# If cos θ + sin θ = m and sec θ + csc θ = n, prove that n(m2 – 1) = 2m.

2
by Deleted account

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]
2016-03-31T06:12:14+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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]