Question

I hope this question is quite easy one for Maths lovers...
I'll surely mark u as brainlist​

Answer 1

Given : tanA = ntanB

$\mathsf{\implies \dfrac{tanA}{tanB} = n}$

$\bigstar \ \ \textsf{We know that : \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{sinA}{cosA}}{\dfrac{sinB}{cosB}} = n}$

$\mathsf{\implies \dfrac{sinA}{cosA} \times \dfrac{cosB}{sinB} = n}$

$\mathsf{\implies \dfrac{sinA}{sinB} \times \dfrac{cosB}{cosA} = n}$

Given : sinA = m sinB

$\mathsf{\implies \dfrac{sinA}{sinB} = m}$

Substituting this value in the above equation, we get :

$\mathsf{\implies m \times \dfrac{cosB}{cosA} = n}$

$\mathsf{\implies \dfrac{cosB}{cosA} = \dfrac{n}{m}}$

$\mathsf{\implies \dfrac{cosA}{cosB} = \dfrac{m}{n}}$

$\mathsf{\implies cosA = \dfrac{m.cosB}{n}}$

Squaring on both sides, We get :

$\mathsf{\implies cos^{2}A = \dfrac{m^2.cos^2B}{n^2} \ \ \ -------[1]}$

Now we need to find the value of cos²B in terms of m and n

Consider : sinA = msinB

Squaring on both sides, we get :

$:\implies$  sin²A = m².sin²B

$\bigstar \ \ \textsf{We know that : \boxed{\mathsf{sin^2\theta = 1 - cos^2\theta}}}$

$:\implies$  1 - cos²A = m²(1 - cos²B)

$:\implies$  1 - cos²A = m² - m².cos²B

$:\implies$  m².cos²B = m² + cos²A - 1

$\mathsf{:\implies cos^2B = \dfrac{m^2 + cos^2A - 1}{m^2}}$

Substituting the value of cos²B in Equation [1], We get :

$\mathsf{\implies cos^2A = \dfrac{m^2}{n^2}\bigg(\dfrac{m^2 + cos^2A - 1}{m^2}\bigg)}}$

$\mathsf{\implies cos^2A = \dfrac{m^2 + cos^2A - 1}{n^2}}}$

$\mathsf{\implies cos^2A = \dfrac{m^2 - 1}{n^2} + \dfrac{cos^2A}{n^2}}$

$\mathsf{\implies cos^2A - \dfrac{cos^2A}{n^2} = \dfrac{m^2 - 1}{n^2}}$

$\mathsf{\implies cos^2A\bigg(1 - \dfrac{1}{n^2}\bigg) = \dfrac{m^2 - 1}{n^2}}$

$\mathsf{\implies cos^2A\bigg(\dfrac{n^2 - 1}{n^2}\bigg) = \dfrac{m^2 - 1}{n^2}}$

$\mathsf{\implies cos^2A(n^2 - 1) = (m^2 - 1)}$

$\mathsf{\implies cos^2A = \dfrac{m^2 - 1}{n^2 - 1}}$