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Question

Accepted Answer

Given: $ p \sin A - \cos A = 1 $

Let's solve for $ \cos A $ in terms of $ p $ and $ \sin A $:

$ p \sin A - \cos A = 1 $
$ \Rightarrow p \sin A = 1 + \cos A $
$ \Rightarrow \sin A = \frac{1 + \cos A}{p} $

Now, recall that $ \sin^2 A + \cos^2 A = 1 $:

Substitute $ \sin A $:
$$
\left(\frac{1 + \cos A}{p}\right)^2 + \cos^2 A = 1
$$
$$
\frac{(1 + \cos A)^2}{p^2} + \cos^2 A = 1
$$
$$
(1 + \cos A)^2 + p^2 \cos^2 A = p^2
$$
$$
1 + 2\cos A + \cos^2 A + p^2 \cos^2 A = p^2
$$
$$
1 + 2\cos A + (1 + p^2)\cos^2 A = p^2
$$
$$
(1 + p^2)\cos^2 A + 2\cos A + (1 - p^2) = 0
$$

This is a quadratic in $ \cos A $:
$$
(1 + p^2)\cos^2 A + 2\cos A + (1 - p^2) = 0
$$

Let $ x = \cos A $:
$$
(1 + p^2)x^2 + 2x + (1 - p^2) = 0
$$

We are to find $ p^2 - (1 + p^2)\cos A $:
Let’s denote this as $ S = p^2 - (1 + p^2)x $.

Let’s try to find $ x $ in terms of $ p $:

From the quadratic:
$$
(1 + p^2)x^2 + 2x + (1 - p^2) = 0
$$

Let’s solve for $ x $:
$$
x = \frac{-2 \pm \sqrt{4 - 4(1 + p^2)(1 - p^2)}}{2(1 + p^2)}
$$
$$
= \frac{-2 \pm \sqrt{4 - 4(1 - p^4)}}{2(1 + p^2)}
$$
$$
= \frac{-2 \pm \sqrt{4 - 4 + 4p^4}}{2(1 + p^2)}
$$
$$
= \frac{-2 \pm 2p^2}{2(1 + p^2)}
$$
$$
= \frac{-1 \pm p^2}{1 + p^2}
$$

So, possible values for $ x = \cos A $ are:
1. $ \frac{-1 + p^2}{1 + p^2} $
2. $ \frac{-1 - p^2}{1 + p^2} $

Let’s check which value is valid. Recall from earlier:
$ \sin A = \frac{1 + \cos A}{p} $
$ \sin A $ must be in [-1, 1].

Try $ \cos A = \frac{-1 + p^2}{1 + p^2} $:
$$
1 + \cos A = 1 + \frac{-1 + p^2}{1 + p^2} = \frac{1 + p^2 - 1 + p^2}{1 + p^2} = \frac{2p^2}{1 + p^2}
$$
$$
\sin A = \frac{2p^2}{p(1 + p^2)} = \frac{2p}{1 + p^2}
$$

This is valid for real $ p $. For the other root:
$ \cos A = \frac{-1 - p^2}{1 + p^2} $
$$
1 + \cos A = 1 + \frac{-1 - p^2}{1 + p^2} = \frac{1 + p^2 - 1 - p^2}{1 + p^2} = 0
$$
$$
\sin A = 0
$$

But if $ \sin A = 0 $, then from the original equation:
$ p \cdot 0 - \cos A = 1 \implies -\cos A = 1 \implies \cos A = -1 $, which matches $ p = 0 $. But for general $ p $, the first root is valid.

Now, calculate $ S = p^2 - (1 + p^2)\cos A $:
$$
S = p^2 - (1 + p^2)\left(\frac{-1 + p^2}{1 + p^2}\right)
$$
$$
= p^2 - (-1 + p^2)
$$
$$
= p^2 + 1 - p^2
$$
$$
= 1
$$

Thus, the answer is 1.

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Answer

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