2015年7月6日 · tan(60°)=sqrt(3) Have a look: How do you use the ordered pairs on a unit circle to evaluate a trigonometric function of any angle?

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?

Range of Trigonometric Ratios from 0 to 90 Degrees. Question. Evaluate each of the following : tan 2 60 ∘ + 4 cos 2 45 ∘ + 3 sec 2 30 ∘ + 5 cos 2 90 ∘ cosec 30 ∘ + sec 60 ∘ − cot 2 30 ∘

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

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

Click here:point_up_2:to get an answer to your question :writing_hand:prove that tan theta tan 60otheta tan 120otheta3 tan 3 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°).

2015年8月27日 · sin(60^@) = sqrt(3)/2color(white)("XX")csc(60^@) = 2/sqrt(3) cos(60^@) = 1/2color(white)("XXXX")sec(60^@) = 2 tan(60^@) = sqrt(3)color(white)("XXX")cot(60^@)=1/sqrt(3) Use the basic trigonometric definitions and the diagram below. Note: only the left half triangle is directly relevant; both sides combine to form …

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