Question

Ok here it goes

Answer 1

\[\tan x \sin x+ \cos x= \sec x\]

Answer 2

how can you rewrite tanx = ?

Answer 3

so tanx = sin x / cos x

yes ?

Answer 4

\[\frac{ \sin x }{ \cos x }\]

Answer 5

than substitute in place of tanx what will get ?

Answer 6

cos x + cos x

Answer 7

no

Answer 8

check it please

Answer 9

\[\frac{ \sin x }{ \cos x } \sin x + \cos x\]

Answer 10

alryt looks like you guys gave up on me

Answer 11

yes but do you can continue it ?

Answer 12

Should you cancel something here?

Answer 13

multiplie sin x *sin x = ?

Answer 14

so than what you get ?

Answer 15

will be sin^2 x / cos x

y

Answer 16

yes ?

Answer 17

Wait how was that?

Answer 18

so

tanxsinx +cosx = secx

because tanx = sinx / cos x

sinx
---- *sinx +cosx = secx
cosx

sin^2 x
------ + cos x = sec x
cos x

yes ?

Answer 19

\[\large tan(x)sin(x) + cos(x) = sec(x)\]

we know \(\large tan(x) = \frac{sin(x)}{cos(x)}\) so

\[\large \frac{sin(x)}{cos(x)}\times sin(x) + cos(x) = sec(x)\]

Multiply the sin/cos times sin

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

we need a common denominator to add those fractions...looks like cos(x) would be the best...so multiply the cos(x) by cos(x)/cos(x)

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

we can now place the fraations over the commondenominator

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

We know that \(\large sin^2(x) + cos^2(x) = 1 \) so

\[\large \frac{1}{cos(x)} = sec(x)\] which is true...so the identity is verified

Answer 20

oh thx