a.)
[tex]\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)[/tex]
Use the identities [tex]\csc x = 1 / \sin x[/tex] and [tex]\tan x = \sin x / \cos x[/tex] on the left-hand side
[tex]\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}[/tex]
Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x
[tex]\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}[/tex]
Use Pythagorean identity for the numerator.
If [tex]\sin^2 (x) + \cos^2(x) = 1[/tex] then subtracting both sides by [tex]\cos^2 (x)[/tex] yields [tex]\sin^2(x) = 1 - \cos^2(x)[/tex]. We can substitute that into the numerator
[tex]\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}[/tex]
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b.)
[tex]\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1[/tex]
For the left-hand side:
By definition, [tex]\sec(x) = 1/\cos(x)[/tex] and [tex]\tan (x) = 1/\cot (x)[/tex]
[tex]\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}[/tex]
Since [tex]\cot (x) = \cos (x) / \sin (x)[/tex]
[tex]\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}[/tex]
Using Pythagorean identity, [tex]\cos^2(x) = 1 - \sin^2(x)[/tex] so
[tex]\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}[/tex]
