Advertisements
Advertisements
प्रश्न
If `cos(tan^-1x + cot^-1 sqrt(3))` = 0, then value of x is ______.
उत्तर
If `cos(tan^-1x + cot^-1 sqrt(3))` = 0, then value of x is `sqrt(3)`.
Explanation:
We have, `cos(tan^-1x + cot^-1 sqrt(3))` = 0
⇒ `tan^-1x + cot^-1 sqrt(3) = cos^-1 0`
⇒ `tan^-1x + cot^-1 sqrt(3) = pi/2`
⇒ `tan^-1x = pi/2 - cot^-1 sqrt(3)`
⇒ `tan^-1x = tan^-1 sqrt3)` .....`(because tan^-1x + cot^-1x = pi/2)`
∴ x = `sqrt(3)`
APPEARS IN
संबंधित प्रश्न
Find the value of `tan^(-1) sqrt3 - cot^(-1) (-sqrt3)`
Find the principal value of the following:
`tan^-1(-1/sqrt3)`
Find the principal value of the following:
`tan^-1(2cos (2pi)/3)`
Find the principal value of the following:
`sec^-1(-sqrt2)`
Find the principal value of the following:
`sec^-1(2sin (3pi)/4)`
Find the principal value of the following:
`cosec^-1(2cos (2pi)/3)`
For the principal value, evaluate the following:
`sec^-1(sqrt2)+2\text{cosec}^-1(-sqrt2)`
Find the principal value of the following:
`cot^-1(-1/sqrt3)`
if sec-1 x = cosec-1 v. show that `1/x^2 + 1/y^2 = 1`
If `sin^-1"x" + tan^-1"x" = pi/2`, prove that `2"x"^2 + 1 = sqrt5`
Find value of tan (cos–1x) and hence evaluate `tan(cos^-1 8/17)`
The value of `sin^-1 (cos((43pi)/5))` is ______.
The domain of sin–1 2x is ______.
The value of sin (2 sin–1 (.6)) is ______.
Find the value of `4tan^-1 1/5 - tan^-1 1/239`
The domain of the function cos–1(2x – 1) is ______.
The set of values of `sec^-1 (1/2)` is ______.
The value of expression `tan((sin^-1x + cos^-1x)/2)`, when x = `sqrt(3)/2` is ______.
The principal value of `sin^-1 [cos(sin^-1 1/2)]` is `pi/3`.
If `5 sin theta = 3 "then", (sec theta + tan theta)/(sec theta - tan theta)` is equal to ____________.
The period of the function f(x) = cos4x + tan3x is ____________.
The general solution of the equation `"cot" theta - "tan" theta = "sec" theta` is ____________ where `(n in I).`
If sin `("sin"^-1 1/5 + "cos"^-1 "x") = 1,` then the value of x is ____________.
`"cos" ["tan"^-1 {"sin" ("cot"^-1 "x")}]` is equal to ____________.
If `"tan"^-1 ("a"/"x") + "tan"^-1 ("b"/"x") = pi/2,` then x is equal to ____________.
