Question : If $\sec ^2 \mathrm{~A}+\tan ^2 \mathrm{~A}=3$, then what is the value of $\cot \mathrm{A}$?
Option 1: $\frac{1}{\sqrt{3}}$
Option 2: $0$
Option 3: $1$
Option 4: $\sqrt{3}$
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Correct Answer: $1$
Solution : Given: $\sec ^2\mathrm{A}+\tan ^2 \mathrm{A}=3$ ⇒ $\frac{1}{\cos^2 \mathrm{A}} + \frac{\sin^2\mathrm{A}}{\cos^2\mathrm{A}} = 3$ ⇒ $1+\sin^2\mathrm{A} = 3\cos^2\mathrm{A}$ ⇒ $\sin^\mathrm{A} + \cos^2\mathrm{A} + \sin^2\mathrm{A} = 3 \cos^2\mathrm{A}$ ⇒ $2\sin^2\mathrm{A} = 2\cos^2\mathrm{A}$ ⇒ $\frac{\sin^2\mathrm{A}}{\cos^2\mathrm{A}} = 1$ ⇒ $\tan^2\mathrm{A} = 1$ ⇒ $\tan\mathrm{A} = 1$ $\therefore$ $\cot\mathrm{A} = 1$ Hence, the correct answer is 1.
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