veify that the equation is a trig identity (cos^2(x)+tan^2(x)-1)/sin^2(x) = tan^2(x)

[andie38290]

this is what I have so far:

(cos^2(x)/sin^2(x))+(tan^2(x)/sin^2(x))-(1/sin^2(x)) = tan^2(x)

cot^2(x)+((1/cot^2(x)/(1/csc^2(x))-csc^2(x) =tan^2(x)

cot^2(x)+(csc^2(x)/cot^2(x))-csc^2(x) = tan^2(x)

cot^2(x)+((1+cot^2(x)/cot^2(x)-(1+cot^2(x)) = tan^2(x)

cot^2(x)+1-1+cot^2(x) = tan^2(x)

2cot^2(x) = tan^2(x)

I can't tell if I did something wrong while working it out, or I just can't see how to get past this step.

 

[soroban]

Hello, andie38290!

\(\displaystyle \text{Prove: }\:\dfrac{\cos^2\!x + \tan^2\!x - 1}{\sin^2\!x} \:=\:\tan^2\!x\)

We have: .\(\displaystyle \displaystyle \frac{\cos^2\!x}{\sin^2\!x} + \frac{\tan^2\!x}{\sin^2\!x} - \frac{1}{\sin^2\!x}\)

. . . . . \(\displaystyle \displaystyle =\; \left(\frac{\cos x}{\sin x}\right)^2 + \frac{\frac{\sin^2\!x}{\cos^2\!x}}{\sin^2\!x} - \left(\frac{1}{\sin x}\right)^2\)

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

. . . . . \(\displaystyle =\;\cot^2\!x + \sec^2\!x - \csc^2\!x\)

. . . . . \(\displaystyle =\;\sec^2\!x - \underbrace{(\csc^2\!x - \cot^2\!x)}_{\text{This is 1}}\)
. . . . . \(\displaystyle =\;\sec^2 - 1 \)

. . . . . \(\displaystyle =\;\tan^2\!x\)

 

[andie38290]

Thank you, thank you, thank you!!!

I've been struggling with that question for over an hour!!