Question

If sin theta+ cos theta = p and sec theta + cosec theta = 9, show that q (p^2- 1) = 2p.​

Answer 1

Go through the pic.

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and next time pps pls write question properly it is q in place of 9 .

Answer 2

$\bf{\Large{\underline{\sf{Correct\:Question\::}}}}}}$

If sinФ + cosФ = p & secФ + cosecФ = q, show that q(p² - 1) = 2p.

$\bf{\Huge{\boxed{\sf{\green{ANSWER\::}}}}}}}$

$\bf{\Large{\underline{\bf{Given\;:}}}}}$

If sinФ + cosФ = p & secФ + cosecФ = q.

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

Show that q(p² - 1) = 2p.

$\bf{\Large{\underline{\sf{\blue{Explanation\::}}}}}$

$\longmapsto\sf{sin\theta+cos\theta=p....................(1)}$

$\longmapsto\sf{sec\theta+cosec\theta=q.......................(2)}$

From equation (2), we get;

$\leadsto\sf{sec\theta+cosec\theta=q}$

$\leadsto\sf{\frac{1}{cos\theta} +\frac{1}{sin\theta} =q}$

$\leadsto\sf{\frac{sin\theta+cos\theta}{cos\theta\:sin\theta} =q}$

$\leadsto\sf{\frac{p}{cos\theta\:sin\theta} =q}\:\:\:\:\:\:\:\:\:\:\:[from\:equation\:(1)]$

$\leadsto\sf{\frac{p}{q} =cos\theta\:sin\theta................(3)}$

from equation (1), we get;

$\leadsto\sf{sin\theta+cos\theta=p}$

Squaring both the sides, on

$\leadsto\sf{(sin\theta+cos\theta)^{2} =(p)^{2} }$

$\leadsto\sf{sin^{2}\theta +cos^{2}\theta +2sin\theta\:cos\theta=p^{2} \:\:\:\:\:\:\:\:[(a+b)^{2}=a^{2}+b^{2} +2ab ]}$

$\leadsto\sf{1+2\frac{p}{q} =p^{2} \:\:\:\:\:\:\:\:[sin^{2} \theta+cos^{2} \theta=1]}$

$\leadsto\sf{\frac{2p}{q} =p^{2} -1}$

$\leadsto\sf{\red{2p=q(p^{2} -1)}}$

Thus,

Proved.