Question

verify that the equations are identities:
tan^2x-sin^2x= tan^2xsin^2x

Answer 1

What is the values of tan in term of sin and cos?

Answer 2

sin/cos

Answer 3

Firstly use the formula:

\[a^2 - b^2 = (a + b)(a - b)\]

can you use it there??

Answer 4

no sorry

Answer 5

In the formula:

put a = tanx and b = sinx

Answer 6

oh ok i see now let me try it

Answer 7

\[
\tan ^2(x)-\sin ^2(x) \tan
^2(x)=\\\left(1-\sin
^2(x)\right) \tan
^2(x)=\\\frac{\sin ^2(x) \cos
^2(x)}{\cos ^2(x)}=\sin
^2(x)
\]

Answer 8

Wait there is something wrong we are doing..

Answer 9

im do not understand what eliassaab done

Answer 10

\[\frac{\sin^2x}{\cos^2x} - \sin^2x \implies \frac{\sin^2x(1 - \cos^2x)}{\cos^2x}\]

Answer 11

\[= \tan^2x .\sin^2x\]

Answer 12

\[1 - \cos^2 x = \sin^2x\]

Answer 13

\[
\tan ^2(x)-\sin ^2(x) \tan
^2(x)=\\\left(1-\sin
^2(x)\right) \tan
^2(x)=\\\frac{\sin ^2(x) \cos
^2(x)}{\cos ^2(x)}=\sin
^2(x)\\

\]

Hence \[ \tan ^2(x)-\sin ^2(x) \tan
^2(x)=\sin^2(x)\\
\tan ^2(x)-\sin^2(x)=\sin ^2(x) \tan
^2(x)
\]

Answer 14

Don't you think you are taking the long route Sir @eliassaab

Answer 15

It is a two line proof.

Answer 16

what about that negative sign sin^2x/cos^2x - (1-cos^2x)

Answer 17

\[\frac{\sin^2x}{\cos^2x} - \sin^2x \implies \frac{(\sin^2x - \sin^2xcos^2x)} {\cos^2x} \implies\frac{\sin^2x(1 - \cos^2x)}{\cos^2x}\]

Answer 18

I have factored out \(sin^2x\)..

Answer 19

@waterineyes Your method is almost the same as mine.

Answer 20

Yes that is what I am seeing...

Answer 21

\[
\frac{\sin^2x}{\cos^2x} - \sin^2x \implies\\ \frac{(\sin^2x - \sin^2xcos^2x)} {\cos^2x} \\
\implies\frac{\sin^2x(1 - \cos^2x)}{\cos^2x}

\implies \frac{\sin^2x(\sin^2(x))}{\cos^2x}=\sin^2(x) \tan^2(x)
\]

Answer 22

Did you got @onmypagrind ??

Answer 23

no i do not... I got sin^2x/cos^2x- 1-cos^2 then im stuck because of that negative sign sorry im being such a pain

Answer 24

Do not write their 1 - cos^2x leave it sin^&2x only..

See above eliassaab has done it for you...

Answer 25

I see that but I dont understand the steps when they went from sin^2x/cos^2x -sin^2x --> sin^2x-sin^2xcos^2x/cos^2x... where did that cos^2x come from in the numerator

Answer 26

THANKS GUYS I FINALLY GOT IT.... COMMON DENOMINATORS