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, Perpendicular2 + Base2 = Hypotenuse2 ⇒ AC2 + AB2 = BC2 ⇒ AC2 = BC2 – AB2 = $1–x^2$ ⇒ 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 $x\cos^{2}30^{\circ}\cdot \sin60^{\circ}=\frac{\tan^{2}45^{\circ}\cdot \sec60^{\circ}}{\operatorname{cosec}60^{\circ}}$, then the value of $x$ is:
Option 1: $\frac{1}{\sqrt{3}}$
Option 2: $\frac{1}{\sqrt{2}}$
Option 3: $2\frac{2}{3}$
Option 4: $\frac{1}{2}$
Question : What is the value of $\frac{\tan 45^{\circ}-\tan 15^{\circ}}{1+\tan 45^{\circ} \tan 15^{\circ}}$?
Option 1: $\sqrt{3}$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: ${\sqrt{2}}$
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:
Option 1: $x^2-1$
Option 2: $\sqrt{x^2-1}$
Option 3: $ \sqrt{1-x^2}$
Option 4: $1-x^2$
Question : If $x\sin ^{2}60^{\circ}-\frac{3}{2}\sec 60^{\circ}\tan^{2}30^{\circ}+\frac{4}{5}\sin ^{2}45^{\circ}\tan ^{2}60^{\circ}=0$, then $x$ is:
Option 1: $-\frac{1}{15}$
Option 2: $–4$
Option 3: $-\frac{4}{15}$
Option 4: $–2$
Question : The value of tan 25° tan 35° tan 45° tan 55° tan 65° is:
Option 1: $2$
Option 2: $1$
Option 3: $\sqrt{3}$
Option 4: $0$
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