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## anonymous one year ago sina-cosatanb/cosa+sinatanb = tan(a+b) Prove the identity

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1. anonymous

divide both the numerator and denominator by cosa

2. welshfella

are you sure the question is correct?

3. anonymous

Yeah, the question is correct

4. anonymous

Why divide cosa?

5. anonymous

What's the next step after that?

6. welshfella

I plugged in 2 values of a and b and didn't get the 2 sides to be equal

7. anonymous

Ohh

8. anonymous

What about this question? 2cos^23-1

9. welshfella

Yes there is a mistake in the first question.

10. anonymous

How do you solve 2cos^23-1

11. welshfella

is that supposed to be an equation? there is no variable

12. anonymous

I think you're meant to simplify it

13. welshfella

is the 3 degrees or radians?

14. anonymous

Woops I just realised I wrote the question wrong

15. anonymous

It's actually meant to be 2cos^23x-1

16. anonymous

Sorry

17. anonymous

X is the variable

18. welshfella

2 cos^2 3x = 1 so cos ^2 3x = 1/2 cos 3x = +/- 1 / sqrt2

19. Hero

Taking a look at your original question

20. Hero

The original question says to Prove The Identity, but the given trig equation is not an identity. Double check all the Variables to make sure they are correct.

21. Hero

Apparently the expression on the left side is equivalent to tan(a-b) not tan (a+b)

22. Hero

We can prove this identity: (sina-cosatanb )/(cosa+sinatanb) = tan(a - b)

23. welshfella

yes one way to do this would be to convert tan b to sin b / cos b and simplify

24. welshfella

and finally convert back to a tangent

25. welshfella

Hint: use compound angle formula for sin and cos

26. Hero

@welshfella give @Cococute a chance to respond first

27. welshfella

ok

28. anonymous

Thank you @welshfella and @Hero

29. Hero

YW

30. anonymous

I changed the tanb to sinb/cosb and got this: sina-cosaxsinb/cosb / cosa+sinaxsinb/cosb

31. anonymous

What's the next step?

32. prizzyjade

$\frac{ \sin \alpha-\cos \alpha \tan \beta}{\cos \alpha+\sin \\[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta }- \tan \beta }{ \cos^2 \alpha -\sin^2 \alpha \tan^2 \beta }$alpha \tan \beta }= \tan(\alpha+\beta)\]$\tan (\alpha+\beta)=\frac{ \sin \alpha -\cos \alpha \tan \beta}{ \cos \alpha+ \sin \alpha \tan \beta }$ $=\frac{ \sin \alpha-\cos \alpha \tan \beta }{ \cos \alpha+ \sin \alpha \tan \beta } \times \frac{ \cos \alpha-\sin \alpha \tan \beta }{ \cos \alpha-\sin \alpha \tan \beta }$ $=\frac{ \sin \alpha \cos \alpha-\sin ^{2}\alpha \tan \beta- \cos^2 \alpha \tan \beta +\cos \alpha \sin \alpha \tan^2 \beta }{ \cos^2 \alpha-\sin^2 \alpha \tan ^2 \beta}$ $=\frac{ \sin \alpha \cos \alpha (1+\tan^2 \beta) -\tan \beta (\sin^2 \alpha + \cos^2 \alpha )}{ \cos^2 \alpha - \sin^2 \alpha \tan ^2 \beta}$$=\frac{ \sin \alpha \cos \alpha(\sec^2 \beta)- \tan \beta (1) }{ \cos^2 \alpha- \sin^2 \alpha \tan^2 \beta}$

33. prizzyjade

$=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta }- \tan \beta }{ \cos^2 \alpha - \sin^2 \alpha \tan^2 \beta }$ $=\frac{\frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta} -\frac{ \sin \beta}{ \cos \beta }}{\cos^2\alpha- \frac{ \sin^ 2 \alpha \sin^2 \beta }{\cos^2 \beta } }$$=\frac{ \frac{ \sin \alpha \cos \alpha- \sin \beta \cos \beta }{ \cos^2 \beta } }{ \frac{ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin ^2 \beta }{\cos^2 \beta } }$$=\frac{ \sin \alpha \cos \alpha -\sin \beta \cos \beta }{ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta }$$=\frac{ \frac{ \tan \alpha }{ 1+\tan^2 \alpha } - \frac{ \tan \beta }{ 1+ \tan^\beta } }{ 2-\sin^2 \alpha - \sin^ 2 \beta + \sin^2 \alpha \sin^2 \beta-\sin^2 \alpha \sin^2 \beta}$

34. anonymous

work from left, sub in the addition formula for tan(A+B)

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