Prove that tan−1x+tan−1y=tan−1x+y1−xy,When(xy<1).
Step 1:Lettan−1x=Aandtan−1y=B. Then,
x=tanAandy=tanBandA,B∈−π2,π2
∴tan(A+B)=tanA+tanB1−tanAtanB=x+y1−xy…(1)
Now, the following cases arise.
Case 1) When x>0,y>0 and xy<1
In this case we have:
x>0,y>0andxy<1
⇒x+y1−xy>0
⇒tan(A+B)>0 [Using (1)]
⇒A+B lies in first or third quadrant.
⇒0<A+B<π2∵x>0⇒0<A<π2y>0⇒0<B<π2⇒0<A+B<π
∴tan(A+B)=x+y1−xy [From (1)]
⇒A+B=tan−1x+y1−xy∵0<A+B<π2
⇒tan−1x+tan−1y=tan−1x+y1−xy
Step 2: Case 2) When x<0,y<0 and xy<1
⇒x+y1−xy<0
⇒tan(A+B)<0 [From (1)]
⇒A+B lies in second or fourth quadrant.
⇒A+B lies in fourth quadrant
∵x<0⇒−π2<A<0y<0⇒−π2<B<0⇒−π<A+B<0
⇒−π2<A+B<0
∴tan(A+B)=x+y1−xy
⇒A+B=tan-1x+y1−xy
Hence, proved.