If sin θ + cos θ = p and sec θ + cosec θ = q, then prove

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

Ans: 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 = $\frac{1}{\cos \theta} + \frac{1}{\sin \theta}$

∴ q = $\frac{(\sin \theta + \cos \theta)}{\sin \theta \cos \theta}$ ……………… (ii)

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

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

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

Since sin² θ + cos ²θ = 1

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

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

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

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

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

= 2 p [Since sin θ + cos θ = p]

Hence Proved !

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