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

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

Vedclass · Accepted Answer

$6$

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(B) Given $y = \frac{7 + 6 	an x - 	an^2 x}{1 + 	an^2 x}$.Since $1 + 	an^2 x = \sec^2 x$,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 trigonometric identities $\cos^2 x = \frac{1 + \cos 2x}{2}$,$\sin^2 x = \frac{1 - \cos 2x}{2}$,and $\sin x \cos x = \frac{\sin 2x}{2}$:$y = 7 \left( \frac{1 + \cos 2x}{2} ight) + 6 \left( \frac{\sin 2x}{2} ight) - \left( \frac{1 - \cos 2x}{2} ight)$.$y = \frac{7 + 7 \cos 2x + 6 \sin 2x - 1 + \cos 2x}{2} = \frac{6 + 8 \cos 2x + 6 \sin 2x}{2} = 3 + 4 \cos 2x + 3 \sin 2x$.The expression $a \sin 	heta + b \cos 	heta$ has a maximum value of $\sqrt{a^2 + b^2}$.Here,$3 \sin 2x + 4 \cos 2x$ has a maximum value of $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.Thus,the maximum value of $y$ is $\lambda = 3 + 5 = 8$.We need to find $\log_{\sqrt{2}}(\lambda) = \log_{\sqrt{2}}(8)$.Since $8 = 2^3 = (\sqrt{2})^6$,we have $\log_{\sqrt{2}}((\sqrt{2})^6) = 6$.

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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