If the maximum value of y = frac{7 + 6 tan x | vedclass

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(B) Given $y = \frac{7 + 6 	an x - 	an^2 x}{1 + 	an^2 x}$.Using $1 + 	an^2 x = \sec^2 x$,we have $y = (7 + 6 	an x - 	an^2 x) \cos^2 x$.$y = 7 \

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$6$

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(B) Given $y = \frac{7 + 6 	an x - 	an^2 x}{1 + 	an^2 x}$.Using $1 + 	an^2 x = \sec^2 x$,we have $y = (7 + 6 	an x - 	an^2 x) \cos^2 x$.$y = 7 \cos^2 x + 6 \sin x \cos x - \sin^2 x$.Using $\cos^2 x = \frac{1 + \cos 2x}{2}$,$\sin^2 x = \frac{1 - \cos 2x}{2}$,and $\sin x \cos x = \frac{\sin 2x}{2}$:$y = 7 \left( \frac{1 + \cos 2x}{2} ight) + 3 \sin 2x - \left( \frac{1 - \cos 2x}{2} ight)$.$y = \frac{7 + 7 \cos 2x - 1 + \cos 2x}{2} + 3 \sin 2x = 4 \cos 2x + 3 \sin 2x + 3$.The maximum value of $a \cos 	heta + b \sin 	heta$ is $\sqrt{a^2 + b^2}$.Here,the maximum value is $\sqrt{4^2 + 3^2} + 3 = 5 + 3 = 8$.Thus,$\lambda = 8$.We need to find $\log_{\sqrt{2}}(8) = \log_{\sqrt{2}}((\sqrt{2})^6) = 6$.

If the maximum value of $y = \frac{7 + 6 \tan x - \tan^2 x}{1 + \tan^2 x}$ is $\lambda$,then the value of $\log_{\sqrt{2}}(\lambda)$ is

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