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## anonymous 5 years ago the most general solution which satisfies tan a=1, cos a=-1/sqrt()

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1. anonymous

$\tan \theta=1, \cos \theta=-1/\sqrt{2}$

2. anonymous

$\theta=\pi/4+\pi+2n \pi={5\pi \over 4}+2n \pi; n=0,1,2,..$

3. anonymous

would you explain it a bit

4. anonymous

First, the angle at which tan is equal 1, and cos is 1/sqrt(2) is pi/4.

5. anonymous

Then we should look for the quadrant that has positive values for tan and negative values for cos, and that would be the third quadrant.

6. anonymous

So the correspondent angle for pi/4 in the third quadrant is pi/4 +pi.

7. anonymous

This angle will occur again after 2n(pi) for n=0,1,2,...

8. anonymous

Does that make sense?!

9. anonymous

$2n \pi +3\pi/4$ is it rite?

10. anonymous

Yeah, if you have both cos and tan negatives.

11. anonymous

nd id i have jus the cos -ve it wud be 5pi/4...is that correct?

12. anonymous

*if

13. anonymous

Exactly.

14. anonymous

kool;-)

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