If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q (p2 – 1) = 2p

Question

Q) If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q (p2  – 1) = 2 p.

Sapling Academy · Accepted Answer

Let's take the components one by one:

Step 1: Given that sin θ + cos θ = p

∴ p = sin θ  + cos θ ..... (i)

Step 2: Given that sec θ + cosec θ = q

∴ q = $\frac{1}{\cos 	heta} + \frac{1}{\sin 	heta}$

∴ q = $\frac{(\sin 	heta + \cos 	heta)}{\sin 	heta \cos 	heta}$ .................. (ii)

Step 3: starting from LHS =  q (p2 – 1)

= $(\frac{(\sin 	heta + \cos 	heta)}{\sin 	heta \cos 	heta}) ( (\sin 	heta  + \cos 	heta)^2 - 1)$

= $(\frac{(\sin 	heta + \cos 	heta)}{\sin 	heta \cos 	heta}) (\sin ^2 	heta  + \cos ^2 	heta + 2 \sin 	heta \cos 	heta - 1)$

Since sin2  θ + cos 2 θ = 1

∴ LHS = $(\frac{(\sin 	heta + \cos 	heta)}{\sin 	heta \cos 	heta}) (1 + 2 \sin 	heta \cos 	heta - 1)$

= $(\frac{(\sin 	heta + \cos 	heta)}{\sin 	heta \cos 	heta}) (\cancel{1} + 2 \sin 	heta \cos 	heta - \cancel{1})$

= $(\frac{(\sin 	heta + \cos 	heta)}{\sin 	heta \cos 	heta}) (2 ~ \sin 	heta \cos 	heta)$

= $(\frac{(\sin 	heta + \cos 	heta)}{\cancel{\sin 	heta \cos 	heta}}) (2 ~ \cancel{\sin 	heta \cos 	heta})$

= $ 2 (\sin 	heta + \cos 	heta)$

= 2 p        [Since sin θ + cos θ = p]

Hence Proved !

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