Question

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

• A. $0$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is B $1$

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

$\Rightarrow 3\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+4\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=5$

$\Rightarrow \frac{6\tan \frac{x}{2}+4-4{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}=5$

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

$\Rightarrow 6\tan \frac{x}{2}+4-4{\tan }^2\frac{x}{2}-5-5{\tan }^2\frac{x}{2}=0$

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

$\Rightarrow {\left(3\tan \frac{x}{2}\right)}^2-2\times 3\tan \frac{x}{2}+1=0$

$\Rightarrow {(3\tan \frac{x}{2}-1)}^2=0$

$\Rightarrow 3\tan \frac{x}{2}=1$

$\Rightarrow \tan \frac{x}{2}=\frac{1}{3}$

Now,$6\tan \frac{x}{2}-9{\tan }^2\frac{x}{2}$

$=6\times \frac{1}{3}-9{\left(\frac{1}{3}\right)}^2$

$=2-1=1$