Question

if sin theta + cos theta=p and sec theta+ cosec theta =q show that Q(P^2-1 )

Answer 1

P=sinθ+cosθ and q=secθ+cosecθ

q(p²-1)

=(secθ+cosecθ)[(sinθ+cosθ)²-1]

=(1/cosθ+1/sinθ)(sin²θ+2sinθcosθ+cos²θ-1)

={(sinθ+cosθ)/sinθcosθ}(2sinθcosθ) [ Since, sin²θ+cos²θ=1]

=2(sinθ+cosθ)

=2p (Proved)

Hope its usefull

Answer 2

$\textsf{\textbf{Question - }}\\\textsf{\hspace{1.5em} If \sin \theta + \cos \theta = p and \sec \theta + \csc \theta = q , then prove that q(p^2 - 1) = 2p .}$

$\textsf{\textbf{Answer:}}$

    ${\bf Given,}\\{\bf \hspace{3em} \sin \theta + \cos \theta = p \hspace{2em} ... (i) \: \: and \: \: \sec \theta + \csc \theta = q \hspace{2em} ... (ii)}\\{\bf \Rightarrow \hspace{1.6em} \dfrac{1}{\cos \theta} + \dfrac{1}{\sin \theta} = q \hspace{1.5em} \Rightarrow \:\: \dfrac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}}\\{ \Rightarrow \hspace{1.6em} \dfrac{p}{\sin \theta \cos \theta} = q \hspace{15em} {\bf (using (i))}}\\{\Rightarrow \hspace{1.6em} \sin \theta \cos \theta = \dfrac{p}{q}}$

                                                                                                           ${\bf ... \:(iii)}$

    $\textsf{On squaring (i), we get}\\{\Rightarrow \hspace{1.6em} (\sin \theta + cos \theta)^2 = p^2 = (\sin^2 \theta + \cos^2 \theta) + 2 \sin \theta \cos \theta = p^2}\\{\Rightarrow \hspace{1.6em} 1 + 2 \dfrac{p}{q} = p^2 \hspace{18em} {\bf (using (iii))}}\\{\Rightarrow \hspace{1.6em} \dfrac{2p}{q} = p^2 - 1 \: \: \: \Rightarrow \: \: \: 2p = q(p^2 + 1)}$