given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90°-Θ)

thank you :)

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- Jun 13th 2008, 10:52 PMtangerine_tomatotrig
given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90°-Θ)

thank you :) - Jun 13th 2008, 11:28 PMChris L T521
You can observe that $\displaystyle \csc(90^o-\vartheta)=\sec(\vartheta)$. But if you didn't know that, how would you go about solving this problem?

Note 3 things:

$\displaystyle \sec(\varphi)=\frac{1}{\cos(\varphi)}$

$\displaystyle \csc(\varphi)=\frac{1}{\sin(\varphi)}$ and $\displaystyle \sin(\varphi-\vartheta)=\sin(\varphi)\cos(\vartheta)-\cos(\varphi)\sin(\vartheta)$

Thus,

$\displaystyle \csc(90^o-\vartheta)=\frac{1}{\sin(90^o-\vartheta)}=\frac{1}{\sin(90^o)\cos(\vartheta)-\cos(90^o)\sin(\vartheta)}$

Simplifying, we get:

$\displaystyle \csc(90^o-\vartheta)=\frac{1}{\cos(\vartheta)}=\sec(\varthet a)$

Since $\displaystyle \tan(\vartheta)=2$, we can conclude that $\displaystyle \cos(\vartheta)=\frac{1}{\sqrt{5}}$

Thus, $\displaystyle \csc(90^o-\vartheta)=\sec(\vartheta)=\color{red}\boxed{\sqrt {5}}$

Does this make sense? :D

--Chris - Jun 13th 2008, 11:30 PMMathstud28
- Jun 14th 2008, 02:57 AMtangerine_tomato
- Jun 14th 2008, 03:22 AMmr fantastic
- Jun 14th 2008, 10:08 AMChris L T521