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प्रश्न
If \[\text{ tan } x = \frac{a}{b}\], then \[b \cos 2x + a \sin 2x\]
विकल्प
a
b
- \[\frac{a}{b}\]
- \[\frac{b}{a}\]
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उत्तर
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:
\[ = b \left( \frac{1 - \tan^2 x}{1 + \tan^2 x} \right) + a\left( \frac{2\text{ tan } x}{1 + \tan^2 x} \right)\]
\[ = b\left( \frac{1 - \frac{a^2}{b^2}}{1 + \frac{a^2}{b^2}} \right) + a\left( \frac{2 \times \frac{a}{b}}{1 + \frac{a^2}{b^2}} \right)\]
\[ = \frac{b\left( b^2 - a^2 \right)}{a^2 + b^2} + \frac{2 a^2 b}{a^2 + b^2}\]
\[= \frac{b^3 - a^2 b + 2 a^2 b}{a^2 + b^2}\]
\[ = \frac{b^3 + a^2 b}{a^2 + b^2}\]
\[ = \frac{b\left( b^2 + a^2 \right)}{a^2 + b^2}\]
\[ = b\]
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