Question

simplify.
sin(-B)tan(-B)+cos(B)

Answer 1

Since both tan and sin are odd functions:
\[
\sin(-B)\tan(-B)+\cos(B)=
\sin(B)\tan(B)+\cos(B)
\]So:
\[
=\sin(B)\frac{\sin(B)}{\cos(B)}+\cos(B)
\]Multiplying out and adding:
\[
=\frac{\sin(B)^2}{\cos(B)}+\cos(B)
\]Whic gives us:
\[
=\frac{\sin(B)^2+\cos(B)^2}{\cos(B)}
\]By our first trigonometric identity, we receive:
\[
=\frac{1}{\cos(B)}
\]
Et voilá.

Answer 2

*Which

Answer 3

thanks so much. could you help me on one other question that is bugging me as well?

Answer 4

Post it as a separate question, and sure.