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Question
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
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Solution
We have,
tan α = n tan β
⇒ `tan β = tan α/n`
⇒ `cot β = n/tan α`
sin α = m sin β
⇒ `sin β = sin α /m`
⇒ `cosec β = m/sin α`
Since, cosec2 β - cot2 β = 1
⇒ `m^2/sin^2 α - n^2/tan^2 α = 1`
⇒ `m^2/sin^2 α - (n^2cos^2α )/sin^2 α = 1`
⇒ m2 - n2cos2 α = sin2 α
⇒ m2 - n2cos2 α = 1 - cos2 α
⇒ m2 - 1 = (n2 - 1)cos2 α
⇒ cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Hence proved.
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