given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90°-Θ) thank you
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Originally Posted by tangerine_tomato given tanΘ = 2, use trigometric identities to find the exact value of cscΘ(90°-Θ) thank you You can observe that . But if you didn't know that, how would you go about solving this problem? Note 3 things: and Thus, Simplifying, we get: Since , we can conclude that Thus, Does this make sense? --Chris
Originally Posted by Chris L T521 You can observe that . But if you didn't know that, how would you go about solving this problem? Note 3 things: and Thus, Simplifying, we get: Since , we can conclude that Thus, Does this make sense? --Chris Two "var"s in one post...you..you've made me proud
Originally Posted by Chris L T521 You can observe that . But if you didn't know that, how would you go about solving this problem? Note 3 things: and Thus, Simplifying, we get: Since , we can conclude that Thus, Does this make sense? --Chris i dont understand that stuff in the 3rd line after the word thus is that cause of the complementary angle theorem and i dont understand anything after that but im too tired ATM to make much effort to understand it
Originally Posted by tangerine_tomato i dont understand that stuff in the 3rd line after the word thus is that cause of the complementary angle theorem and i dont understand anything after that but im too tired ATM to make much effort to understand it It all boils down to finding from . Half the solution is given above. The other half is .
Originally Posted by mr fantastic It all boils down to finding from . Half the solution is given above. The other half is . Shoot...darn negative sign... ...wait...how would we know for sure it was in the first quadrant instead of the third quadrant or vice versa? I didn't see any restriction on the angle... EDIT: Just realized what he meant...never mind...
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