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2014-05-20T13:33:00+05:30
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


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