2017年10月18日 · sqrt3 Use trig identity: tan 2a = (2tan a)/(1 - tan^2 a) In this case: tan (60) = (2tan (30))/( 1 - tan^2 30) Trig table gives --> tan 30 = sqrt3/3 --> tan^2 30 = 3/9 ...

2016年9月3日 · sqrt3 tan60 = sin60/cos60 = (sqrt3/2)/(1/2) = sqrt3. How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle?

tanA+tan(60+A)+tan(60-A)=3tan3A. View Solution. Q5

2015年10月19日 · tan(60^@) = sqrt(3) Consider the equilateral triangle below: with sides of length 2 by bisecting the upper ...

Click here:point_up_2:to get an answer to your question :writing_hand:tan a tan 60a tan 60a is equal to

Prove that sin 60 o + sin 30 o sin 60 o − sin 30 o = tan 60 o + tan 45 o tan 60 o ...

Click here:point_up_2:to get an answer to your question :writing_hand:prove that displaystyle tantheta tan 60circ theta tan60circ theta

2015年10月16日 · 2 cos(30°)/cos(60°) Applying the definition: csc(30°) tan(60°) = 1/sin(30°)* sin(60°)/cos(60°) Since 60 is the double of 30 we can use the double angle formula for sin(60°): sin(60°)=sin(2*30°)=2sin(30°)cos(30°).

Solve: tan x = tan 60 o − tan 30 o 1 + tan 60 o tan 30 o. View Solution. Q4. Solve: 2 tan 57 o cot 33 o ...

Click here:point_up_2:to get an answer to your question :writing_hand:if tan thetatan60circthetatan120circtheta3 then theta