How do you write the expression #(tan140^circ-tan60^circ)/(1+tan140^circtan60^circ)# as the sine, cosine, or tangent of an angle?

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Answer 1 · 2017-07-21T15:28:48

#(tan140^circ-tan60^circ)/(1+tan140^circtan60^circ)=tan80^@#

#(tan140^circ-tan60^circ)/(1+tan140^circtan60^circ)# is of the form #(tanA-tanB)/(1+tanAtanB)#.

This is the expansion of #tan(A-B)#.

Subbing #140# for #A# and #60# for #B# shows that the original expression was #tan(140-60)=tan80#

Answer 2 · 2017-07-21T15:28:50

The expression is #=tan80^@#

We need

#tan(a-b)=sin(a-b)/cos(a-b)=(sinacosb-sinbcosa)/(cosacosb+sinasinb)#

Dividing by #cosacosb#

#tan(a-b)=(tana-tanb)/(1+tanatanb)#

Here,

#a=140^@#

#b=60^@#

Therefore,

#(tan140^@-tan60^@)/(1-tan140^@tan60^@)=tan(140^@-60^@)=tan80^@#

#tan80^@=sin80^@/cos80^@#