Question

If tanA = ntanB &
sinA = msinB
prove:
cos^2A = m^2-1/n^2-1
please use grade 10th icse level steps & not just 11 or 12 or cbse maths

Answer 1

tan A = n tan B
tanA= n . 1/cotB
cot B = n / tanA

sin A = m sin B
sin A =m . 1/cosec B
cosec B =m/sinA

cosec^2 B - cot^2 B. =1
m^2/sin^2A. - n^2/tan^2. =1
m^2 - n^2 cos^2 / sin^A =1
m^2 - n^2 cos^2 A = sin^2 A
m^2-n^2 cos^2 A = 1-cos^2
m^2 - 1. = cos^2A(n^2 - 1)
m^2 - 1/ n^2 - 1 = cos ^2 A

Answer 2

Heya user,

Given, tan A = ntan B
=> sin A cos B = n * sin B cos A
=> m * sin B cos B = n * sin B cos A
=> m * cos B = n * cos A
=> m / n = cos A / cos B --------------> (1)

Now, cos² B = 1 - sin² B
=> cos² B = 1 - sin² A / m² = ( m² - sin² A )/ m²

Putting this in (1) -->
m² / n² = ( cos² A * m² ) / ( m² - sin² A )
=> m² - sin² A = n² cos² A -----------> [ m² gets cancelled ]
=> m² + cos² A - 1 = n² cos² A
=> m² - 1 = ( n² - 1 ) ( cos² A )
=> cos² A = [ (m² - 1) / (n² - 1) ]

Hence, proved...

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