Question

HELP I'LL GIVE A METAL!!

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.
cot x sec4x = cot x + 2 tan x + tan3x

Answer 1

\cot x \sec^4 x = \cot x + 2 \tan x + \tan^3 x

Answer 2

cotx+2tanx+tan3x

1tanx+2tanx+tan3x

1+2tan2x+tan4xtanx

1+tan2x+tan2x+tan4xtanx

sec2x+tan2x+tan4xtanx

sec2x+tan2x(1+tan2x)tanx

sec2x+tan2x(sec2x)tanx

sec2x(1+tan2x)tanx

sec2x(sec2x)tanx

sec4xtanx

cotxsec4x

= LEFT HAND SIDE

Answer 3

That's it?

Answer 4

i dont get some of that

Answer 5

sec^2=tna^1+1

Answer 6

cot x sec4x
= cos x 1 1
---- * --- = -------
sin x cos^4 x sinx cos^3 x

Answer 7

now i've managed to prove that RHS = same thing:-

Answer 8

changing RHS to sin and cos

= cos x + 2 sin x + sin^3 x
----- ----- -------
sin x cos x cos^3 x

= cos^4 x + 2 sin^2 x cos ^2 x + sin^4x
---------------------------------
sin x cos^3 x

= (cos^2 x + sin^2 x)(cos^2 x + sin^2 x)
--------------------------------
sin x cos^3 x

= 1 * 1
----- = LHS
sin x cos^3 x

Answer 9

Ok... Your legit.

Answer 10

Is that it?

Answer 11

yes both LHS and RHS = 1 /sin x cos^3 x so they must be equal

Answer 12

its a bit long winded and there might be an easier bit it is legit.

Answer 13

Same question, different problem.
(sin x)(tan x cos x - cot x cos x) = 1 - 2 cos2x

Answer 14

you know the identity cos^2 x + sin^2 x = 1 right?

Answer 15

I dont get GoArmy's solution . It might be right but i'm not sure...

Answer 16

similar way
convert LHS to sin and cos

= sin x( sin x cos x - cos x cos x)
---- ---
cos x sin x


= sin x( sin x - cos^2 x)
-------
sin x

= sin x( sin^2 x - cos^2 x)
----------------
sin x

= sin^2 x - cos^2 x

Answer 17

now can you finish this off?

Answer 18

use the identity sin^2x + cos^2 x = 1

Answer 19

substitute for sin^2 x