Question

if cos theta + Sin theta =m and sec theta + Coses theta =n


prove n(m²-1)= 2m​

Answer 1

Answer:

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 + cosec\theta)[2sin\theta cos\theta] since {cos^2 \theta + sin^2 \theta = 1}(secθ+cosecθ)[2sinθcosθ]since cos

2

θ+sin

2

θ=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

Step-by-step explanation:

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