Question

If sin theta + cos theta= m, sec theta + cosec
theta = n. Prove that, n(m^2 - 1) = 2m

Answer 1

Given that, 
sin theta + cos theta= m, sec theta + cosec theta = n.
Consider LHS
$n(m^2 - 1)$

$(sec\theta + cosec\theta)[(cos\theta + sin\theta)^2 - 1]$

$(sec\theta + cosec\theta)[cos^2 \theta + sin^2 \theta + 2sin\theta cos\theta - 1]$

$(sec\theta + cosec\theta)[2sin\theta cos\theta] since {cos^2 \theta + sin^2 \theta = 1}$

$sec\theta . 2sin\theta cos\theta + cosec\theta . 2sin\theta cos\theta$

$2sin\theta + 2 cos\theta$

$2[sin\theta + cos\theta]$

= 2m

Answer 2

Given that,

sin theta + cos theta= m, sec theta + cosec theta = n.

Consider LHS

n(m^2 - 1)n(m

2

−1)

(sec\theta + cosec\theta)[(cos\theta + sin\theta)^2 - 1](secθ+cosecθ)[(cosθ+sinθ)

2

−1]

(sec\theta + cosec\theta)[cos^2 \theta + sin^2 \theta + 2sin\theta cos\theta - 1](secθ+cosecθ)[cos

2

θ+sin

2

θ+2sinθcosθ−1]

sec\theta . 2sin\theta cos\theta + cosec\theta . 2sin\theta cos\thetasecθ.2sinθcosθ+cosecθ.2sinθcosθ

2sin\theta + 2 cos\theta2sinθ+2cosθ

2[sin\theta + cos\theta]2[sinθ+cosθ]

= 2m