JEE Main 2021 (Online) 27th August Morning ShiftQuestion 1
Let $\frac{\sin A}{\sin B}=\frac{\sin(A-C)}{\sin(C-B)}$, where $A,B,C$ are angles of a triangle $ABC$. If the lengths of the sides opposite these angles are $a,b,c$ respectively, then :
Trigonometric Ratio & Identities
Solution
Concept. Cross-multiply and replace $\sin A$ by $\sin(B+C)$ and $\sin B$ by $\sin(A+C)$; each side becomes a difference of squares of sines. Formulas used. (1) $\sin(x+y)\sin(x-y)=\sin^2x-\sin^2y$(2) Sine rule: $\sin A:\sin B:\sin C=a:b:c$Step 1 - Cross-multiply. $\sin A\sin(C-B)=\sin B\sin(A-C)$. Step 2 - Use (1). $\sin A=\sin(C+B)$, so the left side is $\sin^2C-\sin^2B$. $\sin B=\sin(A+C)$, so the right side is $\sin^2A-\sin^2C$. Step 3 - Sides. By (2), $c^2-b^2=a^2-c^2$, i.e. $2c^2=a^2+b^2$. So $b^2,c^2,a^2$ are in A.P. Answer: (B)