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

1. SolomonZelman

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

2. SolomonZelman

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

3. anonymous

The first one. :)

4. SolomonZelman

cool.

5. anonymous

So I'm trying to do the right side.

6. anonymous

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

7. anonymous

HI

8. anonymous

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

9. anonymous

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

10. anonymous

And then I'm kinda stuck with it.

11. anonymous

I used the reciprocal and quotient identities by the way.

12. SolomonZelman

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

13. anonymous

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

14. SolomonZelman

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

15. SolomonZelman

I just found the common denominator to each fraction.

16. anonymous

Ohhh. Thank you!

17. anonymous

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. :)

18. SolomonZelman

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

19. anonymous

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

20. SolomonZelman

21. anonymous

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

22. SolomonZelman

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

23. SolomonZelman

then the rest is even easier...

24. anonymous

Why did it become like that?

25. anonymous

The third one I mean.

26. SolomonZelman

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

27. anonymous

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

28. SolomonZelman

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

29. SolomonZelman

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

30. anonymous

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

31. SolomonZelman

yes, because: 1/cos$$^4$$x = ? cos(x)$$/$$sin(x) = ?

32. anonymous

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

33. SolomonZelman

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