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Karnataka Board PUCPUC Science 2nd PUC Class 12

PUC Science 2nd PUC Class 12 - Karnataka Board PUC Question Bank Solutions for Mathematics

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`2tan^-1  3/4-tan^-1  17/31=pi/4`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

`2tan^-1(1/2)+tan^-1(1/7)=tan^-1(31/17)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

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`4tan^-1  1/5-tan^-1  1/239=pi/4`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

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)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Prove that

`tan^-1((1-x^2)/(2x))+cot^-1((1-x^2)/(2x))=pi/2`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Prove that

`sin{tan^-1  (1-x^2)/(2x)+cos^-1  (1-x^2)/(2x)}=1`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

If `sin^-1  (2a)/(1+a^2)+sin^-1  (2b)/(1+b^2)=2tan^-1x,` Prove that  `x=(a+b)/(1-ab).`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Show that `2tan^-1x+sin^-1  (2x)/(1+x^2)` is constant for x ≥ 1, find that constant.

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Find the value of the following:

`tan^-1{2cos(2sin^-1  1/2)}`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Find the value of the following:

`cos(sec^-1x+\text(cosec)^-1x),` | x | ≥ 1

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Solve the following equation for x:

`tan^-1  1/4+2tan^-1  1/5+tan^-1  1/6+tan^-1  1/x=pi/4`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

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)=pi/3`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Solve the following equation for x:

`tan^-1((2x)/(1-x^2))+cot^-1((1-x^2)/(2x))=(2pi)/3,x>0`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Solve the following equation for x:

`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Solve the following equation for x:

`cos^-1((x^2-1)/(x^2+1))+1/2tan^-1((2x)/(1-x^2))=(2x)/3`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Solve the following equation for x:

`tan^-1((x-2)/(x-1))+tan^-1((x+2)/(x+1))=pi/4`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

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

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Prove that:

`tan^-1  (2ab)/(a^2-b^2)+tan^-1  (2xy)/(x^2-y^2)=tan^-1  (2alphabeta)/(alpha^2-beta^2),`   where `alpha=ax-by  and  beta=ay+bx.`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

For any a, b, x, y > 0, prove that:

`2/3tan^-1((3ab^2-a^3)/(b^3-3a^2b))+2/3tan^-1((3xy^2-x^3)/(y^3-3x^2y))=tan^-1  (2alphabeta)/(alpha^2-beta^2)`

`where  alpha =-ax+by, beta=bx+ay`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined

Write the value of `sin^-1((-sqrt3)/2)+cos^-1((-1)/2)`

[2] Inverse Trigonometric Functions
Chapter: [2] Inverse Trigonometric Functions
Concept: undefined >> undefined
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