Math, asked by Surya1690, 8 months ago

if p =sintheta + cos theta and q = sec theta + cosec theta then show that q ( p²-1) =2 p​

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Answered by sandy1816
1

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Answered by Oreki
4

\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)}

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