Verified Solution • 2025–26 Session moderate

QJEEM21Mad2a2: JEE Main 2021 Mathematics - Class 11 JEEM21Mad2a2 Solution: Let , where A, B, C are angles of triangle ABC. If the lengths…

Let sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B){{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}} , where A, B, C are angles of triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then :

Multiple Choice Options

  • A b2 −- a2 = a2 + c2
  • B b2, c2, a2 are in A.P.
  • C c2, a2, b2 are in A.P.
  • D a2, b2, c2 are in A.P.

Step-by-Step Verified Solution

sin⁡Asin⁡B=sin⁡(A−C)sin⁡(C−B){{\sin A} \over {\sin B}} = {{\sin (A - C)} \over {\sin (C - B)}}

As A, B, C are angles of triangle.

A + B + C = π\pi

A = π\pi −- (B + C) ...... (1)

Similarly sinB = sin(A + C) ..... (2)

From (1) and (2)

sin⁡(B+C)sin⁡(A+C)=sin⁡(A−C)sin⁡(C−B){{\sin (B + C)} \over {\sin (A + C)}} = {{\sin (A - C)} \over {\sin (C - B)}}

sin⁡(C+B).sin⁡(C−B)=sin⁡(A−C)sin⁡(A+C)\sin (C + B).\sin (C - B) = \sin (A - C)\sin (A + C)

sin⁡2C−sin⁡2B=sin⁡2A−sin⁡2C{\sin ^2}C - {\sin ^2}B = {\sin ^2}A - {\sin ^2}C

∵\because {sin⁡(x+y)sin⁡(x−y)=sin⁡2x−sin⁡2y}\{ \sin (x + y)\sin (x - y) = {\sin ^2}x - {\sin ^2}y\}

2sin⁡2C=sin⁡2A+sin⁡2B2{\sin ^2}C = {\sin ^2}A + {\sin ^2}B

By sine rule

2c2=a2+b22{c^2} = {a^2} + {b^2}

⇒\Rightarrow b2, c2 and a2 are in A.P.

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