Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

If Tan X = a B , Then B Cos 2 X + a Sin 2 X - Mathematics

MCQ

If $\text{ tan } x = \frac{a}{b}$, then $b \cos 2x + a \sin 2x$

Options

• a

• b

• $\frac{a}{b}$

• $\frac{b}{a}$

Solution

Given: $\text{ tan } x = \frac{a}{b}$

Now,

$b \cos2x + a \sin2x$
$= b \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) + a\left( \frac{2\text{ tan } x}{1 + \tan^2 x} \right)$
$= b\left( \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} \right) + a\left( \frac{2 \times \frac{a}{b}}{1 + \frac{a^2}{b^2}} \right)$
$= \frac{b\left( b^2 - a^2 \right)}{a^2 + b^2} + \frac{2 a^2 b}{a^2 + b^2}$

$= \frac{b^3 - a^2 b + 2 a^2 b}{a^2 + b^2}$
$= \frac{b^3 + a^2 b}{a^2 + b^2}$
$= \frac{b\left( b^2 + a^2 \right)}{a^2 + b^2}$
$= b$

Hence, the correct answer is option B.
Given:

$\text{ tan } x = \frac{a}{b}$
Now,
$b \cos2x + a \sin2x$
$= b \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) + a\left( \frac{2\text{ tan } x}{1 + \tan^2 x} \right)$
$= b\left( \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} \right) + a\left( \frac{2 \times \frac{a}{b}}{1 + \frac{a^2}{b^2}} \right)$
$= \frac{b\left( b^2 - a^2 \right)}{a^2 + b^2} + \frac{2 a^2 b}{a^2 + b^2}$

$= \frac{b^3 - a^2 b + 2 a^2 b}{a^2 + b^2}$

$= \frac{b^3 + a^2 b}{a^2 + b^2}$

$= \frac{b\left( b^2 + a^2 \right)}{a^2 + b^2}$

$= b$

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Q 36 | Page 45