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Question: If \(a\tan\theta = b\), then \(a\cos 2\theta + b\sin 2\theta =\)...

If atan⁡θ=ba\tan\theta = b, then acos⁡2θ+bsin⁡2θ=a\cos 2\theta + b\sin 2\theta =

A

aa

B

bb

C

−a- a

D

−b- b

Answer

aa

Explanation

Solution

Given that tan⁡θ=ba\tan\theta = \frac{b}{a}.

Now, acos⁡2θ+bsin⁡2θ=a(1−tan⁡2θ1+tan⁡2θ)+b(2tan⁡θ1+tan⁡2θ)a\cos 2\theta + b\sin 2\theta = a\left( \frac{1 - \tan^{2}\theta}{1 + \tan^{2}\theta} \right) + b\left( \frac{2\tan\theta}{1 + \tan^{2}\theta} \right)

Putting tan⁡θ=ba\tan\theta = \frac{b}{a}, we get

=a(1−b2a21+b2a2)+b(2ba1+b2a2)=a(a2−b2a2+b2)+b(2baa2+b2)= a\left( \frac{1 - \frac{b^{2}}{a^{2}}}{1 + \frac{b^{2}}{a^{2}}} \right) + b\left( \frac{2\frac{b}{a}}{1 + \frac{b^{2}}{a^{2}}} \right) = a\left( \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \right) + b\left( \frac{2ba}{a^{2} + b^{2}} \right)

=1(a2+b2){a3−ab2+2ab2}=a(a2+b2)a2+b2=a= \frac{1}{(a^{2} + b^{2})}\{ a^{3} - ab^{2} + 2ab^{2}\} = \frac{a(a^{2} + b^{2})}{a^{2} + b^{2}} = a.