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

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Question

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

 

 

Options

  • a

  • b

  • \[\frac{a}{b}\]

     

  • \[\frac{b}{a}\]

     

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

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

\[ = \frac{b^3 + a^2 b}{a^2 + b^2}\]

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

\[ = b\]

 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.5 [Page 45]

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RD Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.5 | Q 36 | Page 45

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