How do you use the sum or difference identities to find the exact value of #tan165^circ#?

1 Answer
Feb 9, 2017

#tan165^@=sqrt3-2#

Explanation:

We know that #tan(180^@-A)=-tanA#, hence to find #tan165^@#, as #tan165^@=tan(180^@-15^@)=-tan15^@#

what we need is #tan15^@#

as #tan(A-B)=(tanA-tanB)/(1+tanAtanB)#

#tan15^@=tan(45^@-30^@)#

= #(tan45^@-tan30^@)/(1+tan45^@tan30^@)#

= #(1-1/sqrt3)/(1+1xx1/sqrt3)#

= #(sqrt3-1)/(sqrt3+1)#

(muliplying numerator and denominator by #sqrt3#)

= #(sqrt3-1)/(sqrt3+1)xx(sqrt3-1)/(sqrt3-1)# - raionalizing

= #(3-2sqrt3+1)/(3-1)#

= #2-sqrt3#

Hence #tan165^@=-tan15^@=sqrt3-2#