If cos θ =$\displaystyle m^2-n^2 / m^2+n^2$ , where m>n>0,and cosec θ <0 , express sin θ and tan θ in terms of m and n.

i don't really know how to even start it.

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- Jul 20th 2010, 04:39 AMtempq1express in m and n
If cos θ =$\displaystyle m^2-n^2 / m^2+n^2$ , where m>n>0,and cosec θ <0 , express sin θ and tan θ in terms of m and n.

i don't really know how to even start it. - Jul 20th 2010, 04:50 AMAlso sprach Zarathustra
Hint(I think):

sin^2(x)+cos^2(x)=1 - Jul 20th 2010, 04:54 AMMathelogician
Use the identity sin^2(x)+cos^2(x)=1 and the definition of tangent.

Just note that cos(x)>0 (why?) - Jul 20th 2010, 05:29 AMtempq1
- Jul 20th 2010, 05:46 AMSoroban
Hello, tempq1!

Quote:

$\displaystyle \text{If }\:\cos\theta \,=\, \dfrac{m^2-n^2}{m^2+n^2}\:\text{ where } m>n>0\:\text{ and }\:\csc\theta < 0$

. . $\displaystyle \text{express }\sin\theta\text{ and }\tan\theta\text{ in terms of }m\text{ and }n.$

$\displaystyle \text}We are given: }\;\cos\theta \:=\:\dfrac{m^2-n^2}{m^2+n^2} \;=\;\dfrac{adj}{hyp}$

$\displaystyle \cs\theta$ is in a right triangle with: .$\displaystyle adj \,=\, m^2-n^2,\;hyp \,=\, m^2+n^2$

Using Pythagorus, we find that: .$\displaystyle opp \,=\,\pm 2mn$

And we can now express $\displaystyle \sin\theta$ and $\displaystyle \tan\theta.$

To apply the correctto our answer, we note that:*signs*

. . $\displaystyle \cos\theta$ is positive: $\displaystyle \cs\theta$ is in Quadrant 1 or 4.

. . $\displaystyle \csc\theta$ is negative: $\displaystyle \cs\theta$ is in Quadrant 3 or 4.

. . . . Hence, $\displaystyle \cs\theta$ is in Quadrant 4.

*Go for it!*

- Jul 22nd 2010, 09:15 PMtempq1