If \( \tan x=\frac{a}{b} \), Find \( b\cos2x+a\sin2x \)
Question
If
\[ \tan x=\frac{a}{b}, \]
then
\[ b\cos2x+a\sin2x \]
is equal to
(a) \(a\)
(b) \(b\)
(c) \(\frac{a}{b}\)
(d) \(\frac{b}{a}\)
Solution
Using
\[ \tan x=\frac{a}{b} \]
we have the standard expressions
\[ \cos2x=\frac{b^2-a^2}{a^2+b^2} \]
and
\[ \sin2x=\frac{2ab}{a^2+b^2} \]
Substituting,
\[ b\cos2x+a\sin2x = b\left(\frac{b^2-a^2}{a^2+b^2}\right) + a\left(\frac{2ab}{a^2+b^2}\right) \]
\[ = \frac{b^3-a^2b+2a^2b}{a^2+b^2} \]
\[ = \frac{b^3+a^2b}{a^2+b^2} \]
\[ = \frac{b(a^2+b^2)}{a^2+b^2} \]
\[ =b \]
Final Answer
\[ \boxed{b} \]
Hence, the correct option is (b) \(b\).