Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
Advertisement Remove all ads

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

  • \[\frac{a}{b}\]

     

  • \[\frac{b}{a}\]

     

Advertisement Remove all ads

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
  Is there an error in this question or solution?
Advertisement Remove all ads

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
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×