If-sin-2A-sin-2B-Prove-that-tan-A-B-tan-A-B-1-1-

Question Number 74966 by ajfour last updated on 05/Dec/19

I f \sin 2 A = λ \sin 2 B

P r o v e t h a t \frac{\tan (A + B)}{\tan (A - B)} = \frac{λ + 1}{λ - 1} .

Commented by Prithwish sen last updated on 05/Dec/19

\frac{\sin 2 A}{\sin 2 B} = λ

\frac{\sin 2 A + \sin 2 B}{\sin 2 A - \sin 2 B} = \frac{λ + 1}{λ - 1}

\frac{\sin (A + B) \cos (A - B)}{\cos (A + B) \sin (A - B)} = \frac{λ + 1}{λ - 1}

\frac{\tan (A + B)}{tam (A - B)} = \frac{λ - 1}{λ + 1} proved .

Commented by ajfour last updated on 05/Dec/19

T h a n k s .

Answered by ajfour last updated on 05/Dec/19

\sin [(A + B) + (A - B)] = λ \sin [A + B - (A - B)]

\Rightarrow

\sin (A + B) \cos (A - B) + \cos (A + B) \sin (A - B)

- λ {\sin (A + B) \cos (A - B) - \cos (A + B) \sin (A - B) = 0

d i v i d i n g b y \cos (A + B) \cos (A - B)

\tan (A + B) + \tan (A - B)

= λ {\tan (A + B) - \tan (A - B)

\Rightarrow \frac{\tan (A + B)}{\tan (A - B)} = \frac{λ + 1}{λ - 1} .

Commented by Prithwish sen last updated on 05/Dec/19

Wow!