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UPSC CDS 2 2020 Math Paper

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Question : 58 of 100

Marks: +1, -0

If

\csc \theta - \sin \theta = p^{3}

\sec \theta - \cos \theta = q^{3}

, then what is the value of

\tan \theta

?

$\dfrac{p}{q}$
$\dfrac{q}{p}$
pq
$p^{2}q^{2}$

Solution:

Concept:

We simplify the given trigonometric expressions using reciprocal identities to find

\tan \theta

in terms of

p

and

q

.

Explanation:

Start with the first equation:

\csc \theta - \sin \theta = p^3

.

Rewrite

\csc \theta

as

\frac{1}{\sin \theta}

, so

\frac{1}{\sin \theta} - \sin \theta = \frac{1 - \sin^2 \theta}{\sin \theta}

.

Since

1 - \sin^2 \theta = \cos^2 \theta

, we get

\frac{\cos^2 \theta}{\sin \theta} = p^3

…(1).

Now the second equation:

\sec \theta - \cos \theta = q^3

.

Rewrite

\sec \theta

as

\frac{1}{\cos \theta}

, giving

\frac{1}{\cos \theta} - \cos \theta = \frac{1 - \cos^2 \theta}{\cos \theta}

.

Using

1 - \cos^2 \theta = \sin^2 \theta

, we have

\frac{\sin^2 \theta}{\cos \theta} = q^3

…(2).

Divide equation (2) by equation (1):

\frac{\sin^2 \theta}{\cos \theta} \div \frac{\cos^2 \theta}{\sin \theta} = \frac{q^3}{p^3}

.

This simplifies to

\frac{\sin^3 \theta}{\cos^3 \theta} = \frac{q^3}{p^3}

.

Therefore

\left(\frac{\sin \theta}{\cos \theta}\right)^3 = \left(\frac{q}{p}\right)^3

, so

\tan \theta = \frac{q}{p}

.

Answer:

\boxed{\frac{q}{p}}

(Option B).

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