Prove the result : tan^-1(1/4) + tan^-1(2/9) = (1/2)cos^-1(3/5) = (1/2)sin^-1(4/5) Watch Solution
Prove the result : tan^-1(2/3) = (1/2)tan^-1(12/5) Watch Solution
Prove the result : tan^-1(1/7) + 2tan^-1(1/3)=π/4 Watch Solution
Prove the result : sin^-1(4/5) + 2tan^-1(1/3) = π/2 Watch Solution
Prove the result : 2sin^-1(3/5) – tan^-1(17/31) = π/4 Watch Solution
Prove the result : 2tan^-1(1/5) + tan^-1(1/8) = tan^-1(4/7) Watch Solution
Prove the result : 2tan^-1(3/4) – tan^-1(17/31) = π/4 Watch Solution
Prove the result : 2tan^-1(1/2) + tan^-1(1/7) = tan^-1(31/17) Watch Solution
Prove the result : 4tan^-1(1/5) – tan^-1(1/239) = π/4 Watch Solution
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+ab) Watch Solution
Prove that : tan^-1{(1-x^2)/2x} + cot^-1{(1-x^2)/2x} = π/2 Watch Solution
Prove that : sin{tan^-1((1-x^2)/2x) + cot^-1((1-x^2)/(1+x^2))} = 1 Watch Solution
If sin^-1(2a/(1+a^2)) + sin^-1(2b/(a+b^2)) = 2tan^-1x, prove that x = (a+b)/(1-ab) Watch Solution
Show that 2tan^-1x + sin^-1(2x/(1+x^2)) is constant for x≥1, find that constant. Watch Solution
Find the value of the tan^-1{2cos(2sin^-1(1/2))} Watch Solution
Find the value of the cos(sec^-1x + cosec^-1x), |x| ≥ 1 Watch Solution
Solve the following equation for x : tan^-1(1/4 + 2tan^-1(1/5) + tan^-1(1/6) + tan^-1(1/x) = π/4 Watch Solution
Solve the following equation for x : 3sin^-1(2x/(1+x^2)) – 4cos^-1((1-x^2)/(1+x^2)) + 2tan^-1(2x/1-x^2) = π/3 Watch Solution
Solve the following equation for x : tan^-1(2x/(1-x^2)) + cot^-1((1-(x^2))/2x) = 2π/3, x > 0. Watch Solution
Solve the following equation for x : 2tan^-1(sin x) = tan^-1(2sec x), x ≠ π/2 Watch Solution
Solve the following equation for x : cos^-1((x^2-1)/(x^2+1)) + 1/2 (tan^-1(2x/(1-x^2))) = 2π/3 Watch Solution
Prove that 2tan^-1((x-2)/(x-1)) + tan^-1((x+2)/(x+1)) = π/4 Watch Solution
Prove that 2tan^-1{√(a-b)/√(a+b)} .tan(θ/2) = cos^-1{(acosθ+b)/(a+bcosθ)} Watch Solution
Prove that: tan^-1{2ab/(a^2-b^2)} + tan^-1{2xy/(x^2-y^2)} = tan^-1{2αβ/(α^2-β^2)} , where α=ax-by and β=ay+bx. Watch Solution
For any a, b, x, y greater than 0, prove that: 2/3 (tan^-1{(3ab^2-a^3)/(b^3-3a^2b)} + 2/3 {(tan^-1(3xy^2-x^3)/(y^3-3x^2y)} = tan^-1{(2αβ/(α^2-β^2)}, where α = -ax+by, β = bx+ay. Watch Solution
INVERSE TRIGONOMETRIC FUNCTIONS – R.D. Sharma Class 12th Math
