## trig: solving tanA=kcotB then cos(A+B)/cos(A-B)

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### trig: solving tanA=kcotB then cos(A+B)/cos(A-B)

tanA=kcotB then cos(A+B)/cos(A-B)
1.1+k/1-k 2.1-k/1+k 3.k+1/k-1 4.k-1/k+1
yamini

Posts: 1
Joined: Mon Aug 10, 2009 1:25 pm

A good start might be to notice that cotangent is the reciprocal of the tangent:

. . . . .$\cot(B)\, =\, \frac{1}{\tan(B)}$

See if this helps with the application of angle-sum identities....

stapel_eliz

Posts: 1797
Joined: Mon Dec 08, 2008 4:22 pm