Let $f:(-1,1) ightarrow I R$ be such that $f(\cos 4 heta)=\frac{2}{2-\sec ^{2} heta}$ for $ heta \in\left(0, \frac{\pi}{4} ight) \cup\left(\frac{\pi}{4}, \frac{\pi}{2} ight)$. Then the value (s) of $f\left(\frac{1}{3} ight)$ is (are)

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Let $f:(-1,1) ightarrow I R$ be such that $f(\cos 4 	heta)=\frac{2}{2-\sec ^{2} 	heta}$ for $	heta \in\left(0, \frac{\pi}{4}ight) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}ight)$. Then the value (s) of $f\left(\frac{1}{3}ight)$ is (are)

Aicharya · Accepted Answer

Given : $f(\cos 4 	heta)=\frac{2}{2-\sec ^{2} 	heta}=\frac{2 \cos ^{2} 	heta}{2 \cos ^{2} 	heta-1}$ $=\frac{1+\cos 2 	heta}{\cos 2 	heta}=1+\frac{1}{\cos 2 	heta}$ Let $\cos 4 	heta=\frac{1}{3} \Rightarrow 2 \cos ^{2} 2 	heta-1=\frac{1}{3} \Rightarrow \cos 2 	heta=\pm \sqrt{\frac{2}{3}}$ $	herefore f(\cos 4 	heta)=1+\frac{1}{\cos 2 	heta}=1 \pm \sqrt{\frac{3}{2}}$ or $f\left(\frac{1}{3}ight)=1 \pm \sqrt{\frac{3}{2}}$

Let $f:(-1,1) \rightarrow I R$ be such that $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}$ for $\theta…

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