Question

a+b+c=pi/2 then sigma cos(b+c)/cosbcosc=
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Answer 1

Given :  a + b + c  = π/2

To find : ∑ Cos(b+c)/cosbCosc

Solution:

a + b + c  = π/2

∑ Cos(b+c)/cosbCosc

= Cos(b+c)/cosbCosc  + Cos(c+a)/coscCosa + Cos(a+b)/cosaCosb

Cos(b+c)/cosbCosc = (CosbCosc  - SinbSinc)/cosbCosc = 1 - tanbtanc

Cos(c+a)/coscCosa  = 1 - tanctana

Cos(a+b)/cosaCosb = 1 - tantanb

=> ∑ Cos(b+c)/cosbCosc = 1 - tanbtanc + 1 - tanctana + 1 - tantanb

=> ∑ Cos(b+c)/cosbCosc = 3  - (tanatanb + tanbtanc + tanctana)

a + b + c  = π/2

=> a + b  =  π/2 - c

taking tan both Sides

=> tan (a + b ) =  tan (π/2 - c)

=>  (tana + tanb)/(1 - tanatanb)  =  Cotc

=>  (tana + tanb)/(1 - tanatanb)  = 1/tanc

=> tanatanc + tanbtanc = 1 - tantanb

=>  tantanb + tanbtanc + tanctana = 1

=>  ∑ Cos(b+c)/cosbCosc = 3  - (1)

=>  ∑ Cos(b+c)/cosbCosc = 2

∑ Cos(b+c)/cosbCosc = 2

Learn more:

If A + B + C = \frac{\pi}{2} and if none of A, B, C is an odd multiple of \frac{\pi}{2}, then prove that

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{cos^{2} A . cos^{2} B}

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Answer 2

Answer:

that is require solution