Question : If $\sec x- \cos x$ = 4, then what will be the value of $\frac{\left(1+\cos ^2x\right)}{\cos x}?$
Option 1: $\frac{9}{4}$
Option 2: $\frac{1}{4}$
Option 3: $2\sqrt{5}$
Option 4: $\sqrt{5}$
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Correct Answer: $2\sqrt{5}$
Solution :
$\sec x- \cos x$ = $4$
⇒ $(\sec x- \cos x)^2$ = $4^2$
⇒ $\sec^2 x- 2\sec x \cos x+\cos^2 x $ = $16$
⇒ $\sec^2 x+\cos^2 x $ = $16+2\sec x \cos x$
⇒ $\sec^2 x+\cos^2 x + 2\sec x \cos x$ = $16+2\sec x \cos x+2\sec x \cos x$
Since $\sec x \cos x$ = 1,
So, $(\sec x+ \cos x)^2$ = $16+2+2$
⇒ $(\frac{1}{\cos x}+ \cos x)^2$ = $20$
⇒ $(\frac{1+ \cos^2 x}{\cos x})$ = $\sqrt{20}$ = $2\sqrt5$
Hence, the correct answer is $2\sqrt5$.
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