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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

If a θ=b θ , where a, b ≠ 0 , then a 2 θ+b 2 θ=

If asin⁡θ=bcos⁡θ\mathrm{a} \sin \theta=\mathrm{b} \cos \theta, where a,b≠0\mathrm{a}, \mathrm{b} \neq 0, then acos⁡2θ+bsin⁡2θ=\mathrm{a} \cos 2 \theta+\mathrm{b} \sin 2 \theta=

Options

  1. Aab
  2. Ba
  3. Cb
  4. Dab\frac{\mathrm{a}}{\mathrm{b}}

Correct answer

B. a

Step-by-step solution

asin⁡θ=bcos⁡θ⇒tan⁡θ=baacos⁡2θ+bsin⁡2θ=a(1−tan⁡2θ1+tan⁡2θ)+b(2tan⁡θ1+tan⁡2θ)=a[1−(b2a2)1+(b2a2)]+b[2(ba)1+(b2a2)]=a(a2−b2a2+b2)+b(2ba×a2a2+b2)=a(a2−b2)a2+b2+b(2ab)a2+b2=a3−ab2+2ab2a2+b2=a3+ab2a2+b2=a(a2+b2)a2+b2=a\begin{aligned} & a \sin \theta=b \cos \theta \Rightarrow \tan \theta=\frac{b}{a} \\ & 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) \\ & =a\left[\frac{1-\left(\frac{b^2}{a^2}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]+b\left[\frac{2\left(\frac{b}{a}\right)}{1+\left(\frac{b^2}{a^2}\right)}\right]=a\left(\frac{a^2-b^2}{a^2+b^2}\right)+b\left(\frac{2 b}{a} \times \frac{a^2}{a^2+b^2}\right) \\ & =\frac{a\left(a^2-b^2\right)}{a^2+b^2}+\frac{b(2 a b)}{a^2+b^2}=\frac{a^3-a b^2+2 a b^2}{a^2+b^2} \\ & =\frac{a^3+a b^2}{a^2+b^2}=\frac{a\left(a^2+b^2\right)}{a^2+b^2}=a\end{aligned}
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