`tan^-1(3/4)+2tan^-1(3/5)=?`

Prove that: tan^(-1)(1)/(7)+tan^(-1)(1)/(13)=tan^(-1)(2)/(9)tan^(-1)+tan^(-1)(1)/(5)+tan^(-1)(1)/(8)=(pi)/(4)tan^(-1)(3)/(4)+tan^(-1)(3)/(5)-tan^(-1)(8)/(19)=(pi)/(4)tan^(-1)(1)/(5)+tan^(-1)(1)/(7)+tan^(-1)(1)/(3)+tan^(-1)(1)/(8)=(pi)/(4)cot^(-1)7+cot^(-1)8+cot^(-1)18=cot^(-1)(1)/(13)

tan^-1 (1/2) + tan^-1 (1/3) = π/4

If: tan^(-1)(1/3) + tan^(-1)( 3/4) - tan^(-1)(x/3) =0 , then: x=

tan^(-1)(x/2)+tan^(-1)(x/3)=(pi)/(4)

Prove that : tan^(-1)(1/2) + tan^(-1)(1/3) = tan^(-1)(3/5) + tan^(-1)(1/4) = pi/4

Pove that i) tan^(-1)1/2+tan^(-1)2/11=tan^(-1)3/4 ii) tan^(-1)2/11+tan^(-1)7/24=tan^(-1)1/2 iii) tan^(-1)1+tan^(-1)1/2+tan^(-1)1/3=pi/2 iv) 2tan^(-1)1/3+tan^(-1)/17=pi/4 v) tan^(-1)2-tan^(-1)1=tan^(-1)1/3 vi) tan^(-1)+tan^(-1)2+tan^(-1)3=pi vii) tan^(-1)1/2+tan^(-1)1/5+tan^(-1)1/8=pi/4 viii) tan^(-1)1/4+tan^(-1)2/9=1/2tan^(-1)4/3

Prove that tan^(-1)3/4+tan^(-1)3/5-tan^(-1)8/19=pi/4 .

tan^(-1)((3)/(4))+tan^(-1)((3)/(5))-tan^(-1)((8)/(19))=

The angle between the normals of ellipse 4x^2 + y^2 = 5 , at the intersection of 2x+y=3 and the ellipse is (A) tan^(-1) (3/5) (B) tan^(-1) (3/4) (C) tan^(-1) (4/3) (D) tan^(-1) (4/5)