Inverse Trigonometry

Q2 Find non zero integral solutions of equation
tan^-112x+1+tan^-114x+1=tan^-12x²

q)2
only one integral solution except 0-
x=3

Answer 4

cot^-1 (2x + 1) + cot^-1(4x+1) = cot^-1 (x²/2)

=> {(8x² +6x)/ (6x +2) } =x² /2......applying the formula for cot(A+B)
=>3x² -7x -6 =0
solving we get , x=-2/3 , 3
....the integral value is 3 .....hence the answer !

tan^-14=4tan^-1x

put 4 = tan\theta

hence \theta= 4tan^-1x

x = tan(\frac{\theta }{4})..(1)

where 4 = tan \theta.........(2)

on simplifying eqn 1 and 2

we get

x^{4}+x^{3}-6x^{2}-x+1=0........ (3)

now,

x^{5}-7x^{3}+5x^{2}+2x+9870

=x(x^{4}-7x^{2}+5x+2)+9870

=x(x^{4}-6x^{2}-x^{2}+6x-x+1+1)+9870

=x(x^{4}+x^{3}-x^{3}-6x^{2}-x^{2}+6x-x+1+1)+9870

=x(x^{4}+x^{3}-6x^{2} -x+1 -x^{2}+6x+1-x^{3} )+9870

=x([x^{4}+x^{3}-6x^{2} -x+1] -x^{2}+6x+1-x^{3} )+9870

=x(0 -x^{2}+6x+1-x^{3} )+9870

= -(x^{4}+ x^{3}-6x^{2}-x)+9870

= -(-1)+9870 =9871

since\; x^{4}+x^{3}-6x^{2}-x+1=0,from\; eqn (3)