A.)
![\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)](/tpl/images/0145/0613/3e5df.png)
Use the identities
![\csc x = 1 / \sin x](/tpl/images/0145/0613/458c7.png)
and
![\tan x = \sin x / \cos x](/tpl/images/0145/0613/f65e1.png)
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}](/tpl/images/0145/0613/ca179.png)
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}](/tpl/images/0145/0613/76cda.png)
Use Pythagorean identity for the numerator.
If
![\sin^2 (x) + \cos^2(x) = 1](/tpl/images/0145/0613/99c7e.png)
then subtracting both sides by
![\cos^2 (x)](/tpl/images/0145/0613/b3afe.png)
yields
![\sin^2(x) = 1 - \cos^2(x)](/tpl/images/0145/0613/29e8b.png)
. 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}](/tpl/images/0145/0613/399ad.png)
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b.)
![\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1](/tpl/images/0145/0613/47fed.png)
For the left-hand side:
By definition,
![\sec(x) = 1/\cos(x)](/tpl/images/0145/0613/9da1c.png)
and
![\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}](/tpl/images/0145/0613/5f5fb.png)
Since
![\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}](/tpl/images/0145/0613/a88ea.png)
Using Pythagorean identity,
![\cos^2(x) = 1 - \sin^2(x)](/tpl/images/0145/0613/f99c1.png)
so
![\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\ &= 1 \\ &= \text{RHS} \end{aligned}](/tpl/images/0145/0613/bfa1c.png)