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anonymous
 one year ago
sinacosatanb/cosa+sinatanb = tan(a+b)
Prove the identity
anonymous
 one year ago
sinacosatanb/cosa+sinatanb = tan(a+b) Prove the identity

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0divide both the numerator and denominator by cosa

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1are you sure the question is correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, the question is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the next step after that?

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1I plugged in 2 values of a and b and didn't get the 2 sides to be equal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What about this question? 2cos^231

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Yes there is a mistake in the first question.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do you solve 2cos^231

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1is that supposed to be an equation? there is no variable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think you're meant to simplify it

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1is the 3 degrees or radians?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Woops I just realised I wrote the question wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's actually meant to be 2cos^23x1

welshfella
 one year ago
Best ResponseYou've already chosen the best response.12 cos^2 3x = 1 so cos ^2 3x = 1/2 cos 3x = +/ 1 / sqrt2

Hero
 one year ago
Best ResponseYou've already chosen the best response.0Taking a look at your original question

Hero
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

Hero
 one year ago
Best ResponseYou've already chosen the best response.0Apparently the expression on the left side is equivalent to tan(ab) not tan (a+b)

Hero
 one year ago
Best ResponseYou've already chosen the best response.0We can prove this identity: (sinacosatanb )/(cosa+sinatanb) = tan(a  b)

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yes one way to do this would be to convert tan b to sin b / cos b and simplify

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1and finally convert back to a tangent

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Hint: use compound angle formula for sin and cos

Hero
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella give @Cococute a chance to respond first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you @welshfella and @Hero

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I changed the tanb to sinb/cosb and got this: sinacosaxsinb/cosb / cosa+sinaxsinb/cosb

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What's the next step?

prizzyjade
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sin \alpha\cos \alpha \tan \beta}{\cos \alpha+\sin \\[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta } \tan \beta }{ \cos^2 \alpha \sin^2 \alpha \tan^2 \beta }\]alpha \tan \beta }= \tan(\alpha+\beta)\]\[\tan (\alpha+\beta)=\frac{ \sin \alpha \cos \alpha \tan \beta}{ \cos \alpha+ \sin \alpha \tan \beta }\] \[=\frac{ \sin \alpha\cos \alpha \tan \beta }{ \cos \alpha+ \sin \alpha \tan \beta } \times \frac{ \cos \alpha\sin \alpha \tan \beta }{ \cos \alpha\sin \alpha \tan \beta }\] \[=\frac{ \sin \alpha \cos \alpha\sin ^{2}\alpha \tan \beta \cos^2 \alpha \tan \beta +\cos \alpha \sin \alpha \tan^2 \beta }{ \cos^2 \alpha\sin^2 \alpha \tan ^2 \beta}\] \[=\frac{ \sin \alpha \cos \alpha (1+\tan^2 \beta) \tan \beta (\sin^2 \alpha + \cos^2 \alpha )}{ \cos^2 \alpha  \sin^2 \alpha \tan ^2 \beta}\]\[=\frac{ \sin \alpha \cos \alpha(\sec^2 \beta) \tan \beta (1) }{ \cos^2 \alpha \sin^2 \alpha \tan^2 \beta}\]

prizzyjade
 one year ago
Best ResponseYou've already chosen the best response.0\[=\frac{ \frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta } \tan \beta }{ \cos^2 \alpha  \sin^2 \alpha \tan^2 \beta }\] \[=\frac{\frac{ \sin \alpha \cos \alpha }{ \cos^2 \beta} \frac{ \sin \beta}{ \cos \beta }}{\cos^2\alpha \frac{ \sin^ 2 \alpha \sin^2 \beta }{\cos^2 \beta } }\]\[=\frac{ \frac{ \sin \alpha \cos \alpha \sin \beta \cos \beta }{ \cos^2 \beta } }{ \frac{ \cos^2 \alpha \cos^2 \beta  \sin^2 \alpha \sin ^2 \beta }{\cos^2 \beta } }\]\[=\frac{ \sin \alpha \cos \alpha \sin \beta \cos \beta }{ \cos^2 \alpha \cos^2 \beta  \sin^2 \alpha \sin^2 \beta }\]\[=\frac{ \frac{ \tan \alpha }{ 1+\tan^2 \alpha }  \frac{ \tan \beta }{ 1+ \tan^\beta } }{ 2\sin^2 \alpha  \sin^ 2 \beta + \sin^2 \alpha \sin^2 \beta\sin^2 \alpha \sin^2 \beta}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0work from left, sub in the addition formula for tan(A+B)
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