Trigonometrical ratios

Question

Q. If sec A + tan A = P., then dins value of sec A tan A in terms of 'P'.

My solution so far - squaring both side of the former equation, transposing tan A squaring, trans posing sec A amd squaring, making relation and substitute them on squared version of the former equation but not any luck so far

Answer 1

Hint:

As $\sec A+\tan A=P$

and $(\sec A+\tan A)(\sec A-\tan A)=1$

$\sec A-\tan A=\dfrac1P$

Can you find $\sec A,\tan A?$

or use $$4ab=(a+b)^2-(a-b)^2$$

Answer 2

$\sec^2A - \tan^2A = 1$

$(\sec A - \tan A)(\sec A + \tan A) = 1$

$$\sec A - \tan A = \frac{1}{\sec A + \tan A}$$

Substitute the value of $\sec A + \tan A$ as $P$ and find the values of $\sec A$ and $\tan A$