Question

Prove that:

Answer 1

A.)

   $\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

   $\begin{aligned} \text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\ &= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\ &= \frac{1}{\cos^2 (x)} - 1 \end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

   $\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\ &= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\ &= \frac{1 - \cos^2 x}{\cos^2 (x)} \end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$. We can substitute that into the numerator

   $\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\ &= \frac{\sin^2 (x)}{\cos^2 (x)} \\ &= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\ &= \text{RHS} \end{aligned}$

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b.)

   $\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

   $\begin{aligned} \text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\ &= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\ &= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} \end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

   $\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

   $\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\ &= 1 \\ &= \text{RHS} \end{aligned}$