Question : If $\sec A - \tan A = p$, then find the value of $\sec A$.
Option 1: $\frac{p^2-1}{p^2+1}$
Option 2: $\frac{\mathrm{p}^2+1}{\mathrm{p}^2-1}$
Option 3: $\frac{\mathrm{p}^2+1}{\mathrm{p}}$
Option 4: $\frac{\mathrm{p}^2+1}{2 \mathrm{p}}$
Correct Answer: $\frac{\mathrm{p}^2+1}{2 \mathrm{p}}$
Solution :
$\sec A - \tan A = p$
$\frac{1}{\cos A} - \frac{\sin A}{\cos A} = p$
$\frac{(1-\sin A)^2}{(\cos A)^2} = p^2$
$\frac{(1-\sin A)^2}{1-\sin^2 A} = p^2$
$\frac{(1-\sin A)^2}{(1-\sin A)(1+\sin A)} = p^2$
$\frac{1-\sin A}{1+\sin A}= p^2$
$1-\sin A = p^2 + p^2\sin A$
$\sin A =\frac{1-p^2}{1+p^2}$
$\cos A = \sqrt{1-\sin^2 A}$
$= \sqrt{1-(\frac{1-p^2}{1+p^2})^2}$
$=\sqrt{\frac{(1+p^2)^2-(1-p^2)^2}{(1+p^2)^2}}$
$=\sqrt{\frac{(1+p^2+1-p^2)(1+p^2-1+p^2)}{(1+p^2)^2}}$
$=\sqrt{\frac{2×2p^2}{(1+p^2)^2}}$
$=\sqrt{\frac{(2p)^2}{(1+p^2)^2}}$
$= \frac{2p}{1+p^2}$
$\sec A = \frac{1+p^2}{2p}$
Hence, the correct answer is $ \frac{1+p^2}{2p}$.
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