1. if tan #= 5/4, find sin^2 -cos^#.2.simplify (25)^ -1/2x(27)^1/3+(121)^1/2x(625)^-1/4.?

Answer 1

answer is ########################################

Answer 2

I can assit on the first problem...the second one is confusing.

Using trig principles, tangent = oposite/adjacent sides
From here you get the hypothenuse = sqrt(5^2+4^2) = sqrt(41)

so, now you know, for example, cos # = 4/sqrt(41)

After some manipulation you get,

sin^2# - cos^2# = cos^2#(tan^2-1)

= (4/sqrt(41))^2((5/4)^2-1)

= 0.219

Hope your answer compares well to this one.

Answer 3

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Answer 4

if tan = 5/4 then x=4, y = 5 and z = squ. root of 41 in 90 deg.
angle triangle

sin = x/z = 4/squ.root of 41
cos = y/z = 5/squ. root of 41

i don't quite understand the second problem.

Answer 5

you are a jerk does this answer it or you want the best answer