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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Question : If $\sin17°=\frac{x}{y}$, then the value of $(\sec17°–\sin73°)$ is:
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Question : What is the value of $\frac{\tan 45^{\circ}-\tan 15^{\circ}}{1+\tan 45^{\circ} \tan 15^{\circ}}$?
Question : If ${\operatorname{cosec} 39^{\circ}} = x$, then the value of $ \frac{1}{\operatorname{cosec}^2 51^{\circ}} +\sin^239^{\circ} +\tan ^251^{\circ} -\frac{1}{\sin ^2 51^{\circ} \sec ^2 39^{\circ}}$ is:
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