Would be the answer of 13th? Please send me the explanation Share with your friends Share 0 Anand Kumar Pandey answered this Dear Student ∆=-1cosCcosBcosC-1cosAcosBcosA-1=aa-1cosCcosBcosC-1cosAcosBcosA-1=1a-acosCcosBa cosC-1cosAa cosBcosA-1C1→C1+bC2+cC3∆=1a-a+bcosC+c cosBcosCcosBa cosC-b+c cosA-1cosAa cosB+bcosA-ccosA-1=1a-a+bcosC+ccosBcosCcosBa cosC+c cosA-b-1cosAa cosB+bcosA-ccosA-1For any triangle ABC, we havea=bcosC+ccosBb=a cosC+c cosAc=a cosB+bcosA∆=1a-a+acosCcosBb-b-1cosAc-ccosA-1=1a0cosCcosB0-1cosA0cosA-1=0⇒-1cosCcosBcosC-1cosAcosBcosA-1=0 Hence, option (A) is correct. Regards 0 View Full Answer
