Question

tanA = ntanB, SinA =msinB prove that cos^2 A = m^2 -1 / n^2-1

Answer 1

Solution -

Given:

SinA = mSinB 

m/SinA = 1/SinB

As 1/Sinθ = Cosecθ

Therefore, CosecB = m/SinA --------------1

Similarly,

TanA = nTanB

1/TanB = n/TanA

As Cotθ = 1/Tanθ

Therefore,  CotB = n/TanA  ---------------2

We know that,

Cosec^2θ - Cot^2θ = 1

Hence, Cosec^2B - Cot^2B = 1

Substitute the value of CosecB and CotB from equation 1 and equation 2

(m/SinA)^2 - (n/TanA)^2 = 1

m^2/Sin^2A - n^2 Cos^2A/Sin^2A = 1    (As Tan^2A = Sin^2A/Cos^2A)

m^2 - n^2 Cos^2A = Sin^2A

m^2 - n^2 Cos^2A = 1 - Cos^2A          (Sin^2A = 1 - Cos^2A)

n^2Cos^2A - Cos^2A = m^2 - 1

Cos^2A (n^2 - 1) = m^2 - 1

Cos^2A = (m^2 - 1) / (n^2 - 1)

Hence proved.

Answer 2

Tan A = n Tan B  ---(1)
Sin A = m Sin B  --- (2)
(2) ÷ (1) gives:  CosA = (m/n) CosB    --(3)

From (2)
    Cos² A = 1 - m² Sin²B
                 = 1 - m² [1 - cos²B]
                 = 1 - m² + m² Cos²B
                 = 1 - m² + n² Cos²A      using (3)

=> (n²-1) Cos²A = m² -1 
=>  Cos²A = (m² - 1) / (n² - 1)