sec^2xcotx-cotx=tanx (1/cos)^2 times (1/tan)-(1/tan)=tan (1/cos^2) times (-2/tan)=tan (-2/cos^2tan)times tan=tan(tan) sq. root of (-2/cos^2)= sq. root of (tan^2) sq. root of (2i)/cos=tan I'm not sure if I did this right. If I
Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C. I tried drawing perpendiculars and stuff but it doesn't seem to work? For me, the trig identities don't seem to plug in
4tan^2u-1=tan^2u 3tan^2u=1 tan^2u=1/3 sq. root of (tan^2u)=sq. root of (1/3) tanu=sq. root of (3)/3 tanu=sq. root of 1 Did I do this one right? If not, could you please show me the correct method?? Thanks!
Determine all triangles ABC for which tan(A-B)+tan(B-C)+tan(C-A)=0. There's a hint: "Can you relate A-B to B-C and C-A?" Should I apply the tangent difference formula (tan(x-y))? Help would be appreciated, thanks.
h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d (def of tan theta = a^-1 o)a = o opposite = adjacent tan
Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1 - tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1 - tan theta)^2]
Two radar stations A and B, are separated by the distance a = 500 m, track the plane C by recording the angles α and β at 1-second intervals. If three successive readings are t (s) 9 10 11 α 54.80◦ 54.06◦ 53.34◦ β
Use an Addition or Subtraction Formula to write the expression as a trigonometric function of one number. (tan(74°)-tan(14°))/(1+tan(74°)tan(14°)) =__________ Find it's exact value (not decimal form) =__________ Note: A