Question

sina-cosatanb/cosa+sinatanb = tan(a+b)

Prove the identity

Answer 1

divide both the numerator and denominator by cosa

Answer 2

are you sure the question is correct?

Answer 3

Yeah, the question is correct

Answer 4

Why divide cosa?

Answer 5

What's the next step after that?

Answer 6

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

Answer 7

Ohh

Answer 8

What about this question?
2cos^23-1

Answer 9

Yes there is a mistake in the first question.

Answer 10

How do you solve 2cos^23-1

Answer 11

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

Answer 12

I think you're meant to simplify it

Answer 13

is the 3 degrees or radians?

Answer 14

Woops I just realised I wrote the question wrong

Answer 15

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

Answer 16

Sorry

Answer 17

X is the variable

Answer 18

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

Answer 19

Taking a look at your original question

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

and finally convert back to a tangent

Answer 25

Hint: use compound angle formula for sin and cos

Answer 26

@welshfella give @Cococute a chance to respond first

Answer 27

ok

Answer 28

Thank you @welshfella and @Hero

Answer 29

YW

Answer 30

I changed the tanb to sinb/cosb and got this:

sina-cosaxsinb/cosb / cosa+sinaxsinb/cosb

Answer 31

What's the next step?

Answer 32

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

Answer 33

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

Answer 34

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