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Question
If tan A = n tan B and sin A = m sin B, prove that:
`cos^2A = (m^2 - 1)/(n^2 - 1)`
Solution
Given that, tan A = n tan B and sin A = m sin B.
`=> n = tanA/tanB` and `m = sinA/sinB`
∴ `(m^2 - 1)/(n^2 - 1) = ((sinA/sinB)^2 - 1)/((tanA/tanB)^2 - 1)`
= `(sin^2A/sin^2B - 1/1)/(tan^2A/(tan^2B) - 1)`
= `((sin^2A - sin^2B)/sin^2B)/((tan^2A - tan^2B)/tan^2B)`
= `((sin^2A - sin^2B)/tan^2B)/((tan^2A - tan^2B)/sin^2B)`
= `((sin^2A - sin^2B)sin^2B)/((sin^2A/cos^2A-sin^2B/cos^2B)cos^2Bsin^2B)`
= `(sin^2A - sin^2B)/(((sin^2A.cos^2B - sin^2B.cos^2A)/(cos^2A.cos^2B)) cos^2B)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A.cos^2B - sin^2B.cos^2A)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A(1 - sin^2B) - sin^2B (1 - sin^2A))`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A - sin^2A.sin^2B - sin^2B + sin^2B.sin^2A)`
= `((sin^2A - sin^2B)cos^2A)/(sin^2A -sin^2B)`
= cos2 A
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