Commerce (English Medium) इयत्ता १२ - CBSE Important Questions

Prove that: `tan^(-1)(1/2)+tan^(-1)(1/5)+tan^(-1)(1/8)=pi/4`

Solve the equation for x:sin⁻¹x+sin⁻¹(1−x)=cos⁻¹x

If `cos^-1( x/a) +cos^-1 (y/b)=alpha` , prove that `x^2/a^2-2(xy)/(ab) cos alpha +y^2/b^2=sin^2alpha`

Solve for x : tan^-1 (x - 1) + tan^-1x + tan^-1 (x + 1) = tan^-1 3x

Prove that `tan^(-1)((6x-8x^3)/(1-12x^2))-tan^(-1)((4x)/(1-4x^2))=tan^(-1)2x;|2x|<1/sqrt3`

Prove that :

`2 tan^-1 (sqrt((a-b)/(a+b))tan(x/2))=cos^-1 ((a cos x+b)/(a+b cosx))`

Solve the following for x :

`tan^(-1)((x-2)/(x-3))+tan^(-1)((x+2)/(x+3))=pi/4,|x|<1`

Solve the following for x:

`sin^(-1)(1-x)-2sin^-1 x=pi/2`

Show that:

`2 sin^-1 (3/5)-tan^-1 (17/31)=pi/4`

if `sin(sin^(-1)  1/5 + cos^(-1) x)  = 1` then find the value of x

Solve the following equation for x:  `cos (tan^(-1) x) = sin (cot^(-1)  3/4)`

If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find \[\frac{dy}{dx}\] When  \[\theta = \frac{\pi}{3}\] .

If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]

Find : \[\int\frac{2 \cos x}{\left( 1 - \sin x \right) \left( 1 + \sin^2 x \right)}dx\] .

Prove that : \[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} + \sqrt{1 - x^2}}{\sqrt{1 + x^2} - \sqrt{1 - x^2}} \right) = \frac{\pi}{4} + \frac{1}{2} \cos^{- 1} x^2 ;  1 < x < 1\].

Prove that : \[\cot^{- 1} \frac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}} = \frac{x}{2}, 0 < x < \frac{\pi}{2}\] .

Solve for x : \[\tan^{- 1} \left( \frac{x - 2}{x - 1} \right) + \tan^{- 1} \left( \frac{x + 2}{x + 1} \right) = \frac{\pi}{4}\] .

If 2 tan⁻¹ (cos θ) = tan⁻¹ (2 cosec θ), (θ ≠ 0), then find the value of θ.

If \[\tan^{- 1} \left( \frac{1}{1 + 1 . 2} \right) + \tan^{- 1} \left( \frac{1}{1 + 2 . 3} \right) + . . . + \tan^{- 1} \left( \frac{1}{1 + n . \left( n + 1 \right)} \right) = \tan^{- 1} \theta\] , then find the value of θ.

Prove that

\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .