Prove each identity using a t-chart.
Left hand side Right hand side
a) csc^{2} x(1 - cos^{2} x)=1
b) (cot x + tan x)/sec x = csc x
Hello, alejandro!
I'm not familiar with a t-chart.
$\displaystyle \text{(a) Prove: }\:\csc^2x(1-\cos^2x) \:=\:1$
Since $\displaystyle \csc x = \tfrac{1}{\sin x},\,\text{then: }\:\csc^2\!x = \tfrac{1}{\sin^2\!x}$
Since $\displaystyle \sin^2\!x + \cos^2\!x \:=\:1,\,\text{ then: }\:1 - \cos^2\!x \:=\:\sin^2\!x$
The left side becomes: .$\displaystyle \tfrac{1}{\sin^2\!x}\cdot\sin^2\!x \;=\;1$
$\displaystyle \text{(b) Prove: }\:\frac{\cot x + \tan x}{\sec x} \:=\:\csc x$
We have: .$\displaystyle \frac{\dfrac{\cos x}{\sin x} + \dfrac{\sin x}{\cos x}}{\dfrac{1}{\cos x}}$
Multiply by $\displaystyle \frac{\sin x\cos x}{\sin x\cos x}:$
. . $\displaystyle \frac{\sin x\cos x\left(\dfrac{\cos x}{\sin x} + \dfrac{\sin x}{\cos x}\right)}{\sin x\cos x \left(\dfrac{1}{\cos x}\right)} \;=\;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\sin x} \;=\;\frac{1}{\sin x} \;=\;\csc x $