tan^2x sin^2x
= (sin^2 x)(sin^2 x)/cos^2x
= (1-cos^2 x)(1 - cos^2 x)/cos^2 x
= (1 - 2cos^2 x + cos^4 x)/cos^2x
= 1/cos^2 x - 2 +cos^2 x
now remember that cos (2A) = 2cos^2 A - 1
which leads to cos^2 A = (cos (2A) + 1)/2
so above
= 2/(cos (2x) + 1) - 2 + (cos (2x) + 1)/2
which contains only cosines and the highest degree is 1
For the 2nd, cot^2 (x/4)
= cos^2 (x/4) / sin^2 (x/4)
use the same as above after you replace the sin^2 with (1 - cos^2 )
Rewrite tan^2x sin^2x in terms of cosine without power greater than one.
Rewrite cot^2〖x/4〗 in terms of cosine without power greater than one. Assume that x is in QI.
2 answers
tan^2x sin^2x
(sec^2x - 1)(1-cos^2x)
sec^2x - 1 - 1 + cos^2x
1/cos^2x - 2 + cos^2x
(1/cosx - cosx)^2
God luck with that. Or use your double-angle formula a couple of times.
tan^2x sin^2x
cos^2x = (1 + cos2x)/2
sin^2x = (1 - cos2x)/2
You wind up with
(cos4x - 4cos2x + 3) / 4(cos2x + 1)
Use your half-angle formulas for the other one.
(sec^2x - 1)(1-cos^2x)
sec^2x - 1 - 1 + cos^2x
1/cos^2x - 2 + cos^2x
(1/cosx - cosx)^2
God luck with that. Or use your double-angle formula a couple of times.
tan^2x sin^2x
cos^2x = (1 + cos2x)/2
sin^2x = (1 - cos2x)/2
You wind up with
(cos4x - 4cos2x + 3) / 4(cos2x + 1)
Use your half-angle formulas for the other one.