Question

If cos(A+B)sin(C+D)=cos(A-B)sin(C-D) then cotAcotBcotCis equal to???

Answer 1

Answer:

Step-by-step explanation:

If cos(a+b) sin(c+d) =cos(a-b) sin(c-d) then prove cot(a) cot(b) cos(c) =cos(d)?

As

cos(a+b)=cos(a)cos(b)−sin(a)sin(b)

cos(a−b)=cos(a)cos(b)+sin(a)sin(b)

sin(c+d)=sin(c)cos(d)+cos(c)sin(d)

sin(a+b)=sin(c)cos(d)−cos(c)sin(d)

we have

cos(a+b)∗sin(c+d)=cos(a−b)∗sin(c−d)

=>sin(c+d)sin(c−d)=cos(a−b)cos(a+b)

=>sin(c)cos(d)+cos(c)sin(d))sin(c)cos(d)−cos(c)sin(d))=cos(a)cos(b)+sin(a)sin(b)cos(a)cos(b)+sin(a)sin(b)

=>sin(c)cos(d)cos(c)sin(d)=cos(a)cos(b)sin(a)sin(b)

=>tan(c)∗cot(d)=cot(a)∗cot(b)

=>cot(a)∗cot(b)∗cot(c)=cot(d)

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Answer 2

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