Question

How do you evaluate `tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)`?

Answer 1

Using identities for tan(A+B) and cos(A+B)

Explanation:

So tan(A-B)= `"tanA-tanB"/"(1+tanAtanB)"`
So cot(A-B) will be the reciprocal - `"(1+tanAtanB)"/"tanA-tanB"`
So A- `pi` and B is `-pi/6`
So =

"(tan(`2pi/3`)" x "(cos(`pi/3`))"x ("(1+tan(`pi)tan(`-pi/6`))")/"tan(`pi`)-tan(`-pi/6`)"`

= `-sqrt3` x `1/2` x `(\frac{1}{\sqrt{3}}0+1)`/`0-\frac{1}{\sqrt{3}}`

Solving will give 1.5