Question

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

Answer 1

`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`