Q.

If tan(A-B)tanA+sin2Csin2A=1, A,B,C(0,π2), then      [2026]

1 tanA,tanC,tanB are in G.P.  
2 tanA,tanB,tanC are in A.P.  
3 tanA,tanB,tanC are in G.P.  
4 tanA,tanC,tanB are in A.P.  

Ans.

(1)

tanA-tanB(1+tanAtanB)tanA+1+cot2A1+cot2C=1

Put tanA=x, tanB=y, tanC=z

 x-y(1+xy)x+(x2+1)z2x2(z2+1)=1

 x(x-y)(z2+1)+z2(1+x2)(1+xy)=(1+xy)x2(1+z2)

after solving we get

z2=xy      1+x20

 tan2C=tanA·tanB

 tanA, tanC, tanB are in G.P.