \sec \theta

, then what is the value of

k^{2}

?

11
22
77
99

Solution:

Concept:

Use the trigonometric identity

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

and eliminate

\sec \theta

and

\tan \theta

from the given equations.

Explanation:

Given equations:

6 + 8 \tan \theta = \sec \theta

…(1)

8 - 6 \tan \theta = k \sec \theta

…(2)

Multiply (1) by

6

:

36 + 48 \tan \theta = 6 \sec \theta

…(3)

Multiply (2) by

8

:

64 - 48 \tan \theta = 8k \sec \theta

…(4)

Add (3) and (4):

100 = \sec \theta (6 + 8k)

\Rightarrow \sec \theta = \frac{100}{6 + 8k}

…(5)

From (1):

8 \tan \theta = \sec \theta - 6

Substitute (5):

8 \tan \theta = \frac{100}{6 + 8k} - 6 = \frac{100 - 6(6 + 8k)}{6 + 8k} = \frac{64 - 48k}{6 + 8k}

\Rightarrow \tan \theta = \frac{8 - 6k}{6 + 8k}

…(6)

Use identity

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

with (5) and (6):

\left(\frac{100}{6 + 8k}\right)^2 - \left(\frac{8 - 6k}{6 + 8k}\right)^2 = 1

\Rightarrow \frac{10000 - (64 - 96k + 36k^2)}{(6 + 8k)^2} = 1

\Rightarrow 10000 - 64 + 96k - 36k^2 = (6 + 8k)^2

\Rightarrow 9936 + 96k - 36k^2 = 36 + 96k + 64k^2

Cancel

96k

both sides:

9936 - 36k^2 = 36 + 64k^2

\Rightarrow 9936 - 36 = 64k^2 + 36k^2

\Rightarrow 9900 = 100k^2

\Rightarrow k^2 = 99

Answer:

k^{2} = 99

(Option D)

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