# If ##cos(50)=a##, then how do you express ##tan(130)## in terms of a?

You can use some identities to make this easier.

##cos(50^o) = cos(-50^o) = -cos(130^o)##

due to

##cos(x) = cos(-x)##

##cos(x) = -cos(x+pi)##

Thus, you have:

##tan(130^o) = (sin(130^o))/(cos(130^o)) = (-sin(130^o))/(-cos(130^o))##

Now, as for determining ##-sin(130^o)##… Note that ##sin^2(130^o) + cos^2(130^o) = 1##. Therefore:

##sin(130^o) = sqrt(1-cos^2(130^o))##

##=> -sin(130^o) = -sqrt(1-cos^2(130^o))##

Finally, you get:

##color(blue)(tan(130^o)) = (-sin(130^o))/(-cos(130^o)) = (-sqrt(1-(cos(130^o))^2))/(-cos(130^o))##

##= (-sqrt(1-(-cos(50^o))^2))/(cos(50^o))##

##= (-sqrt(1-cos^2(50^o)))/(cos(50^o))##

##= color(blue)((-sqrt(1-a^2))/(a))##

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