Question

Prove that $\frac{\cos (A + B + C) + \cos (- A + B + C) + \cos (A - B + C) + \cos (A + B - C)}{\sin (A + B + C) + \sin (- A + B + C) - \sin (A - B + C) + \sin (A + B - C)} = \cot B$ .

Accepted Answer

$LHS = \frac{\cos (A + B + C) + \cos (- A + B + C) + \cos (A - B + C) + \cos (A + B - C)}{\sin (A + B + C) + \sin (- A + B + C) - \sin (A - B + C) + \sin (A + B - C)}$

Apply the identity,

$\cos (C) + \cos (D) = 2 \cos (\frac{C + D}{2}) \cos (\frac{C - D}{2}) & \sin (C) \pm \sin (D) = 2 \sin (\frac{C \pm D}{2}) \cos (\frac{C \mp D}{2})$

$= \frac{2 \cos \frac{(A + B + C - A + B + C)}{2} \cos \frac{(A + B + C + A - B - C)}{2} + 2 \cos \frac{(A - B + C + A + B - C)}{2} \cos \frac{(A - B + C - A - B + C)}{2}}{2 \sin \frac{(A + B + C - A + B + C)}{2} \cos \frac{(A + B + C + A - B - C)}{2} + 2 \cos \frac{(A + B - C + A - B + C)}{2} \sin \frac{(A + B - C - A + B - C)}{2}}$

$= \frac{2 \cos (B + C) \cos A + 2 \cos A \cos (C - B)}{2 \sin (B + C) \cos A - 2 \cos A \sin (C - B)}$

$= \frac{2 \cos A \{\cos (B + C) + \cos (C - B)\}}{2 \cos A \{\sin (B + C) - \sin (C - B)\}}$

$\cos (C) + \cos (D) = 2 \cos (\frac{C + D}{2}) \cos (\frac{C - D}{2}) & \sin (C) - \sin (D) = 2 \sin (\frac{C - D}{2}) \cos (\frac{C + D}{2})$

$= \frac{2 \cos B \cos C}{2 \cos C \sin B}$

$= \frac{\cos B}{\sin B}$

$= \cot B$

$= RHS$

Proved.

S L Loney Solutions for Chapter: Trigonometrical Ratios of The Sum and Difference of Two Angles, Exercise 2: EXAMPLES XIV

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: Trigonometrical Ratios of The Sum and Difference of Two Angles, Exercise 2: EXAMPLES XIV

Questions from S L Loney Solutions for Chapter: Trigonometrical Ratios of The Sum and Difference of Two Angles, Exercise 2: EXAMPLES XIV with Hints & Solutions

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