Question

If sinA=asinB and tanA=btanB then prove that cos^2A=a^2-1÷b^2-1

Answer 1

a = sin A / sin B

$a^{2}$ = $sin^{2}$A / $sin^{2}$B

b = tan A / tan B

$b^{2}$ = $tan^{2}$A / $tan^{2}$B

Now we compute RHS

$a^{2}$ - 1 / $b^{2}$ - 1

{[$sin^{2}$A -  $sin^{2}$B]/ [$tan^{2}$A -  $tan^{2}$A]} $tan^{2}$B/ $sin^{2}$B

On simplification,

{[$sin^{2}$A -  $sin^{2}$B]/ [$tan^{2}$A -  $tan^{2}$A]} $cos^{2}$B

Now we get the numerator and denominator to the same form.

sin(A+B)sin(A-B) ----> denominator part

sin A + sin B ------> numerator part

$cos^{2}$a$cos^{4}$B ------> numerator multiplicant

On simplification, we get

$cos^{2}$A