Question

1)Given tan(A+B)=1, tan(A-B)=1/7. Find tan A and tan B.
 2)Compute cosec(13pie/12). given in radians...Answer is -sqrt(6)-sqrt(2).Pls explain..

Answer 1

1) 
Tan ( A + B ) = $\frac{Tan A + Tan B}{1 - Tan A * Tan B}$ = 1

Tan ( A - B ) =  $\frac{Tan A - Tan B}{1 + Tan A * Tan B}$ = $\frac{1}{7}$

on adding Tan ( A + B )  and  Tan ( A - B ) we get 8 Tan A + 6 Tan B = 2

4 Tan A + 3 Tan B = 1

on substituting Tan A in Tan ( A + B ) we get

3 $Tan^{2} B$ + 8 Tan B - 3 = 0

then Tan B = $\frac{1}{3}$ or -3

If Tan B is considered as $\frac{1}{3}$ Tan A is $\frac{3}{4}$

If Tan B is considered as -3 , Tan A = -2

If Tan B is considered as $\frac{1}{3}$ Tan A is $\frac{3}{4}$