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anonymous
 one year ago
Verify the equation by substituting identities to match the right side to the left side.
cot x sec4x = cot x + 2 tan x + tan3x
anonymous
 one year ago
Verify the equation by substituting identities to match the right side to the left side. cot x sec4x = cot x + 2 tan x + tan3x

This Question is Closed

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Or, \(\large\color{black}{ \displaystyle \cot x ~\sec(4x) = \cot x + 2 \tan x + \tan(3x ) }\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I'm trying to do the right side.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And first I changed cot x to 1/tan x, and then cos x/sin x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll try to write what I did first. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0RIGHT SIDE: \[\frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then I'm kinda stuck with it.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I used the reciprocal and quotient identities by the way.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \cot x \sec^4x = \cot x + 2 \tan x + \tan^3x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos x }{ \sin x }+\frac{ 2 \sin x }{ \cos x }+\frac{ \sin ^{3}x }{ \cos ^{3}x } }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I may sound dumb to you, but can you explain how you get that?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1then combine the fractions and use (a+b)²=a²+2ab+b²

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I just found the common denominator to each fraction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I will work on it right now. So I will close this question. I'll just ask another question if I'm stuck with something. :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x }{ \sin x \cos^3x}+\frac{ 2 \cos^2x\sin^2x }{ \cos^3x\sin x }+\frac{ \sin ^{4}x }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \cos^4x + 2 \cos^2x\sin^2x + \sin ^{4}x }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(\cos^2x+\sin^2x\right)^2 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ \left(1\right)^2 }{ \cos ^{3}x \sin x} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Please don't tell me the answer. :( I'm trying to know how to do it haha.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, sorry, go ahead.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cot x \sec ^{4}x=\frac{ 1 }{ \cos ^{3}xsin x }\] @SolomonZelman What's next?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1then the rest is even easier...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why did it become like that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The third one I mean.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1i multiplyied by cos(x)/cos(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why? Sorry if I keep on bugging you. Because what I did was I turned 1/sin x to csc x.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, don't .... and it's ok

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x \sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x }\cdot\frac{1}{\sin x} \cdot \frac{\cos x}{\cos x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{3}x\cdot \cos x }\cdot\frac{1\cdot \cos x }{\sin x} }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x = \frac{ 1 }{ \cos ^{4}x }\cdot\frac{\cos x}{\sin x} }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And from the third one, it will become the equation on the left side, right?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, because: 1/cos\(^4\)x = ? cos(x)\(/\)sin(x) = ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sec^4x and cot x :) Thank you so much for guiding me! I can now do the other ones.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Yeah.... 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
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