Question

Solve:

$(a+2)\sin \alpha +(2a-1)\cos \alpha =(2a+1)$ if $\tan \alpha$ is :

• 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 $\tan \alpha =t$ so that ${\sec }^2\alpha =1+t^2$

Given : $(a+2)\sin \alpha +(2a-1)\cos \alpha =(2a+1)$
dividing by $\cos \alpha$ and squaring both sides, we get
$\Rightarrow {[(a+2)t+(2a-1)]}^2=\left[{(2a+1)}^2(1+t^2)\right]$
$\Rightarrow t^2[{(a+2)}^2-{(2a+1)}^2]+2(a+2)(2a-1)t+[{(2a-1)}^2-{(2a+1)}^2]=0$

$\Rightarrow 3(1-a^2)t^2+2(2a^2+3a-2)t-4\times 2a=0$

$\Rightarrow 3(1-a^2)t^2-4(1-a^2)t+6at-8a=0$

$\Rightarrow t(1-a^2)(3t-4)+2a(3t-4)=0$

$\Rightarrow (3t-4)[(1-a^2)t+2a]=0$

$\Rightarrow 3t-4=0$ or $(1-a^2)t+2a=0$

$\Rightarrow t=\frac{4}{3}$ or $t=-\frac{2a}{1-a^2}=\frac{2a}{a^2-1}$

Hence, options $B$ and $D$ are correct