Question

If $(a+2)\sin \alpha +(2a-1)\cos \alpha =(2a+1)$ , then $\tan \alpha =$

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{2a}{a^2+1}$
• D. $\frac{2a}{a^2-1}$

Answer 1

Answer: (B), (D)

$\frac{4}{3}$
$\frac{2a}{a^2-1}$

Let $t=\tan \frac{\theta }{2}$
$(a+2)\sin \theta +(2a-1)\cos \theta =2a+1\Rightarrow (a+2)\frac{2t}{1+t^2}+(2a-1)\frac{1-t^2}{1+t^2}=2a+1\Rightarrow (2t-1)(at-1)=0$

Gives $t=\frac{1}{2}$ or $t=\frac{1}{a}$
For $t={\tan }^2\frac{\theta }{2}=\frac{1}{2}\Rightarrow \tan \theta =\frac{2t}{1-t^2}=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}$
For $t={\tan }^2\frac{\theta }{2}=\frac{1}{a}\Rightarrow \tan \theta =\frac{2t}{1-t^2}=\frac{2\frac{1}{a}}{1-\frac{1}{a^2}}=\frac{2a}{(a^2-1)}$