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cot(α)tan(β) – formula’s demonstration

Posted in Math Theory by Lia on 04/05/2013

For a list with trigonometric identities click here

 

We know that:

[1]  tan{alpha} + cot{beta} =   {{cos({alpha} - {beta})}/{cos{alpha}sin{beta}}}

(look for  tan{alpha} + cot{beta} demonstration  here)

 

therefore:

tan{beta} + cot{alpha} = cot{alpha} + tan{beta} =  {{cos({beta} - {alpha})}/{cos{beta}sin{alpha}}}

but  cos(- alpha) =   cos{alpha}    therefore cos(beta - alpha) =  cos(alpha - beta)

 

and  we have:   

[2]   tan{beta} + cot{alpha} = {{cos({alpha} - {beta})}/{cos{beta}sin{alpha}}}

 

Now we divide [2] by [1]:

 

{{cot{alpha} + tan{beta}}/{tan{alpha} + cot{beta}}} = {{{{cos ({alpha} - {beta})}/{cos{beta}sin{alpha}}}}/{{{cos({alpha} - {beta})}/{cos{alpha}sin{beta}}}}}    {}{right}{}    {{cot{alpha} + tan{beta}}/{tan{alpha} + cot{beta}}} = {{{cos({alpha} - {beta})}/{cos{beta}sin{alpha}}}*{{cos{alpha}sin{beta}}/{cos({alpha} - {beta})}}}     {}{right}{}

 

{{cot{alpha} + tan{beta}}/{tan{alpha} + cot{beta}}} = {{cos{alpha}sin{beta}}/{sin{alpha}cos{beta}}}    {}{right}{}       {{cot{alpha} + tan{beta}}/{tan{alpha} + cot{beta}}} = {{cos{alpha}}/{sin{alpha}}} *{{sin{beta}}/{cos{beta}}}

↓↓

cot{alpha}tan{beta} = {{cot{alpha} +tan{beta}}/{tan{alpha} +cot{beta}}}

 

 

 

 

 

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