If r2 = x2 + y2 + z2 and tan-1yzxr + tan-1xzyr = π2 - tan-1ϕ, then
ϕ = zrxy
ϕ = xyzr
ϕ = x + yzr
ϕ = yzxr + xzyr
B.
We have, tan-1yzxr + tan-1xzyr = π2 - tan-1ϕNow, tan-1yzxr + xzyr1 - yzxrxzyr = π2 - tan-1ϕ ∵ tan-1x + tan-1y = tan-1x + y1 - xy⇒ tan-1y2zr + x2zrxyr2 - xyz2 = π2 - tan-1ϕ⇒ tan-1zry2 + x2xyr2 - z2 = π2 - tan-1ϕ⇒ tan-1zr x2 + y2xyx2 + y2 = π2 - tan-1ϕ ∵ x2 + y2 = r2 - z2⇒ tan-1zrxy = cot-1ϕ ∵ tan-1ϕ + cot-1ϕ = π2⇒ tan-1zrxy = tan-11ϕ ∵ tan-11θ = cot-1θ⇒ 1ϕ = zrxy⇒ ϕ = xyzr
If α = sin-1cossin-1x and β = cos-1sincos-1x, then tanα . tanβ is equal to
1
- 1
2
12
The value of 2tan-1csctan-1x - tancot-1x is
tan-1x
tan(x)
cot(x)
csc-1x
If tan-1x2 + cot-1x2 = 5π28, then x is equal to
0
None of these
α32csc212tan-1αβ + β32sec21212tan-1βα is equal to
α - βα2 + β2
α + βα2 - β2
α + βα2 + β2
None of the above
tanπ4 + 12cos-1ab + tanπ4 - 12cos-1ab is equal to
2ab
2ba
ab
ba
The solution of equation
sin-11 - x2 - 2sin-1x2 = π2 is
x = 0
x = 12
x = 0 and 12
The value of sinπ2 - sin-1- 32 is
- 12
If a ≤ sin-1x + cos-1x + tan-1x ≤ b, then
a = 0, b = π
a = 0, b = π2
a = π2, b = π
If sin-1x + sin-1y = 2π3 and cos-1x + cos-1y = π3. Then, (x, y) is equal to
(0, 1)
(1/2, 1)
(1, 1/2)
3/2, 1