मराठी

Write the following function in the simplest form: tan^(–1)  (sqrt(1 + x^2) – 1)/x, x ≠ 0

Advertisements
Advertisements

प्रश्न

Write the following function in the simplest form:

`tan^(-1)  (sqrt(1 + x^2) - 1)/x, x ≠ 0`

सोपे रूप द्या
Advertisements

उत्तर

`tan^(-1)  (sqrt(1 + x^2) - 1)/x`

Put x = tan θ ⇒ θ = tan–1 x

∴ `tan^(-1)  (sqrt(1 + x^2) - 1)/x`

= `tan^(-1) ((sqrt(1 + tan^2 θ) - 1)/tan θ)`

= `tan^(-1) ((sec θ - 1)/tan θ)`

= `tan^(-1) ((1 - cos θ)/sin θ)`

= `tan^(-1) ((2 sin^2  θ/2)/(2 sin  θ/2  cos  θ/2))`

= `tan^(-1) (tan  θ/2)`

= `θ/2`

= `1/2 tan^(-1) x`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Inverse Trigonometric Functions - EXERCISE 2.2 [पृष्ठ २९]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 2 Inverse Trigonometric Functions
EXERCISE 2.2 | Q 3. | पृष्ठ २९

संबंधित प्रश्‍न

Prove the following:

3 sin−1 x = sin−1 (3x − 4x3), `x ∈ [-1/2, 1/2]`


Write the function in the simplest form: `tan^(-1)  1/(sqrt(x^2 - 1)), |x| > 1`


Write the function in the simplest form:  `tan^(-1)  ((cos x - sin x)/(cos x + sin x)) `,` 0 < x < pi`


Write the following function in the simplest form:

`tan^(-1) ((3a^2 x - x^3)/(a^3 - 3ax^2)), a > 0; (-a)/sqrt(3) < x < a/sqrt(3)`


if `tan^(-1)  (x-1)/(x - 2) + tan^(-1)  (x + 1)/(x + 2) = pi/4` then find the value of x.


Prove that `tan^(-1) sqrt(x) = 1/2 cos^(-1)  (1 - x)/(1 + x), x ∈ [0, 1]`.


Prove `(9pi)/8 - 9/4  sin^(-1)  1/3 = 9/4 sin^(-1)  (2sqrt2)/3`


Prove that `tan {pi/4 + 1/2 cos^(-1)  a/b} + tan {pi/4 - 1/2 cos^(-1)  a/b} = (2b)/a`


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


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 y = `(x sin^-1 x)/sqrt(1 -x^2)`, prove that: `(1 - x^2)dy/dx = x + y/x`


If tan-1 x - cot-1 x = tan-1 `(1/sqrt(3)),`x> 0 then find the value of x and hence find the value of sec-1 `(2/x)`.


Find: ∫ sin x · log cos x dx


Find the value of the expression in terms of x, with the help of a reference triangle

sin (cos–1(1 – x))


Prove that `tan^-1  2/11 + tan^-1  7/24 = tan^-1  1/2`


Solve: `sin^-1  5/x + sin^-1  12/x = pi/2`


Choose the correct alternative:

`sin^-1  3/5 - cos^-1  13/13 + sec^-1  5/3 - "cosec"^-1  13/12` is equal to


Choose the correct alternative:

sin–1(2 cos2x – 1) + cos1(1 – 2 sin2x) =


Evaluate: `sin^-1 [cos(sin^-1 sqrt(3)/2)]`


Prove that `2sin^-1  3/5 - tan^-1  17/31 = pi/4`


Evaluate `cos[cos^-1 ((-sqrt(3))/2) + pi/6]`


If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.

`tan[tan^-1("d"/(1 + "a"_1 "a"_2)) + tan^-1("d"/(21 + "a"_2 "a"_3)) + tan^-1("d"/(1 + "a"_3 "a"_4)) + ... + tan^-1("d"/(1 + "a"_("n" - 1) "a""n"))]`


The number of real solutions of the equation `sqrt(1 + cos 2x) = sqrt(2) cos^-1 (cos x)` in `[pi/2, pi]` is ______.


If cos–1x > sin–1x, then ______.


The maximum value of sinx + cosx is ____________.


The value of `"tan"^ -1 (3/4) + "tan"^-1 (1/7)` is ____________.


If `"tan"^-1 ("cot"  theta) = 2theta, "then"  theta` is equal to ____________.


The value of sin (2tan-1 (0.75)) is equal to ____________.


`"cot" ("cosec"^-1  5/3 + "tan"^-1  2/3) =` ____________.


`"cos" (2  "tan"^-1 1/7) - "sin" (4  "sin"^-1 1/3) =` ____________.


If tan-1 2x + tan-1 3x = `pi/4,` then x is ____________.


The value of cot-1 9 + cosec-1 `(sqrt41/4)` is given by ____________.


The value of `"tan"^-1 (1/2) + "tan"^-1(1/3) + "tan"^-1(7/8)` is ____________.


If `3  "sin"^-1 ((2"x")/(1 + "x"^2)) - 4  "cos"^-1 ((1 - "x"^2)/(1 + "x"^2)) + 2 "tan"^-1 ((2"x")/(1 - "x"^2)) = pi/3` then x is equal to ____________.


Solve for x : `{"x cos" ("cot"^-1 "x") + "sin" ("cot"^-1 "x")}^2` = `51/50


`tan^-1 sqrt3 - cot^-1 (- sqrt3)` is equal to ______.


Solve for x: `sin^-1(x/2) + cos^-1x = π/6`


Solve:

sin–1 (x) + sin–1 (1 – x) = cos–1 x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×