Question

If $3\sin x+\cos x=8$ then $6\tan \frac{x}{2}-9{\tan }^2\frac{x}{2}=$

• A. $0$
• B. $7$
• C. $3$
• D. $4$

Answer 1

The correct option is B $7$

Given

$3\sin x+\cos x=8\cdots \left(1\right)$

$\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$

$\cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$

Put the equations in $\left(1\right)$

$⟹3\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)=8$

$⟹6\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}=8+8{\tan }^2\frac{x}{2}$

$⟹6\tan \frac{x}{2}-9{\tan }^2\frac{x}{2}=7$