Question

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

1. cot x sec^4x = cot x + 2 tan x + tan^3x





2. (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2x





3. 1 + sec^2x sin^2x = sec^2x

Answer 1

The first one:
\[LS=cot x sec^4x \]\[=\frac{cosx}{sinx}\times \frac{1}{cos^4x}\]\[=\frac{1}{sinxcos^3x}\]
\[RS= cot x + 2 tan x + tan^3x\]\[=\frac{cosx}{sinx}+\frac{2sinx}{cosx}+\frac{sin^3x}{cos^3x}\]\[=\frac{cos^4x+2sin^2xcos^2x+cos^4x}{sinxcos^3x}\]\[=\frac{(cos^2x+sin^2x)^2}{sinxcos^3x}\]\[=...\]

Answer 2

\[ \frac{ 1 }{\sin x \cos^3x }\]

Answer 3

So, you can verify the first one. Btw, do you understand my work?

Answer 4

yes I understand it

Answer 5

Okay. So, let's move on to the next question.
\[ (sin x)(tan x cos x - cot x cos x) = 1 - 2 cos^2x\]
\[LS= (sin x)(tan x cos x - cot x cos x) \]\[=sinx(\frac{sinx}{cosx})cosx-sinx(\frac{cosx}{sinx})cosx\]Can you simplify it?

Answer 6

I need some help

Answer 7

Cancel the common factors

Answer 8

okay so sinx-cosx?

Answer 9

No.
\[sinx(\frac{sinx}{cosx})cosx=sinx(\frac{sinx}{\cancel{cosx}})\cancel{cosx}=sin^2x\]

Answer 10

Can you please explain it a little more?

Answer 11

Sure, which part?

Answer 12

the simplifying

Answer 13

Okay.
\[sinx(\frac{sinx}{cosx})cosx \]Multiply\[=\frac{(sinx)(sinx)(cosx)}{cosx}\]\[=\frac{(sin^2x)(cosx)}{cosx}\]Cancel common factor (cosx)\[=\frac{(sin^2x)\cancel{(cosx)}}{\cancel{cosx}}\]\[=sin^2x\]

Answer 14

For \(sinx(\frac{sinx}{cosx})cosx-sinx(\frac{cosx}{sinx})cosx\) , I've only shown you the first term. You need to work out (to simplify) the second term!

Answer 15

okay

Answer 16

would it be \[\cos^2x\]

Answer 17

Yes!

Answer 18

So,
\[LS= (sin x)(tan x cos x - cot x cos x)\]\[=sinx(\frac{sinx}{cosx})cosx-sinx(\frac{cosx}{sinx})cosx\]\[=sin^2x - cos^2x\]Now, rewrite \(sin^2x= 1-cos^2x\). See what you get!

Answer 19

\[\sin^2x+ \cos^2x=1\]

Answer 20

Sorry :( I meant for
\[sin^2x-cos^2x\] Replace \(sin^2x\) by \(1-cos^2x\) :S

Answer 21

im a little confused

Answer 22

Using the identity \(sin^2x = 1-cos^2x\)
\[sin^2x-cos^2x\]\[=(1-cos^2x)-cos^2x\]\[=...\]

Answer 23

1-2 cos^2x?

Answer 24

Yes!!