If 3 sin x + 4 cos x = 5,then 6 tan frac{x}{2 | vedclass

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(B) Given,$3 \sin x + 4 \cos x = 5$. Using the half-angle formulas $\sin x = \frac{2 	an \frac{x}{2}}{1 + 	an^2 \frac{x}{2}}$ and $\cos x = \frac{1

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(B) Given,$3 \sin x + 4 \cos x = 5$. Using the half-angle formulas $\sin x = \frac{2 	an \frac{x}{2}}{1 + 	an^2 \frac{x}{2}}$ and $\cos x = \frac{1 - 	an^2 \frac{x}{2}}{1 + 	an^2 \frac{x}{2}}$,we have: $3 \left( \frac{2 	an \frac{x}{2}}{1 + 	an^2 \frac{x}{2}} ight) + 4 \left( \frac{1 - 	an^2 \frac{x}{2}}{1 + 	an^2 \frac{x}{2}} ight) = 5$. Multiplying both sides by $(1 + 	an^2 \frac{x}{2})$: $6 	an \frac{x}{2} + 4 - 4 	an^2 \frac{x}{2} = 5(1 + 	an^2 \frac{x}{2})$. $6 	an \frac{x}{2} + 4 - 4 	an^2 \frac{x}{2} = 5 + 5 	an^2 \frac{x}{2}$. Rearranging the terms: $6 	an \frac{x}{2} - 9 	an^2 \frac{x}{2} = 5 - 4$. $6 	an \frac{x}{2} - 9 	an^2 \frac{x}{2} = 1$.

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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