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If $3 \sin x+4 \cos x=5$, then $6 \tan \frac{x}{2}-9 \tan ^2 \frac{x}{2}$ is equal to
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Given, $3 \sin x+4 \cos x=5$
$\begin{aligned} & \Rightarrow 3\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+4\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)=5 \\ & \Rightarrow 6 \tan \frac{x}{2}+4-4 \tan ^2 \frac{x}{2}=5\left(1+\tan ^2 \frac{x}{2}\right) \\ & \Rightarrow 6 \tan \frac{x}{2}-9 \tan ^2 \frac{x}{2}=1\end{aligned}$
$\begin{aligned} & \Rightarrow 3\left(\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)+4\left(\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)=5 \\ & \Rightarrow 6 \tan \frac{x}{2}+4-4 \tan ^2 \frac{x}{2}=5\left(1+\tan ^2 \frac{x}{2}\right) \\ & \Rightarrow 6 \tan \frac{x}{2}-9 \tan ^2 \frac{x}{2}=1\end{aligned}$
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