# Thread: Help with solving these Trig Identities... Part 2

1. ## Help with solving these Trig Identities... Part 2

How do I solve:

Using Double angle identities

1. Find tan2x when secx = sq.root(65); sin x > 0

Using Pythagorean identities

1. [(1+ cot^3 x) / ( 1 + cot x)] + cot x = csc^2 x

Thank you very much for your time! =)

2. Hello, coolacu!

Using Double angle identities

$\displaystyle 1.\;\;\text{Find }\tan2x\text{ when }\sec x = \sqrt{65},\;\sin x > 0$
We have: .$\displaystyle \tan2x \:=\:\frac{\sin2x}{\cos2x} \;=\;\frac{2\sin x\cos x}{\cos^2\!x - \sin^2\!x}$ .[1]

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

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

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

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

Using Pythagorean identities

$\displaystyle 1.\;\text{Prove: }\:\frac{1+ \cot^3\!x}{1 + \cot x} + \cot x \;= \;\csc^2\!x$
The numerator is the sum of two cubes . . . factor it.

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