Advertisements
Advertisements
प्रश्न
Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`
उत्तर
`cot^-1 x - cot^-1 (x + 2) = pi/12`
`tan^-1[1/x] - tan^-1 [1/(x + 2)] = pi/12`
⇒ `tan^-1 [(1/x - 1/(x + 2))/(1 + (1/x)(1/(x + 2)))] = pi/12`
⇒ `(x + 2 - x)/(x(x + 2) + 1) = tan pi/12`
⇒ `2/(x^2 + 2x + 1) = tan15^circ`
We know that, tan 15° = `2 - sqrt(3)`
⇒ `2/(x^2 + 2x + 1) = 2 - sqrt(3)`
⇒ `x^2 + 2x + 1 = 2/(2 - sqrt(3)`
⇒ `(x + 1)^2 = 2/(2 - sqrt(3)) xx [(2 + sqrt(3))/(2 + sqrt(3))]`
⇒ `(x + 1)^2 = (2(2 + sqrt(3)))/(4 - 3)`
⇒ (x + 1)2 = `4 + 2sqrt(3)`
⇒ (x + 1)2 = `1 + 3 + 2sqrt(3)`
⇒ (x + 1)2 = `(1 + sqrt(3))^2`
⇒ x + 1 = `1 + sqrt(3)`
∴ x = `sqrt(3)`
APPEARS IN
संबंधित प्रश्न
Prove that:
`tan^(-1)""1/5+tan^(-1)""1/7+tan^(-1)""1/3+tan^(-1)""1/8=pi/4`
Find the value of following:
`tan 1/2 [sin^(-1) (2x)/(1+ x^2) + cos^(-1) (1-y^2)/(1+y^2)], |x| < 1, y> 0 and xy < 1`
Solve the following equation:
`2 tan^(-1) (cos x) = tan^(-1) (2 cosec x)`
Prove that
\[2 \tan^{- 1} \left( \frac{1}{5} \right) + \sec^{- 1} \left( \frac{5\sqrt{2}}{7} \right) + 2 \tan^{- 1} \left( \frac{1}{8} \right) = \frac{\pi}{4}\] .
Solve for x : \[\cos \left( \tan^{- 1} x \right) = \sin \left( \cot^{- 1} \frac{3}{4} \right)\] .
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 the value, if it exists. If not, give the reason for non-existence
`sin^-1 (cos pi)`
Find the value of the expression in terms of x, with the help of a reference triangle
sin (cos–1(1 – x))
Simplify: `tan^-1 x/y - tan^-1 (x - y)/(x + y)`
Choose the correct alternative:
`tan^-1 (1/4) + tan^-1 (2/9)` is equal to
If α ≤ 2 sin–1x + cos–1x ≤ β, then ______.
If `sin^-1 ((2"a")/(1 + "a"^2)) + cos^-1 ((1 - "a"^2)/(1 + "a"^2)) = tan^-1 ((2x)/(1 - x^2))`. where a, x ∈ ] 0, 1, then the value of x is ______.
The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.
The minimum value of sinx - cosx is ____________.
`"tan"^-1 1/3 + "tan"^-1 1/5 + "tan"^-1 1/7 = "tan"^-1 1/8 =` ____________.
The value of cot-1 9 + cosec-1 `(sqrt41/4)` is given by ____________.
`"tan"^-1 1 + "cos"^-1 ((-1)/2) + "sin"^-1 ((-1)/2)`
The value of `tan^-1 (x/y) - tan^-1 (x - y)/(x + y)` is equal to
`"tan" ^-1 sqrt3 - "cot"^-1 (- sqrt3)` is equal to ______.
