Question

if tan A = n tan B and sin A = m sin B prove that cos2A = (m2-1)/(n2-1)​

Answer 1

answer :-

In this question we have to find cos²A in terms of m and n , so we have to eliminate ∠B  from the given relations.

tan A = n tan B

tan B = 1/n tan A

Cot B = n /tan A  [ cot B = 1/tan B]

sin A = m sinB

sin B = 1/m sinA

cosec B = m / sinA     [sinB = 1/cosecB]

cosec²A - cot²B =1

Substitute the value of cot B and cosec B in the above relation.

(m / sinA)² - (n /tan A)²

(m² / sin²A) - (n² /tan² A)

(m² / sin²A) - (n² /(sin²A / cos²A))

[ tan A = sinA / cosA]

(m² / sin²A) - n²cos²A / sin²A = 1

m² - n²cos²A  = sin²A

m² - n²cos²A  = 1-  cos²A

[sin²A = 1-  cos²A]

m² -1  = n²cos²A -  cos²A

m² - 1 = cos ²A(n² -1)

cos²A = m² -1/ n²-1

Answer 2

Answer:

Here is step by step solution for your question.

This is solved

Using identities:-