Question

[Pre-Calc] Please Help! I don’t know where to start. How do I do this?

Answer 1

Answer:

See Below.

Step-by-step explanation:

Problem A)

We have:

$\displaystyle \csc^2\theta \tan^2\theta -1=\tan^2\theta$

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:

$\displaystyle \left(\frac{1}{\sin^2\theta}\right)\left(\frac{\sin^2\theta}{\cos^2\theta}\right)-1=\tan^2\theta$

Cancel:

$\displaystyle \frac{1}{\cos^2\theta}-1=\tan^2\theta$

Let 1/cosθ = secθ:

$\sec^2\theta -1=\tan^2\theta$

From the Pythagorean Identity, we know that tan²θ + 1 = sec²θ. Hence, sec²θ - 1 = tan²θ:

$\tan^2\theta =\tan^2\theta$

Problem B)

We have:

$\sin^3x=\sin x-\sin x \cos^2 x$

Factor out a sine:

$\sin x(\sin^2 x)=\sin x-\sin x\cos^2 x$

From the Pythagorean Identity, sin²θ + cos²θ = 1. Hence, sin²θ = 1 - cos²θ:

$\sin x(1-\cos^2 x)=\sin x-\sin x\cos^2x$

Distribute:

$\sin x- \sin x \cos^2 x=\sin x-\sin x\cos^2 x$

Problem C)

We have:

$\displaystyle \frac{\cos 2x+1}{\sin 2x}=\cot x$

Recall that cos2θ = cos²θ - sin²θ and that sin2θ = 2sinθcosθ. Hence:

$\displaystyle \frac{\cos^2 x-\sin^2 x+1}{2\sin x\cos x}=\cot x$

From the Pythagorean Identity, sin²θ + cos²θ = 1 so cos²θ = 1 - sin²θ:

$\displaystyle \frac{2\cos^2 x}{2\sin x\cos x}=\cot x$

Cancel:

$\displaystyle \frac{\cos x}{\sin x}=\cot x$

By definition:

$\cot x = \cot x$