在三角形ABC中,cosA=1/3,求tan^2[(B+C)/2]+sin^2(A/2)的值?

因为是A,B,C是三角形的内角,故
tan^2[(B+C)/2]+sin^2(A/2)
= tan^2[(180-A)/2]+sin^2(A/2) 【A+B+C=180】
= tan^2(90-A/2)+sin^2(A/2)
= ctg^2(A/2)+sin^2(A/2) 【诱导公式】
= [cos^2(A/2)] / [sin^2(A/2)] + sin^2(A/2)
= [(1+cosA)/2] / [(1-cosA)/2] + [(1-cosA)/2] 【倍角公式】
= {[1+(1/3)]/2} / {[1-(1/3)]/2} + [1-(1/3)]/2 【代入cosA=1/3】
= 2 + (1/3)
= 7/3

7/3.