Exams › SSC CGL (Prelims) › General

If $\tan A=\frac{x}{x+1}$, find $\sec^2 A$.

1. $\frac{x^2+(x+1)^2}{(x+1)^2}$
2. $\frac{x^2+1}{x^2}$
3. $\frac{2x^2+2x+1}{x^2+1}$
4. 1

Correct answer: $\frac{x^2+(x+1)^2}{(x+1)^2}$

Solution

We know $\sec^2 A=1+\tan^2 A$. Substituting $\tan A=\frac{x}{x+1}$ gives $\sec^2 A=1+\frac{x^2}{(x+1)^2}=\frac{(x+1)^2+x^2}{(x+1)^2}$. This matches the first option.

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