(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I'm currently in cal 1 and we are doing trigonometry identities for review. I can't recall ever seeing a sum/difference formula for secant in precal

(1) Prove secant² (4θ) / 2 - secant² (4θ) = secant (8θ)

(2) Prove tan(x)/(1-cot(x)) + cot(x)/(1-tan(x)) = sec(x)csc(x) + 1

3. The attempt at a solution

(1) secant² (4θ) / 2 - secant² (4θ) = -1/2 *(sec²(4θ))

(2) tan(x)/(1-cot(x)) + cot(x)/(1-tan(x))

=[ tan(x)(1-tan(x)) + cot(x)(1-cot(x))] /(1-cot(x))(1-tan(x))

= [tan(x) - tan²(x) + cot(x) - cot²(x)]/ (1-tan(x)-cot(x)+cot(x)tan(x))

= sin(x)/cos(x) - sin²(x)/cos²(x) + cos(x)/sin(x) - cos²(x)/sin²(x) / 2-tan(x)-cot(x)

= [sin^{3}(x)cos(x)-sin^{4}(x) +cos^{3}(x)sin(x)-cos^{4}(x)]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

= [sin^{3}(x)cos(x) + cos^{3}(x)sin(x) - (sin^{4}(x) + cos^{4}(x))]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

= [sin(x)cos(x)[sin^{2}(x)+cos^{2}(x)] -1]/cos²(x)sin²(x)/ 2-tan(x)-cot(x)

= sin(x)cos(x) -1 / cos²(x)sin²(x)(2-sin(x)/cos(x)-cos(x)/sin(x))

= sin(x)cos(x) -1/ 2cos²(x)sin²(x) - sin^{3}(x)cos(x) -cos^{3}(x)sin(x)

= sin(x)cos(x) -1/ 2cos²(x)sin²(x) - sin(x)cos(x)(sin²(x) + cos²(x))

= sin(x)cos(x) -1 /sin(x)cos(x)(2sin(x)cos(x)-1)

= 1/2*csc(x)sec(x)

This is where I'm stuck I can't find a way for it to look like what is trying to be proven here

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# Homework Help: Trigonometry Refresher

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