Verified answer

1.

Remember: sin^2 x + cos^2 x = 1, so sin^2 x = 1 - cos^2 x

cos^2 β - sin^2 β = cos^2 β - (1 - cos^2 β) = 2 cos^2 β - 1

2.

cos[pi/2 - x] = sin x and sin[pi/2 - x] = cos x

You can show this easily in two ways:

i) Expand by using cos(a+b) = cos a cos b - sin a sin b, and sin(a+b) = sin a cos b + cos a sin b

ii) Imagine a right-angled triangle, where the angles are x, pi/2 and pi/2 - x. Then you can easily see that cos[pi/2 - x] = sin x and sin[pi/2 - x] = cos x.

So then, cos[(pi/2)-x]/sin[(pi/2)-x] = sin x / cos x = tan x by definition.

3.

sec^2[pi/2 - x] - 1 = 1/cos^2[pi/2 - x] - 1 = 1/sin^2[x] - 1 = (1 - sin^2[x]) / sin^2[x] = cos^2[x] / sin^2[x] = cot^2[x].

4.

sin x (1 - 2cos^2 x + cos^4 x) = sin x (1 - cos^2 x)(1 - cos^2 x) = sin x (sin^2 x) (sin^2 x) = sin^5 x

5.

Remember that sec x = 1/cos x, and that ln(a/b) = ln a - ln b.

ln(sec θ) = ln(1/cos θ) = ln 1 - ln(cos θ) = -ln(cos θ) [because ln 1 = 0].

Hope this helps :)

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