Question

Simplifying complex trig identities pls help

Answer 1

[Cot(y)-tan(y)]÷[sin(y)cos(y)] +1

Answer 2

rewrite cot and tan in terms of sin cos

Answer 3

Omg i messed up the plus 1 should be +sec^2(y)

Answer 4

Ok

Answer 5

Cos(y)/sin(y) - sin(y)/cos(y)÷ sin(y)cos(y) + 1/cos^2(y)

Answer 6

Sorry im typing this on my phone

Answer 7

\[\large\rm \frac{ \cot(y) -\tan(y) }{\sin(y)\cos(y) } +\sec^2(y)\] like this

Answer 8

?

Answer 9

Yea exactly

Answer 10

\[\frac{ \color{Red}{ \frac{ \cos(y) }{ \sin(y) }-\frac{ \sin(y) }{ \cos(y)} }}{ \cos(y)\sin(y) } +\frac{ 1 }{\cos^2(y) }\] first simplify the red part
find common denominator

Answer 11

Cos^2-sin^

Answer 12

Whoops so its cos^2+sin^2÷sincos?

Answer 13

negative sign should be there..
let me think something easy ..

Answer 14

yes that's correct
now you can change division to multiplication \[\frac{ \frac{ \cos^2(y) -\sin^2(y) }{ \sin(y)\cos(y) } }{ \sin(y)\cos(y) } +\frac{1}{\cos^2(y)}\]

Answer 15

\[\frac{ \frac{ \cos^2-\sin^2 }{ \cos(y)\sin(y) } }{ \color{red}{\frac{\sin(y)\cos(y) }{1}}}+\frac{1}{\cos^2(y)}\] is same as \[\frac{ \cos^2(y)-\sin^2 (y)}{ \cos(y)\sin(y) } \cdot \frac{1}{\color{Red}{\sin(y)\cos(y)}} + \frac{1}{\cos^2(y)}\]
multiply the top fraction with the reciprocal of the bottom fraction (to change division to multiplication )

Answer 16

multiply the denominators you will get \[\frac{ \cos^2(y)-\sin^2(y) }{ \sin^2(y)\cos^2(y) } +\frac{1}{\cos^2(y)}\]
now once again find the common denominator and then simplify