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 $\sin x = \frac{2 	an(x/2)}{1 + 	an^2(x/2)}$ and $\cos x = \frac{1 - 	an^2(x/2)}

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

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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