सिद्ध कीजिए कि `2tan^(-1)sqrt((b)/(a))=cos^(-1)((a-b)/(a+b))`

If a , b , c >0 and s=(a+b+c)/2,p rov et h a t tan^(-1)sqrt((2a s)/(b c))+tan^(-1)sqrt((2b s)/(c a))+tan^(-1)sqrt((2c s)/(a b))=pi

If a , b , c >0 and s=(a+b+c)/2 , prove that tan^(-1)sqrt((2a s)/(b c))+tan^(-1)sqrt((2b s)/(c a))+tan^(-1)sqrt((2c s)/(a b))=pi

Prove that 2tan^(-1)(sqrt((a-b)/(a+b))tan(theta/2))=cos^(-1)((acostheta+b)/(a+bcostheta))

Write the value of tan^(-1)(a/b)-tan^(-1)((a-b)/(a+b))

Tangents are drawn to the ellipse (x^2)/(a^2)+(y^2)/(b^2)=1,(a > b), and the circle x^2+y^2=a^2 at the points where a common ordinate cuts them (on the same side of the x-axis). Then the greatest acute angle between these tangents is given by (A) tan^(-1)((a-b)/(2sqrt(a b))) (B) tan^(-1)((a+b)/(2sqrt(a b))) (C) tan^(-1)((2a b)/(sqrt(a-b))) (D) tan^(-1)((2a b)/(sqrt(a+b)))

cos^(- 1)sqrt((a-x)/(a-b))=sin^(- 1)sqrt((x-b)/(a-b)) is possible ,if

Solve : sin^(-1)((2a)/(1+a^2)) + cos^(-1)( (1-b^2)/(1+b^2)) = 2 tan^(-1) x,|a|le1, |b|ge 0 .

prove that 2tan^-1(sqrt((a-b)/(a+b))tan(theta/2)) = cos^-1((acostheta+b)/(a+bcostheta))

If sin^(-1)((2a)/(1+a^2))-cos^(-1)((1-b^2)/(1+b^2))=tan^(-1)((2x)/(1-x^2)) , then prove that x=(a-b)/(1+a b)

Let a, b and c be positive real numbers. Then prove that tan^(-1) sqrt((a(a + b + c))/(bc)) + tan^(-1) sqrt((b (a + b + c))/(ca)) + tan^(-1) sqrt((c(a + b+ c))/(ab)) =' pi'