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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 75-76

Consider the following for the next items that follow:

Let

p=\frac{1}{3}-\frac{tan3x}{tanx}

and q = 1 - 3 tan2 x, 0 < x < π,

x≠\frac{\pi }{2}

Question : 75 of 120

Marks: +1, -0

What is pq equal to ?

Solution:

Concept:

Use the triple-angle identity:

\tan 3x = \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}

Explanation:

Given

q = 1 - 3\tan^2 x

(for

0 < x < \pi

x \neq \frac{\pi}{2}

Also

p = \frac{1}{3} - \frac{\tan 3x}{\tan x}

Substitute the triple-angle formula into

p

p = \frac{1}{3} - \frac{ \frac{3\tan x - \tan^3 x}{1 - 3\tan^2 x} }{\tan x}

Simplify:

p = \frac{1}{3} - \frac{3\tan x - \tan^3 x}{\tan x (1 - 3\tan^2 x)}

Combine over a common denominator

3\tan x (1 - 3\tan^2 x)

p = \frac{\tan x (1 - 3\tan^2 x) - 3(3\tan x - \tan^3 x)}{3\tan x (1 - 3\tan^2 x)}

Expand numerator:

\tan x - 3\tan^3 x - 9\tan x + 3\tan^3 x = -8\tan x

Thus

p = \frac{-8\tan x}{3\tan x (1 - 3\tan^2 x)} = \frac{-8}{3(1 - 3\tan^2 x)}

Now compute

pq = \frac{-8}{3(1 - 3\tan^2 x)} \times (1 - 3\tan^2 x) = -\frac{8}{3}

Answer:

-\frac{8}{3}

(Option D).

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