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July 25, 2014

July 25, 2014

Posted by **Mark** on Tuesday, March 23, 2010 at 6:12pm.

sec Q = 5

tan Q = 2sqrt6

a) cos Q

b) cotQ

c) cot(90 degrees - Q)

d) sin Q

- Math(Correction) -
**Reiny**, Tuesday, March 23, 2010 at 6:21pmOk, finally , now it makes sense

And now it is easy.

given: secQ = 5, then cosQ = 1/5

So we have a right angles triangle, where the hypotenuse is 5, the side adjacent to angle Q is 1, and the opposite to angle Q is 2√6

Notice that the second function, tanQ = 2√6 was not necessary, since we can see that from our triangle.

a) cosQ = 1/5

b) cotQ = 1/tanQ = 1/(2√6)

c) cot(90 - Q) = tan Q = 2√6

d) sinQ = 2√6/5

- Math(Correction) -
**DIKSHA**, Saturday, July 7, 2012 at 7:32amSIN(90-Q)/COSEC(90-Q)=COS(90-Q)/COSEC(90-Q)

- Math(Correction) -
**DIKSHA**, Saturday, July 7, 2012 at 7:33amSIN(90-Q)/COSEC(90-Q)+COS(90-Q)/COSEC(90-Q)=1

- Math(Correction) -
**DIKSHA**, Saturday, July 7, 2012 at 7:34amSIN(90-Q)/COSEC(90-Q)+COS(90-Q)/COSEC(90-Q)=1

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