## anonymous one year ago -tan^2x + sec^2x=1

1. mathstudent55

Do you need to prove this identity?

2. anonymous

Yes

3. mathstudent55

Rewrite tan x and sec x in terms of sin x and cos x. Then multiply both sides by $$\cos^2 x$$

4. anonymous

Sec^2x=1/cos^2x

5. mathstudent55

yes. what about tan? what is tan in terms of sin and cos?

6. anonymous

I'm not sure😓

7. anonymous

NO

8. anonymous

U GAYY

9. triciaal

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10. triciaal

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11. mathstudent55

Right. tan x = sin x / cos x

12. mathstudent55

$$-\tan^2x + \sec^2x=1$$ $$-\dfrac{\sin^2x}{\cos^2 x} + \dfrac{1}{\cos^2x}=1$$ Now multiply both sides by $$\cos^2 x$$

13. anonymous

ok :)

14. anonymous

-sin^2x/cos^2x=cos^2x?

15. mathstudent55

No

16. mathstudent55

$$-\dfrac{\sin^2x}{\cos^2 x} + \dfrac{1}{\cos^2x}=1$$ Now we multiply both sides by $$\cos ^2 x$$ $$\cos^2 x(-\dfrac{\sin^2x}{\cos^2 x} + \dfrac{1}{\cos^2x})=1 \times \cos^2 x$$ $$-\dfrac{\sin^2x \cancel{\cos^2 x}}{\cancel{\cos^2 x}~~1} + \dfrac{\cancel{\cos^2 x}~~1}{\cancel{\cos^2x}~~1})= \cos^2 x$$ $$- \sin^2 x + 1 = \cos^2 x$$ $$1 = \sin^2 x + \cos^2 x$$ $$\sin^2 x + \cos^2 x = 1$$ The last equation is a well-known trig identity.

17. anonymous

thank you >.< ill work harder to memorize the identities