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2016-03-31T05:57:10+05:30

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Solution


Given sec θ + csc θ = n ………………… (A) 


⇒ 1/cos θ + 1/sin θ = n


⇒ (sin θ + cos θ)/cos θ sin θ = n


cos θ sin θ = (sin θ + cos θ)/n


sin θ = m/n, [using (A)] ………………… (B) 


Now cos θ + sin θ = m


⇒ (cos θ + sin θ)2 = m2


⇒ cos2 θ + sin2 θ + 2 sin θ cos θ = m


⇒ 1 + 2 sin θ cos θ = m2


⇒ 1 + 2 ∙ (m/n) = m2, [Using (B)] 


⇒ 2 (m/n) = m2 - 1


⇒ 2m = n(m2 - 1), [Proved] 
1 5 1
good one deserves brainiest answer......................................
2016-03-31T06:12:14+05:30

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Given sec θ + csc θ = n ………………… (i) 
⇒ 1/cos θ + 1/sin θ = n
⇒ (sin θ + cos θ)/cos θ sin θ = n
cos θ sin θ = (sin θ + cos θ)/n
sin θ = m/n, [using (i)] ………………… (ii) 
Now cos θ + sin θ = m
⇒ (cos θ + sin θ)2 = m2
⇒ cos2 θ + sin2 θ + 2 sin θ cos θ = m
⇒ 1 + 2 sin θ cos θ = m2
⇒ 1 + 2 ∙ (m/n) = m2, [Using (ii)] 
⇒ 2 (m/n) = m2 - 1
⇒ 2m = n(m2 - 1), [Proved] 
1 5 1
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