Question

sintheta+costheta=m,sectheta+cosectheta=n, prove n(m2-1)=2m​

Answer 1

Given-

$\sin \theta + \cos \theta = m.....(i) \\ \sec \theta + cosec \theta = n......(ii)$

To prove-

$\implies \: n( {m}^{2} - 1) = 2m$

Proof:-

L. H. S. -

$\large\sf{ = n( {m}^{2} - 1) }$

$\large\sf{ = \sec \theta+ cosec \theta \: [{{(\sin \theta+ \cos \theta })^{2} } - 1] }$

$\large\sf{ = \sec \theta \: + cosec \theta \: [( { \sin}^{2} \theta+ { \cos}^{2} \theta + 2 \sin \theta \times \cos \theta) - 1] }$

$\large\sf{ = \sec \theta \: + cosec \theta \: ( 1 + 2 \sin \theta \times \cos \theta) - 1 }$

$\large\sf{ = \sec \theta\: + cosec \theta \: ( 1 + 2 \sin \theta \times \cos \theta - 1 ) }$

$\large\sf{ = \sec \theta \: + cosec \theta \:(2 \sin \theta \times \cos \theta ) }$

$\large\sf{ = 2 \sin \theta \cos \theta sec \theta + 2 \sin \theta \cos\theta cosec \theta}$

$\large\sf{ = 2 \sin \theta \cos \theta \frac{1}{ \cos \theta } + 2 \cos \theta \sin \theta \frac{1}{ \sin \theta } }$

$\large\sf{ = 2 \sin \theta+ 2 \cos \theta }$

$\large\sf{ = 2(\sin \theta+ \cos \theta) }$

$\large\sf{ = 2m........using\:(i) }$

$\large\sf{L.H.S</p><p>=R. H. S</p><p>}$

HENCE, PROVED.