Question

$\sin( \alpha + \beta) =4 \div 5 and \sin( \alpha - \beta) = 5 \div 13 what is the result of \tan(2 \alpha )$


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Answer 1

Answer:

tan2α = 63 / 16

Step-by-step explanation:

1) plzz refer the attachment

2) First we find value of Cos( α + β )

and Cos(α - β ) , by using a formula

Cos²θ = 1 - Sin²θ

3) Now we find values of tan(α + β ) and

tan ( α - β ) , by using a formula given below

tanθ = Sinθ / Cosθ

4) Then we apply a formula,

tan( θ₁ + θ₂ ) = ( tanθ₁ + tanθ₂) / ( 1 - tanθ₁ tanθ₂ )

and we get value of tan2α

Additional information--->

1) tan ( x - y ) = ( tanx - tany ) /( 1 + tanx tany )

2) tan2x = 2tanx / ( 1 - tan²x )

3) Sin2x = 2tanx / ( 1 - tan²x )

4) Cos2x = ( 1 - tan²x ) / ( 1 + tan²x )

Answer 2

Answer:

$tan 2\alpha = \frac{63}{16}$

Step-by-step explanation:

$sin(\alpha + \beta ) = \frac{4}{5}$

$sin(\alpha - \beta ) = \frac{5}{13}$

$cos(\alpha + \beta ) = \sqrt{1 - sin^{2} (\alpha + \beta ) }$

$= \sqrt{1 - \frac{16}{25} }$

$= \sqrt{\frac{25-16}{25} }$

$= \sqrt{\frac{9}{25} } = \frac{3}{5}$

Similarly, we can find $cos (\alpha - \beta) = \frac{12}{13}$

Now,

$tan (\alpha + \beta) = \frac{\frac{4}{5} }{\frac{3}{5} } = \frac{4}{3}$

$tan (\alpha - \beta) = \frac{\frac{5}{13} }{\frac{12}{13} } = \frac{5}{12}$

$tan 2\alpha = tan[(\alpha + \beta) + (\alpha - \beta)]$

$= \frac{tan (\alpha + \beta) + tan (\alpha - \beta)}{1 - tan (\alpha + \beta) tan (\alpha - \beta)}$

$= \frac{\frac{4}{3} + \frac{5}{12}}{1 - (\frac{4}{3})(\frac{5}{12} )}$

$= \frac{\frac{21}{12} }{\frac{16}{36} }$

$= \frac{21 (3)}{16} = \frac{63}{16}$