Question : If $\cos27^{\circ}$ = $x$, then the value of $\tan 63°$ is:
Option 1: $\frac{x}{\sqrt{1–x^2}}$
Option 2: $\frac{x}{\sqrt{1+x^2}}$
Option 3: $\frac{\sqrt{1–x^2}}{x}$
Option 4: $\frac{\sqrt{1+x^2}}{x}$
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Correct Answer: $\frac{x}{\sqrt{1–x^2}}$
Solution :
In $\triangle ABC$,
⇒ $\cos\ 27°$ = $x$
⇒ $\frac{AB}{BC}$ = $\frac{x}{1}$
By Pythagoras theorem,
Perpendicular
2
+ Base
2
= Hypotenuse
2
⇒ AC
2
+ AB
2
= BC
2
⇒ AC
2
= BC
2
– AB
2
= $1–x^2$
⇒ AC = $\sqrt{1–x^2}$
$\tan 63°$ = $\frac{AB}{AC}$
= $\frac{x}{\sqrt{1–x^2}}$
Hence, the correct answer is $\frac{x}{\sqrt{1–x^2}}$.
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