Question

Tan (A-B) + tan B/ 1- tan( A-B ) . tan B = tan A​

Answer 1

Answer:

$let \: \alpha - \beta = \gamma \\ \frac{ \tan( \gamma ) + \tan( \beta ) }{1 - \tan( \gamma ) \tan( \beta ) } = \tan( \gamma + \beta ) \\ = \tan( \alpha - \beta + \beta ) \\ = \tan( \alpha )$

Answer 2

Simplify:

Step-by-step explanation:

$\ Given \ value:\\\\\tan (A-B) + tan B/ 1- tan( A-B ) .\tan B = \tan A\\\\\ Solution:\\\\\ formula:\\\\\tan(A-B)= \frac{\tan A-\tan B}{1+\tan A \tan B}\\\\\ put\ the \ value \ in\ given \ equation \\\\\ Equation:\\\\\tan (A-B) + tan B/ 1- tan( A-B ) .\tan B = \tan A\\\\\ L.H.S \\\\\rightarrow \frac{\frac{\tan A-\tan B}{1+\tan A \tan B} + \tan B}{1- \frac{\tan A-\tan B}{1+\tan A \tan B}\times \tan B}$

$\rightarrow \frac{\frac{\tan A-\tan B+ \tan B(1+\tan A \tan B)}{1+\tan A \tan B}} { \frac{(1+\tan A \tan B)-\tan A \tan B-\tan^2 B)}{1+\tan A \tan B}}\\\\\rightarrow \frac{\frac{\tan A-\tan B+ \tan B+\tan A \tan^2 B}{1+\tan A \tan B}} { \frac{1-\tan^2 B}{1+\tan A \tan B}}\\\\\rightarrow \frac{\tan A+\tan A \tan^2 B}{1+\tan A \tan B} \times \frac {1+\tan A \tan B} {1-\tan^2 B}\\\\\rightarrow \frac{\tan A(1+\tan^2 B)}{1+\tan A \tan B} \times \frac {1+\tan A \tan B} {1-\tan^2 B}\\\\\rightarrow \tan A$

$\ L.H.S= R.H.S$

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