Question

$(a+2)\sin \alpha +(2a-1)\cos \alpha =(2a+1)$ 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)

The correct options are
B $\frac{4}{3}$
D $\frac{2a}{(a^2-1)}$

$(a+2)\sin \alpha +(2a-1)\cos \alpha =(2a+1)$ ....(1)
We know $a\sin x+b\cos x\le \sqrt{a^2+b^2}$
So here, $(a+2)\sin \alpha +(2a-1)\cos \alpha \le \sqrt{5a^2+5}$
$\Rightarrow 2a+1\le \sqrt{5a^2+5}$
$\Rightarrow (a-2{)}^2\le 0(\because a<b\Rightarrow a^2\le b^2$)
So, a must be 2
Now, put this value in (1)
$4\sin \alpha +3\cos \alpha =5$
Substitute $4=r\cos \beta ,3=r\sin \beta$
$\Rightarrow r=5,\tan \beta =\frac{3}{4}$
$\Rightarrow 5\sin (\beta +\alpha )=5$
$\Rightarrow \sin (\alpha +\beta )=1$
$\Rightarrow \alpha =\frac{\pi }{2}-\beta$
$\Rightarrow \tan \alpha =\tan (\frac{\pi }{2}-\beta )$
$\Rightarrow \tan \alpha =\cot \beta =\frac{4}{3}$
Also, option D satisfies the value of $\tan \alpha$