JEE Mains · Maths · STD 11 - Trigonometrical equations
Let \(\frac{\sin \mathrm{A}}{\sin \mathrm{B}}=\frac{\sin (\mathrm{A}-\mathrm{C})}{\sin (\mathrm{C}-\mathrm{B})}\), where \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are angles f a triangle \(\mathrm{ABC}\). If the lengths of the sides pposite these angles are \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) respectively, then :
- A \(b^{2}-a^{2}=a^{2}+c^{2}\)
- B \(b^{2}, c^{2}, a^{2}\) are in \(A.P.\)
- C \(\mathrm{c}^{2}, \mathrm{a}^{2}, \mathrm{~b}^{2}\) are in \(A.P.\)
- D \(a^{2}, b^{2}, c^{2}\) are in \(A.P.\)
Answer & Solution
Correct Answer
(B) \(b^{2}, c^{2}, a^{2}\) are in \(A.P.\)
Step-by-step Solution
Detailed explanation
\(\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}\) As \(A, B, C\) are angles of triangle \(A+B+C=\pi\) \(A=\pi-(B+C)\) So, \(\sin A=\sin (B+C) \ldots(1)\) \(\text { Similarly } \sin B=\sin (A+C) \ldots(2)\) \(\text { From (1) and (2) }\)…
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