Help with solving these Trig Identities... Part 2

Hello, coolacu!

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Using Double angle identities

$1.\;\;\text{Find }\tan2x\text{ when }\sec x = \sqrt{65},\;\sin x > 0$

We have: . $\tan2x \:=\:\frac{\sin2x}{\cos2x} \;=\;\frac{2\sin x\cos x}{\cos^2\!x - \sin^2\!x}$ .[1]

We are given: . $\sec x \:=\:\frac{\sqrt{65}}{1} \:=\:\frac{hyp}{adj}$

From Pythagorus, we have: . $opp \:=\:8$

Hence: . $\sin x \:=\:\frac{8}{\sqrt{65}},\;\cos x \:=\:\frac{1}{\sqrt{65}}$

Substitute into [1]: . $\tan2x \;=\;\frac{2\left(\frac{8}{\sqrt{65}}\right)\left( \frac{1}{\sqrt{65}}\right)} {\left(\frac{1}{\sqrt{65}}\right)^2 - \left(\frac{8}{\sqrt{65}}\right)^2}$ $=\;\frac{\frac{16}{65}}{\frac{1}{65} - \frac{64}{65}} \;=\;\frac{\frac{16}{65}}{\text{-}\frac{63}{65}} \;=\;-\frac{16}{63}$

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Using Pythagorean identities

$1.\;\text{Prove: }\:\frac{1+ \cot^3\!x}{1 + \cot x} + \cot x \;= \;\csc^2\!x$

The numerator is the sum of two cubes . . . factor it.

$\frac{(1+\cot x)(1 - \cot x + \cot^2\!x)}{1 + \cot x} + \cot x \;=\; (1-\cot x + \cot^2\!x) + \cot x \;=\;$ $1 + \cot^2\!x \;=\;\csc^2\!x$