If $\sec \theta+\tan \theta=p, 0^{\circ}<\theta<90^{\circ}$, then $\frac{p^2-1}{p^2+1}$ is equal to:
Answer & explanation
Correct answer: option 2
A detailed explanation for this question is coming soon.
If $\sec \theta+\tan \theta=p, 0^{\circ}<\theta<90^{\circ}$, then $\frac{p^2-1}{p^2+1}$ is equal to:
Correct answer: option 2
A detailed explanation for this question is coming soon.