Advertisements
Advertisements
प्रश्न
Find the value of the expression in terms of x, with the help of a reference triangle
`tan(sin^-1(x + 1/2))`
उत्तर
Let θ = `tan(sin^-1(x + 1/2))`
⇒ sin θ = `x + 1/2`
1 – sin2θ = `1 - (x + 1/2)^2`
cos2θ = `1 - x^2 + x - 1/4`
cos2θ = `3/4 + x - x^2`
cos2θ = `(3 + 4x - 4x^2)/4`
cos θ = `sqrt(3 + 4x + 4x^2)/2`
tan θ = `((x + 1/2) xx 2)/sqrt(3 + 4x + 4x^2)`
tan θ = `(2x + 1)/sqrt(3 + 4x - 4x^2)`
APPEARS IN
संबंधित प्रश्न
If `sin (sin^(−1)(1/5)+cos^(−1) x)=1`, then find the value of x.
If `tan^-1(2x)+tan^-1(3x)=pi/4`, then find the value of ‘x’.
Write the following function in the simplest form:
`tan^(-1) x/(sqrt(a^2 - x^2))`, |x| < a
`cos^(-1) (cos (7pi)/6)` is equal to ______.
Prove that:
`tan^(-1) 63/16 = sin^(-1) 5/13 + cos^(-1) 3/5`
Prove that:
`cot^(-1) ((sqrt(1+sin x) + sqrt(1-sinx))/(sqrt(1+sin x) - sqrt(1- sinx))) = x/2`, `x in (0, pi/4)`
Solve `tan^(-1) - tan^(-1) (x - y)/(x+y)` is equal to
(A) `pi/2`
(B). `pi/3`
(C) `pi/4`
(D) `(-3pi)/4`
Find the value of `sin^-1[cos(sin^-1 (sqrt(3)/2))]`
Simplify: `tan^-1 x/y - tan^-1 (x - y)/(x + y)`
Solve: `sin^-1 5/x + sin^-1 12/x = pi/2`
Evaluate tan (tan–1(– 4)).
If `tan^-1x = pi/10` for some x ∈ R, then the value of cot–1x is ______.
Prove that `tan^-1 ((sqrt(1 + x^2) + sqrt(1 - x^2))/((1 + x^2) - sqrt(1 - x^2))) = pi/2 + 1/2 cos^-1x^2`
The value of cot–1(–x) for all x ∈ R in terms of cot–1x is ______.
If `"cot"^-1 (sqrt"cos" alpha) - "tan"^-1 (sqrt"cos" alpha) = "x",` the sinx is equal to ____________.
The value of `"tan"^-1 (3/4) + "tan"^-1 (1/7)` is ____________.
If `"tan"^-1 2 "x + tan"^-1 3 "x" = pi/4`, then x is ____________.
`tan^-1 1/2 + tan^-1 2/11` is equal to
`sin^-1(1 - x) - 2sin^-1 x = pi/2`, tan 'x' is equal to
`"tan" ^-1 sqrt3 - "cot"^-1 (- sqrt3)` is equal to ______.
