Answer:
See Below.
Step-by-step explanation:
Problem A)
We have:
[tex]\displaystyle \csc^2\theta \tan^2\theta -1=\tan^2\theta[/tex]
When in doubt, convert all reciprocal trig functions and tangent into terms of sine and cosine.
So, let cscθ = 1/sinθ and tanθ = sinθ/cosθ. Hence:
[tex]\displaystyle \left(\frac{1}{\sin^2\theta}\right)\left(\frac{\sin^2\theta}{\cos^2\theta}\right)-1=\tan^2\theta[/tex]
Cancel:
[tex]\displaystyle \frac{1}{\cos^2\theta}-1=\tan^2\theta[/tex]
Let 1/cosθ = secθ:
[tex]\sec^2\theta -1=\tan^2\theta[/tex]
From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:
[tex]\tan^2\theta =\tan^2\theta[/tex]
Problem B)
We have:
[tex]\sin^3x=\sin x-\sin x \cos^2 x[/tex]
Factor out a sine:
[tex]\sin x(\sin^2 x)=\sin x-\sin x\cos^2 x[/tex]
From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:
[tex]\sin x(1-\cos^2 x)=\sin x-\sin x\cos^2x[/tex]
Distribute:
[tex]\sin x- \sin x \cos^2 x=\sin x-\sin x\cos^2 x[/tex]
Problem C)
We have:
[tex]\displaystyle \frac{\cos 2x+1}{\sin 2x}=\cot x[/tex]
Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:
[tex]\displaystyle \frac{\cos^2 x-\sin^2 x+1}{2\sin x\cos x}=\cot x[/tex]
From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:
[tex]\displaystyle \frac{2\cos^2 x}{2\sin x\cos x}=\cot x[/tex]
Cancel:
[tex]\displaystyle \frac{\cos x}{\sin x}=\cot x[/tex]
By definition:
[tex]\cot x = \cot x[/tex]
