# Thread: Some analytic trig questions

1. ## Some analytic trig questions

first is 1+tan^2x/csc^2x.. I get it down to 1/cos^2x* sin^2x/1. Answer is supposed to be tan^2(x)

also
(sec^2x+csc^2x)-(tan^2x+cot^2x)
I got this down to
(1/cos^2*sin^2x)-(tan^2x*tan^2x+1/tan^2x)
answer books says it all equals 2

2. Originally Posted by bilbobaggins
first is 1+tan^2x/csc^2x.. I get it down to 1/cos^2x* sin^2x/1. Answer is supposed to be tan^2(x)
(sigh) Please use parenthesis. You meant (1 + tan^2(x))/csc^2(x).

$\frac{1 + tan^2(x)}{csc^2(x)} = \frac{sec^2(x)}{csc^2(x)}$

Can you continue?

-Dan

3. Originally Posted by bilbobaggins
first is 1+tan^2x/csc^2x.. I get it down to 1/cos^2x* sin^2x/1. Answer is supposed to be tan^2(x)

also
(sec^2x+csc^2x)-(tan^2x+cot^2x)
I got this down to
(1/cos^2*sin^2x)-(tan^2x*tan^2x+1/tan^2x)
answer books says it all equals 2

$tan^2(x) + 1 = sec^2(x)$

and there is a similar relation for the csc and cot functions.

-Dan

4. Hello, Bilbo!

First is:. $\frac{1+\tan^2\!x}{\csc^2\!x}$
I get it down to:. $\frac{1}{\cos^2\!x}\cdot\frac{\sin^2\!x}{1}$

Answer is supposed to be: . $\tan^2\!x$

You have: . $\frac{\sin^2\!x}{\cos^2\!x}\;=\;\left(\frac{\sin x}{\cos x}\right)^2$ . . . Got it?

$(\sec^2\!x+\csc^2\!x)-(\tan^2\!x+\cot^2\!x)$

Book says it all equals $2.$

$\text{We have: }\;\underbrace{(\sec^2\!x-\tan^2\!x)}_{\text{This is 1}} + \underbrace{(\csc^2\!x - \cot^2\!x)}_{\text{This is 1}}$ . . . Okay?

5. Originally Posted by Soroban
Hello, Bilbo!

You have: . $\frac{\sin^2\!x}{\cos^2\!x}\;=\;\left(\frac{\sin x}{\cos x}\right)^2$ . . . Got it?

$\text{We have: }\;\underbrace{(\sec^2\!x-\tan^2\!x)}_{\text{This is 1}} + \underbrace{(\csc^2\!x - \cot^2\!x)}_{\text{This is 1}}$ . . . Okay?

What did you do for the second one?