(cos+1)/sin^3 = csc/(1-cos)
multiply by (1-cos)
(1+cos)(1-cos)/sin^3 = csc = 1/sin
(1-cos^2)/sin^2 = 1
sin^2/sin^2 = 1
1 = 1
(sin+cos)(tan+cot) = sec+csc
sin*tan + cos*tan + sin*cot + cos*cot = sec+csc
sin^2/cos + sin + cos + cos^2/sin = sec+csc
(sin^2+cos)/cos + (sin^2+cos^2)/sin = sec+csc
1/cos + 1/sin = sec+csc
sec+csc = sec+csc
I can't seem to prove these trig identities and would really appreciate help:
1. cosx + 1/sin^3x = cscx/1 - cosx
I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx
Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx
I don't know where to go from there.
2. (sinx + cosx)(tanx + cotx) = secx + cscx
I made this into (sinx + cosx)(sinx/cosx + cosx/sinx) = secx + cscx
FOILed: sin^2x/cosx + sinxcosx/sinx + sinxcosx/cosx + cos^2x/sinx
Did I go wrong somewhere? What do I do next?
1 answer
1 answer