Dear Student,Solution)LHS, sin2A+sin2600+A+sin2600-A=sin2A+121-cos2600+A+121-cos2600-A {because sin2θ=121-cos2θ=sin2A+1-12cos1200+2A+cos1200-2A=sin2A+1-12×2cos1200+2A+1200-2A2 cos1200+2A-1200-2A2{because cosC+cosD=2cosC+D2 cosC-D2}=sin2A+1-cos1200 cos2A=sin2A+1+12(1-2sin2A)=sin2A+1+12-sin2A=32=RHSHENCE PROVED Regards!

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D e a r S t u d e n t, S o l u t i o n) L H S, \sin^{2} A + s {in}^{2} (60^{0} + A) + \sin^{2} (60^{0} - A) = \sin^{2} A + \frac{1}{2} [1 - \cos 2 (60^{0} + A)] + \frac{1}{2} [1 - \cos 2 (60^{0} - A)] {b e c a u s e \sin^{2} θ = \frac{1}{2} [1 - \cos 2 θ] = \sin^{2} A + 1 - \frac{1}{2} [\cos (120^{0} + 2 A) + \cos (120^{0} - 2 A)] = \sin^{2} A + 1 - \frac{1}{2} \times 2 \cos \frac{(120^{0} + 2 A) + (120^{0} - 2 A)}{2} . \cos \frac{(120^{0} + 2 A) - (120^{0} - 2 A)}{2} {b e c a u s e \cos C + \cos D = 2 \cos \frac{C + D}{2} . \cos \frac{C - D}{2}} = \sin^{2} A + 1 - \cos 120^{0} . \cos 2 A = \sin^{2} A + 1 + \frac{1}{2} (1 - 2 \sin^{2} A) = \sin^{2} A + 1 + \frac{1}{2} - \sin^{2} A = \frac{3}{2} = R H S H E N C E P R O V E D . R e g a r d s!

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