Question

Verify the equation by substituting identities to match the right side to the left side.

cot x sec4x = cot x + 2 tan x + tan3x

Answer 1

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) ?

Answer 2

Or, \(\large\color{black}{ \displaystyle \cot x ~\sec(4x) = \cot x + 2 \tan x + \tan(3x ) }\) ?

Answer 3

The first one. :)

Answer 4

cool.

Answer 5

So I'm trying to do the right side.

Answer 6

And first I changed cot x to 1/tan x, and then cos x/sin x

Answer 7

HI

Answer 8

I'll try to write what I did first. :)

Answer 9

RIGHT SIDE: \[\frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x }\]

Answer 10

And then I'm kinda stuck with it.

Answer 11

I used the reciprocal and quotient identities by the way.

Answer 12

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\)


\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x } }\)


\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\)

Answer 13

I may sound dumb to you, but can you explain how you get that?

Answer 14

then combine the fractions and use (a+b)²=a²+2ab+b²

Answer 15

I just found the common denominator to each fraction.

Answer 16

Ohhh. Thank you!

Answer 17

I will work on it right now. So I will close this question. I'll just ask another question if I'm stuck with something. :)

Answer 18

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x + 2 \cos^2x\sin^2x + \sin ^{4}x }{ \cos ^{3}x \sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(\cos^2x+\sin^2x\right)^2 }{ \cos ^{3}x \sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(1\right)^2 }{ \cos ^{3}x \sin x} }\)

Answer 19

Please don't tell me the answer. :( I'm trying to know how to do it haha.

Answer 20

oh, sorry, go ahead.

Answer 21

\[\cot x \sec ^{4}x=\frac{ 1 }{ \cos ^{3}xsin x }\]
@SolomonZelman What's next?

Answer 22

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)

Answer 23

then the rest is even easier...

Answer 24

Why did it become like that?

Answer 25

The third one I mean.

Answer 26

i multiplyied by cos(x)/cos(x)

Answer 27

Why? Sorry if I keep on bugging you. Because what I did was I turned 1/sin x to csc x.

Answer 28

oh, don't .... and it's ok

Answer 29

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} \cdot \frac{\cos x}{\cos x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x\cdot \cos x }\cdot\frac{1\cdot \cos x }{\sin x} }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)

Answer 30

And from the third one, it will become the equation on the left side, right?

Answer 31

yes, because:

1/cos\(^4\)x = ?
cos(x)\(/\)sin(x) = ?

Answer 32

sec^4x and cot x :) Thank you so much for guiding me! I can now do the other ones.

Answer 33

Yeah.... the main trick was in finding the common denominator and noticing a²+2ab+b²=(a+b)²

you welcome, and I got a prayer to attend right now. cu