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

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)

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)))

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

If A,B,C are the angles of a triangle, show that, (ii) tan ((A - B)/2) = cot (C/2 + B) .

If A, B, C are three angles of /_\ABC , then (i) tan ((A - B)/2) =

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

int e^(tan^(-1)x)(1+x+x^2)d(cot^(-1)x) is equal to (a) -e^(tan^(-1)x)+c (b) e^(tan^(-1)x)+c (c) -xe^(tan^(-1)x)+c (d) xe^(tan^(-1)x)+c

If A, B, C are the angles of a triangle then tan ((B+C)/2)=