Tan A = n tan Bsin A=m sinBsin B = sin A / m --- (1)tan A=sinA/cosAcos A = sin A / tan A = m sin B / n tan B = m cos B / ncos B = n cos A / m --- (2)squaring and adding (1) and (2)[sin^2 B + cos^2 B] = sin^2 A / m^2 + n^2 cos^2 A / m^21 = [1 - cos^2 A]/m^2 + n^2 cos^2 A / m^2m^2 = 1 - cos^2 A + n^2 cos^2 Acos^2 A [n^2 - 1] = [m^2 -1]cos^2 A = [m^2 -1] / [n^2 - 1].

Problem Statement:
If tanA = n and tanB = m*sinB, prove that cos^2A = (m^2-1)/(n^2-1).

Solution:

Given:
tanA = n
tanB = m*sinB

We are required to prove that cos^2A = (m^2-1)/(n^2-1).

Proof:
Step 1: Expressing tanA and tanB in terms of sin and cos:

We know that tanA = sinA/cosA and tanB = sinB/cosB.

Using these identities, we can rewrite the given equations as follows:

sinA/cosA = n
sinB/cosB = m*sinB

Step 2: Solving for sinA and sinB:

Multiplying both sides of the equation sinA/cosA = n by cosA, we get:

sinA = n*cosA

Similarly, multiplying both sides of the equation sinB/cosB = m*sinB by cosB, we get:

sinB = m*cosB*sinB

Step 3: Solving for cosA and cosB:

Using the Pythagorean identity sin^2A + cos^2A = 1, we can express cosA in terms of sinA:

cos^2A = 1 - sin^2A
cos^2A = 1 - (n*cosA)^2
cos^2A = 1 - n^2*cos^2A
cos^2A + n^2*cos^2A = 1
(1+n^2)*cos^2A = 1
cos^2A = 1/(1+n^2)

Similarly, using the Pythagorean identity sin^2B + cos^2B = 1, we can express cosB in terms of sinB:

cos^2B = 1 - sin^2B
cos^2B = 1 - (m*cosB*sinB)^2
cos^2B = 1 - m^2*cos^2B*sin^2B
cos^2B + m^2*cos^2B*sin^2B = 1
(1 + m^2*sin^2B)*cos^2B = 1
cos^2B = 1/(1 + m^2*sin^2B)

Step 4: Substituting cos^2A and cos^2B in the given equation:

Substituting the expressions for cos^2A and cos^2B from Step 3 into the given equation:

(m^2-1)/(n^2-1) = (1/(1 + m^2*sin^2B) - 1)/(1/(1+n^2) - 1)

Simplifying this expression further:

(m^2-1)/(n^2-1) = (1/(1 + m^2*sin^2B) - 1)*((1+n^2)/(1+n^2))

(m^2-1)/(n^2-1) = (1 - (1 + m^2*sin^2B))/((1 + m^2*sin^2B)*(1+n^2))