Question

If tan A = n tan B and sin A = msin B, prove that cos^2 A =
m² - 1
12
n-1​

Answer 1

★Question:-

If tan A = ntan B and sin A = msin B , then prove that cos ² A = m²-1 / n²-1.

★Answer:-

We have,

• SinA = msinB

→SinB = sinA/m.......(1)

Also,

• tanA = ntanB.....(2)

We know:

$\leadsto \underline{\boxed{\sf Tan\theta = \frac{sin\theta }{cos\theta } }}$

Putting this in equation (2),

$\displaystyle :\implies \sf \frac{sinA}{cosA} = n \frac{sinB}{cosB}$

Substituting the value of sinB from equation (1),

$\displaystyle :\implies \sf \frac{sinA}{cosA} =n\frac{(sinA/m)}{cosB} \\\\:\implies \sf cosB=\frac{n}{m} \times \cancel{sinA} \times \frac{cosA}{\cancel{sinA}}\\\\:\implies \sf cosB = \frac{n}{m} cosA.....(3)$

We have,

→sinA = msinB

Squaring on both sides,

⇒ sin²A = m²sin²B

Using the formula,

$\leadsto \underline{\boxed{\sf sin^2\theta = 1-cos^2\theta }}$

$:\implies \sf 1 - cos^2 A = m^2 (1-cos^2 B)$

Now, substituting the value of cosB from equation (3),

$\displaystyle :\implies \sf 1 - cos^2 A=m^2(1-\frac{n^2}{m^2} cos^2A)\\\\\\:\implies \sf \frac{1-cos^2A}{m^2} +\frac{n^2 cos^2A}{m^2} =1\\\\\\:\implies \sf \frac{cos^2A(n^2-1)+1}{m^2}= 1\\\\\\:\implies \sf \frac{cos^2A+1}{m^2} =\frac{1}{n^2 -1} \\\\\\:\implies \sf \frac{cos^2A+1-1}{m^2 -1} =\frac{1}{n^2 -1} \\\\\\:\implies \underline{\boxed{\sf cos^2A = \frac{m^2 -1}{n^2-1} }}$

Hence proved!

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