Question

$3\sin x+4\cos x=5\Rightarrow 6\tan \frac{x}{2}-9{\tan }^2\frac{x}{2}=$

• A. $0$
• B. $-1$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$-1$

$3\sin x+4\cos x=5$

$\Rightarrow 6\tan \frac{x}{2}-9{\tan }^2\frac{x}{2}=?$

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

$3\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+4\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)=5$

$\Rightarrow 6\tan \frac{x}{2}+4-4{\tan }^2\frac{x}{2}=5+5{\tan }^2\frac{x}{2}$

$\Rightarrow 9{\tan }^2\frac{x}{2}-6\tan \frac{x}{2}-1=0$

$\Rightarrow 9{\tan }^2\frac{x}{2}-6\tan \frac{x}{2}=1$