Question : The value of $\tan1° \tan2° \tan3°...\tan89°$ is:
Option 1: $1$
Option 2: $0$
Option 3: $\sqrt{3}$
Option 4: $\frac{1}{\sqrt{3}}$
Correct Answer: $1$
Solution : $\tan1° \tan2° \tan3°...\tan89°$ $= \tan1° \tan2° \tan3°...\tan45°...\tan(90-3)°\tan(90-2)°\tan(90-1)°$ $=\tan1° \tan2° \tan3°...\tan45°...\cot3°\cot2°\cot1°$ As we know, $\tan\theta=\frac{1}{\cot\theta}$ So, it will cancel out all the terms except $\tan45°$. $=\tan45°$ $= 1$ Hence, the correct answer is $1$.
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Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$, what is the value of $(x^3+\frac{1}{x^3})$?
Option 1: $\frac{3\sqrt{3}}{5}$
Option 2: $\frac{3\sqrt{15}}{5}$
Option 3: $\frac{3\sqrt{15}}{8}$
Option 4: $\frac{3\sqrt{5}}{8}$
Question : The value of $(\tan35° \tan45° \tan55°)$ is:
Option 1: $\frac{1}{2}$
Option 2: $2$
Option 3: $0$
Option 4: $1$
Question : If $x^2+\frac{1}{x^2}=\frac{7}{4}$ for $x>0$; then what is the value of $x+\frac{1}{x}$?
Option 1: $2$
Option 2: $\frac{\sqrt{15}}{2}$
Option 3: $\sqrt{5}$
Option 4: $\sqrt{3}$
Question : If $\operatorname{cosec} 39°=x$, then the value of $\frac{1}{\operatorname{cosec}^{2}51°}+\sin^{2}39°+\tan^{2}51°-\frac{1}{\sin^{2}51°\sec^{2}39°}$ is:
Option 1: $\sqrt{x^{2}-1}$
Option 2: $\sqrt{1-x^{2}}$
Option 3: $x^{2}-1$
Option 4: $1-x^{2}$
Question : If $\sin A=\frac{1}{2}$, then the value of $(\tan A+\cos A)$ is:
Option 1: $\frac{2}{3 \sqrt{3}}$
Option 2: $\frac{3}{2 \sqrt{3}}$
Option 3: $\frac{5}{2 \sqrt{3}}$
Option 4: $\frac{5}{3 \sqrt{3}}$
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