Question

More trig @ParthKohli

Answer 1

\[\tan(\theta+\phi)=(2\cos \alpha \cos \beta) \div (\sin^2\alpha+\sin^2\beta)\]

Answer 2

where:\[\tan \theta = \cos(\alpha+\beta), \tan \phi = \cos(\alpha-\beta)\]

Answer 3

\[\tan(\theta + \phi) = \dfrac{\tan\theta + \tan\phi }{1 - \tan\theta \tan \phi}\]

Answer 4

Simply plug in the values you're given.

Answer 5

tan(theta+pi)=(tan(theta)+tanpi)/(1-tan(theta)(tanpi)

Answer 6

@ParthKohli how would i simplify :
\[1-\cos(\alpha+\beta)\cos(\alpha-\beta) = \sin^2\alpha+\sin^2B\]

Answer 7

Remember how\[\cos(\alpha + \beta)\cos(\alpha - \beta) = \cos^2 \alpha - \sin^2 \beta\]And so,\[1-\cos(\alpha + \beta)\cos( \alpha - \beta) = 1 - \cos^2 \alpha + \sin^2 \beta = \cdots\]

Answer 8

ohhh thanks again mate!

Answer 9

no problem :)