Question

Prove the identity:
tan x sin x + cos x = sec x

Answer 1

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

Answer 2

Express everything in terms of sin and cos:\[[(sinx)/(cosx)](sinx)+cosx=1/cosx\] go from there

Answer 3

well what do u do next? i dont know.

Answer 4

@myinaya..what u wrote looks kinda confusing

Answer 5

myin worked it out correctly

Answer 6

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

Answer 7

in teh first step what does sec x equal to?

Answer 8

the first step ?

Answer 9

yes

Answer 10

i put everything in terms of sin(x) and cos(x)
because its easy to think in terms of those

Answer 11

the whole thing is sec(x)
that is what we are trying to prove

Answer 12

myininaya provided you with all the steps

Answer 13

never mind..i get it now

Answer 14

i have anotehr one: 2tan x / (1+tan^2 x ) = sin 2x

Answer 15

i believe you can use one of your identities for the denominator

Answer 16

well not only do i believe i know

Answer 17

and then after that put it in terms of cos(x) and sin(x)

Answer 18

and then you will do some flipping and woolah

Answer 19

somce canceling will also be involved

Answer 20

some*

Answer 21

what does \[1+\tan^2(x)=?\]

Answer 22

1/ sec^2 x

Answer 23

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

Answer 24

\[\frac{\frac{2\sin(x)}{\cos(x)}}{\frac{1}{\cos^2(x)}} \cdot \frac{\cos^2(x)}{\cos^2(x)}\]

Answer 25

\[\frac{2 \frac{\sin(x)}{\cos(x)} \cdot \cos^2(x)}{1}=2\sin(x)\cos(x)=\sin(2x)\]

Answer 26

thanks