How do you evaluate tan(pi/2+pi/6) . cot (pi-pi/6) . cos (pi/2-pi/6)tan(π2+π6).cot(ππ6).cos(π2π6)?

1 Answer
Aug 13, 2016

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

Explanation:

So tan(A-B)= "tanA-tanB"/"(1+tanAtanB)"tanA-tanB(1+tanAtanB)
So cot(A-B) will be the reciprocal - "(1+tanAtanB)"/"tanA-tanB"(1+tanAtanB)tanA-tanB
So A- piπ and B is -pi/6π6
So =

"(tan(2pi/32π3)" x "(cos(pi/3π3))"x ("(1+tan(pi)tan(π)tan(-pi/6))")/"tan()))/tan(pi)-tan()tan(-pi/6)")

= -sqrt33 x 1/212 x (\frac{1}{\sqrt{3}}0+1)(130+1)/0-\frac{1}{\sqrt{3}}013

Solving will give 1.5