Question

If p sin theta +q cos theta =a and p cos theta -q cos theta =b then p+a/q+b+q-b/p-a is equal to?

Answer 1

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Here is your answer in the attachment


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Answer 2

Hi friends

Here is your answer
I

$a = p \: sin \alpha + q \: cos \alpha \\ {a}^{2} = {p}^{2} sin {}^{2} \alpha + {q}^{2} cos \alpha + 2pqsin \alpha \cos( \alpha )$
II
$b = p \cos( \alpha ) - q \sin( \alpha ) \\ on \: squaring \\ {b}^{2} = {p}^{2} \cos {}^{2} ( \alpha ) + {q}^{2} \sin {}^{2} ( \alpha ) - 2pq \sin( \alpha ) \cos( \alpha )$
On Adding both equation
${a}^{2} + {b}^{2} = {p}^{2} ( \sin {}^{2} ( \alpha ) + q {}^{2} \cos {}^{2} ( \alpha ) ) + {q}^{2} ( \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) ) \\ \\ {a}^{2} + {b}^{2} = {p}^{2} + {q}^{2} \\ as \: \sin {}^{2} ( \alpha ) + \cos {}^{2} ( \alpha ) = 1$
Now we find the value
$\frac{p + a}{q + b} + \frac{q - b}{p - a} \\ \\ = \frac{ {p}^{2} - {a}^{2} + {q}^{2} - {b}^{2} }{(q + b)(p - a)} \\ \\ = \frac{ {a}^{2} + {b}^{2} - {a}^{2} - {b}^{2} }{(q + b)(p - a)} \\ \\ = 0$
Hope it helps you

@ MSD