Question : If $\tan^{2}\alpha=1+2\tan^{2}\beta$ ($\alpha,\beta$ are positive acute angles), then $\sqrt2\: \cos\alpha -\cos\beta$ is equal to:
Option 1: $0$
Option 2: $\sqrt2$
Option 3: $1$
Option 4: $–1$
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Correct Answer: $0$
Solution : $\tan^{2}\alpha=1+2\tan^{2}\beta$ $⇒\sec^{2}\alpha-1=1+2(\sec^{2}\beta-1)$ $⇒\sec^{2}\alpha-1=2\sec^{2}\beta-1$ $⇒\frac{1}{\cos^{2}\alpha}=\frac{2}{\cos^{2}\beta}$ Taking square root on both sides, we get, $⇒\sqrt2\cos\alpha=\cos\beta$ $⇒\sqrt2\cos\alpha-\cos\beta=0$ Hence, the correct answer is $0$.
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