If sin B-C-A- sin C-A-B- sin A-B-C are in A-P- then A- B- C are inA- APB- None of theseC- GPD- HP

The correct option is D HP Given: sin(B+C A), sin(C+A B), sin(A+B C) are in A.P ⇒ sin(C+A B) sin(B+C A)=sin(A+B C) sin(C+A B) ⇒ 2sin ( C+A B B C+A/2 )COS ( C+A ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

If sin B+C-A,...

Question

If $s i n (B + C - A), s i n (C + A - B), s i n (A + B - C)$ are in A.P., then cot A, cotB, cotC are in

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None of these

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Solution

The correct option is B
HP

Given:
$s i n (B + C - A), s i n (C + A - B), s i n (A + B - C) a r e i n A . P \Rightarrow s i n (C + A - B) - s i n (B + C - A) = s i n (A + B - C) - s i n (C + A - B) \Rightarrow 2 s i n (\frac{C + A - B - B - C + A}{2}) C O S (\frac{C + A - B + B + C - A}{2}) = 2 s i n (\frac{A + B - C - C - A + B}{2}) c o s (\frac{A + B - C + C + A - B}{2}) \Rightarrow s i n (A - B) c o s C = s i n (B - C) c o s A \Rightarrow s i n A c o s B c o s C - c o s A s i n B c o s C \Rightarrow s i n B c o s C c o s A - c o s B s i n C c o s A \Rightarrow 2 s i n B c o s A c o s C = s i n A c o s B c o s C + c o s A c o s B s i n C D i v i d i n g b o t h s i d e s b y c o s A c o s B c o s C : 2 t a n B = t a n A + t a n C \Rightarrow \frac{2}{c o t B} = \frac{1}{c o t A} + \frac{1}{c o t C}$

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