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Question: Solve for: \[\dfrac{\sin \left( A-B \right)}{\sin A\sin B}+\dfrac{\sin \left( B-C \right)}{\sin B\si...

Solve for: sin⁡(A−B)sin⁡Asin⁡B+sin⁡(B−C)sin⁡Bsin⁡C+sin⁡(C−A)sin⁡Csin⁡A=0\dfrac{\sin \left( A-B \right)}{\sin A\sin B}+\dfrac{\sin \left( B-C \right)}{\sin B\sin C}+\dfrac{\sin \left( C-A \right)}{\sin C\sin A}=0.

Explanation

Solution

Hint: For solving this problem we use some trigonometric formulas like sin(A-B) for expanding the left-hand side. By using this methodology, we easily solved our problem.

Complete step-by-step answer:
Some of the useful trigonometric formulas used in solving this problem.
sin (A + B) = sin A cos B + cos A sin B
sin (A − B) = sin A cos B − cos A sin B
According to the problem statement, we consider the left-hand side of the equation for proving equivalence of both sides. First, we expand the numerator of the left-hand side using the above-mentioned formulas.
⇒sin⁡(A−B)sin⁡Asin⁡B+sin⁡(B−C)sin⁡Bsin⁡C+sin⁡(C−A)sin⁡Csin⁡A\Rightarrow \dfrac{\sin \left( A-B \right)}{\sin A\sin B}+\dfrac{\sin \left( B-C \right)}{\sin B\sin C}+\dfrac{\sin \left( C-A \right)}{\sin C\sin A}
Expand the numerator of the above expression using, sin (x − y) = sin x cos y − cos x sin y formula, we get:
⇒sin⁡Acos⁡B−cos⁡Asin⁡Bsin⁡Asin⁡B+sin⁡Bcos⁡C−cos⁡Bsin⁡Csin⁡Bsin⁡C+sin⁡Ccos⁡A−cos⁡Csin⁡Asin⁡Csin⁡A\Rightarrow \dfrac{\sin A\cos B-\cos A\sin B}{\sin A\sin B}+\dfrac{\sin B\cos C-\cos B\sin C}{\sin B\sin C}+\dfrac{\sin C\cos A-\cos C\sin A}{\sin C\sin A}
Now, separating the numerator and writing all terms individually, we get
⇒sin⁡Acos⁡Bsin⁡Asin⁡B−cos⁡Asin⁡Bsin⁡Asin⁡B+sin⁡Bcos⁡Csin⁡Bsin⁡C−cos⁡Bsin⁡Csin⁡Bsin⁡C+sin⁡Ccos⁡Asin⁡Csin⁡A−cos⁡Csin⁡Asin⁡Csin⁡A\Rightarrow \dfrac{\sin A\cos B}{\sin A\sin B}-\dfrac{\cos A\sin B}{\sin A\sin B}+\dfrac{\sin B\cos C}{\sin B\sin C}-\dfrac{\cos B\sin C}{\sin B\sin C}+\dfrac{\sin C\cos A}{\sin C\sin A}-\dfrac{\cos C\sin A}{\sin C\sin A}
Now, similar terms in the numerator and denominator cancel out each other. On doing so, we get
⇒cos⁡Bsin⁡B−cos⁡Asin⁡A+cos⁡Csin⁡C−cos⁡Bsin⁡B+cos⁡Asin⁡A−cos⁡Csin⁡C\Rightarrow \dfrac{\cos B}{\sin B}-\dfrac{\cos A}{\sin A}+\dfrac{\cos C}{\sin C}-\dfrac{\cos B}{\sin B}+\dfrac{\cos A}{\sin A}-\dfrac{\cos C}{\sin C}
We know that the cos⁡θsin⁡θ=cot⁡θ\dfrac{\cos \theta }{\sin \theta }=\cot \theta . Using this identity, we get
⇒cot⁡B−cot⁡A+cot⁡C−cot⁡B+cot⁡A−cot⁡C ⇒0 \begin{aligned} & \Rightarrow \cot B-\cot A+\cot C-\cot B+\cot A-\cot C \\\ & \Rightarrow 0 \\\ \end{aligned}
All the terms are the same and opposite. So, they cancel out each other. Finally, the result obtained is 0. Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric formulas associated with different functions. The conversion of the respective function should be done carefully, and the magnitude of the required quantity must be copied correctly in the final expression for avoiding calculation error.