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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