# Thread: express in m and n

1. ## express in m and n

If cos θ = $m^2-n^2 / m^2+n^2$ , 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.

2. Hint(I think):

sin^2(x)+cos^2(x)=1

3. Use the identity sin^2(x)+cos^2(x)=1 and the definition of tangent.
Just note that cos(x)>0 (why?)

4. Originally Posted by Mathelogician
Use the identity sin^2(x)+cos^2(x)=1 and the definition of tangent.
Just note that cos(x)>0 (why?)
emm can u explain how to use that identity.as so far i learned i didn't learned this identity,but when i fliped to the later part of the chpt,i saw it.

5. Hello, tempq1!

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

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

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

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

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

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

To apply the correct signs to our answer, we note that:

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

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

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

Go for it!

6. Originally Posted by Soroban
Hello, tempq1!

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

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

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

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

To apply the correct signs to our answer, we note that:

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

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

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

Go for it!

oh ya my teacher taught me how to do the next day already.thanks anyway.i understand how to do already.