Question

Pythagorean Identities Help Please

Answer 1

\[2-\tan \Theta \div2\csc \Theta-\sec\]

Answer 2

Also I don't really understand the concepts of the identities and how to use them so if someone could explain that, I would love that.

Answer 3

\[\large \bf 2-\frac{\tan \theta}{2 cosec \theta}-\sec \theta\]
we know that,
\[\large \bf cosec \theta=\frac{1}{\sin \theta}~~and~~\sec \theta = \frac{1}{\cos \theta}\]
\[\large \bf 2-\frac{\frac{\sin \theta}{\cos \theta}}{\frac{2}{\sin \theta}}-\frac{1}{\cos \theta}\]

Answer 4

now,simplify the expression.

Answer 5

Actually the 2 is included with the numerator and the sec is included with the denominator, sorry for not being clear about that.

Answer 6

\[(2-\tan \Theta)/2\csc \Theta-\sec \Theta)\]

Answer 7

\[\large \bf \tan \theta=\frac{\sin \theta}{\cos \theta}\]
\[\large \bf cosec \theta=\frac{1}{\sin \theta}\]
\[\large \bf \sec \theta=\frac{1}{\cos \theta}\]
just plug the values in your equation

Answer 8

ok will try

Answer 9

sure

Answer 10

so i got
\[2-(\sin/\cos \Theta)\div2(1/\cos \Theta)-(1/\cos \Theta)\]

Answer 11

do you multiply by sincos/sincos now?

Answer 12

\[\large \bf \frac{2-\frac{\sin \theta}{\cos \theta}}{\frac{2}{\sin \theta}-\frac{1}{\cos \theta}}\]

Answer 13

so do you multiply by the common denominator at this point?

Answer 14

nope ! i just plug the values in the equation

Answer 15

now,second step is to take LCM

Answer 16

sin*cos?

Answer 17

correct

Answer 18

\[\large \bf \frac{\frac{2 \cos \theta -\sin \theta}{\cos \theta}}{\frac{2\cos \theta-\sin \theta}{\sin \theta \cos \theta}}\]

Answer 19

so you multiplied sin*cos/sin*cos?

Answer 20

i took LCM

Answer 21

***so you multiplied sin*cos/sin*cos?***

in the denominator, you multiply the first term 2/sin by cos/cos to get 2 cos/(sin cos)
multiply the 2nd term -1/cos by sin/sin to get -sin/(sin cos)
now that both have the same bottoms you can combine the tops to get
(2 cos - sin)/(sin cos)

to finish you problem, replace the division of fraction a/b by *multplying* by b/a
i.e. "flip" the bottom fraction and multiply