tan ( pi/2- theta) tan theta=1
ok.. .thanks for sharing that
thank you they're asking to verify the identity.
notice -> http://www.gradeamathhelp.com/image-files/trigonometric-identities-pythagorean.gif <--- the 1st pythagorean identity there
ahemm rather the 2nd one listed there
ohh smokes... .wait a sec... I got the wrong set
http://1.bp.blogspot.com/-57vrQkzCnVA/TxfkOuTBs6I/AAAAAAAABWU/R91jwBewUyE/s1600/cofunction+identities.png <--- there rather, we'll use the 3rd one listed there, from the trig cofunctions
is this how the answer should be Cot theta tan theta=1 cos theta / sin theta times sin theta / cos theta=1 1=1
well... .kinda.. one sec
sure
\(\bf {\color{brown}{ tan\left(\frac{\pi}{2}-\theta\right)}}=cot(\theta)\qquad {\color{olive}{ cot(\theta)}}=\cfrac{cos(\theta)}{sin(\theta)} \\ \quad \\ \quad \\ tan\left(\frac{\pi}{2}-\theta\right)tan(\theta)=1\implies {\color{brown}{ cot(\theta)}}tan(\theta)=1 \\ \quad \\ {\color{olive}{ \cfrac{\cancel{ cos(\theta) }}{\cancel{ sin(\theta) }}}} \cdot \cfrac{\cancel{ sin(\theta) }}{\cancel{ cos(\theta) }}=1\)
wht is tht ?
hmmm anything you find confusing? there's an assumption you know trig btw
if you are a beginner,then \[\tan \left( \frac{ \pi }{ 2 } - \theta \right)=\frac{ \sin \left( \frac{ \pi }{ 2}-\theta \right) }{ \cos \left( \frac{ \pi }{ 2 } -\theta \right) }\] \[=\frac{ \sin \frac{ \pi }{ 2 }\cos \theta-\cos \frac{ \pi }{ 2} \sin \theta }{ \cos \frac{ \pi }{ 2 }\cos \theta+\sin \frac{ \pi }{ 2 } \sin \theta }\] \[(\sin \frac{ \pi }{ 2 }=1,\cos \frac{ \pi }{ 2 }=0)\] \[=\frac{ \cos \theta }{ \sin \theta }\]
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