Question

Sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p 2 – 1) = 2p.

Answer 1

$\sin( \alpha ) + \cos( \alpha ) = x \\ \sec( \alpha ) + \csc( \alpha ) = y \\ \frac{ \sin( \alpha ) + \cos( \alpha ) }{ \sin( \alpha ) \cos( \alpha ) } = y \\ \sin( \alpha ) \cos( \alpha ) = \frac{x}{y}$
Now consider this
$\sin( \alpha ) + \cos( \alpha ) = x \\ \ \sin {}^{2} \alpha + { \cos }^{2} \alpha + 2 \sin( \alpha ) \cos( \alpha ) = {x}^{2} \\ 2 \sin( \alpha ) \cos( \alpha ) = {x }^{2} - 1 \\ \sin( \alpha ) \cos( \alpha ) = \frac{ {x}^{2} - 1 }{2}$
Now substitute this in the previous equation to get
$\frac{x}{y} = \sin( \alpha ) \cos( \alpha ) \\ \frac{x}{y} = \frac{ {x}^{2} - 1}{2} \\ 2x = y( {x}^{2} - 1) \\ or \\ 2p = q( {p}^{2} - 1)$