Prove that cos(A-B)-cos(A+B)=2sinAsinB

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Solution

Cos(A-B)=cosAcosB+sinA+sinB
and
cos(A+B)=cosAcosB-sinAsinB

cos(A-B)-cos(A+B)=cosAcosB+sinAsinB-(cosAcosB-sinAsinB)
=cosAcosB+sinAsinB-cosAcosB+sinAsinB
=2sinAsinB
(Since Cos(A)Cos(B) -Cos(A)Cos(B) =0)
Hence Proved...

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Prove that $\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} = tan \frac{A}{2} cot (\frac{B}{2})$ .

x = sin (α - β) sin (γ - δ)

y = sin (β - γ) sin (α - δ)

z = sin (γ - α) sin (β - δ)

x + y + z = 0

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