A particle just clears a wall of height b at distance a and strikes the ground at a distance c from the point of projection. The angle of projection is (1) `tan^(-1)(b/(a c))` (2) `"45"^o` (3) `tan^(-1)((b c)/(a(c-a)))` (4) `tan^(-1)((b c)/a)`

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

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)))

prove that , tan ^(-1)""(b-c)/(1+bc)+tan^(-1)""(c-a)/(1+ca)+tan ^(-1 )""(a-b )/(1+ab)=0 .

If sqrt(3)+i=(a+i b)(c+i d) , then find the value of tan^(-1)(b//a)+tan^(-1)(d//c)

Prove that, "tan"^(-1)(a)/(b)-"tan"^(-1)((a-b)/(a+b))=(pi)/(4) .

If C is a right angle in the triangle ABC, show that, tan^(-1) 'a/(b+c)' + tan^(-1) 'b/(c+a)' = pi/4 .

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(a,b,c gt 0)

If a,b,c are real positive numbers and theta =tan^(-1)[(a(a+b+c))/(bc)]^(1/2)+tan^(-1)[(b(a+b+c))/(ca)]^(1/2)+tan^(-1)[(c(a+b+c))/(ab)]^(1/2) , then tantheta equals

Prove that "tan"^(-1)(ap-q)/(aq+p)+"tan"^(-1)(b-a)/(ab+1)+"tan"^(-1)(c-b)/(bc+1)="tan"^(-1)(p)/(q)-"tan"^(-1)(1)/(c )

The number of solutions of the equation tan^(-1)(1+x)+tan^(-1)(1-x)=pi/2 is 2 (b) 3 (c) 1 (d) 0