Question

Verify: csc^2xsecx = secx + cscxcotx

Answer 1

can you type in the right hand side properly?

Answer 2

It is?

Answer 3

is 2xsecx all in the exponent form?

Answer 4

\[\csc^2xsecx = secx + cscxcotx\]

Answer 5

can you express in sine cosine?

Answer 6

\[(1+\cot^2)secx = \frac{ \cos^2x }{ \sin^2x } \times \frac{ 1 }{ cosx }\]

Answer 7

how about consistently including the left side?

Answer 8

cotangent as cosine/sine, cosecant as 1/sine and secant as 1/cosine

Answer 9

\[(1+\frac{ \cos^2x }{ \sin^2x })secx = secx + \frac{ \cos^2x }{ \sin^2x } \times \frac{ 1 }{ cosx }\]

Answer 10

\[\frac{ 1 }{ \cos (x) } \left( \frac{ 1 }{ \sin (x) } \right)^2= \frac{ 1 }{ \cos(x) }+\frac{ 1 }{ \sin (x) }\frac{ \cos (x) }{ \sin(x) }\]

Answer 11

How did you get that?

Answer 12

trig identity

Answer 13

I understand secx -> 1/cosx, but not the 1/sinx^2

Answer 14

csc = 1/sin -> csc^2 = (1/sin)^2

Answer 15

Oh, right. I see. Thanks!

Answer 16

And then the rest simplifies out

Answer 17

I made a booboo laughing out loud

Answer 18

\[\frac{ cosx }{ sinx^2 } = \frac{ 1 }{ cosxsinx^2 }\]

Answer 19

\[\frac{ 1}{ \cos x \sin x^2 }=\frac{ cosx^2 + \sin x^2}{ \cos x \sin x^2 }\]

Answer 20

\[\frac{ 1 }{ cosxsinx^2 } = \frac{ 1 }{ cosxsinx^2 }\]

Answer 21

now the denominator looks the same

Answer 22

And the numerator simplifies to 1 :)

Answer 23

you can cancel it out and then use pythagorean sinx^2+cos x^2 = 1
1 = 1

Answer 24

Awesome, thanks :D

Answer 25

np