Equations Involving Inverse Trigonometric Functions
Ex.1 Solve
Sol.
Given:
...(1)
or cos–1 x√3 = π/2 – cos–1 x
or cos cos–1 x√3 = cos (π/2 – cos–1x)
or x√3 = sin cos–1x or
or x√3 =
Squaring we get:
3x2 = 1 – x2 or 4x2 = 1 = x = ± 1/2
Verification : When x = 1/2
L.H.S. of equation = cos1 ( 3 /2) + cos–1 (1/2) = π/6 + π/3 +π/2 = R.H.S. of equation
When x = –1/2
L.H.S. of equation = cos–1 (– 3 /2) + cos–1 (–1/2) = π – cos–1 ( 3 /2) + π – cos–1 (1/2)
= π – π/6 + π – π/3 = 3p/2 ≠ R.H.S. of equation
∴ x = 1/2 is the only solution
Ex.2 Solve for x : (tan-1 x)2 + (cot-1 x)2 =
Sol.
tan–1 x = - π/4, 3 π/4 = tan–1 x = – π/4; x = –1
Ex.3 Determine the integral values of ' k ' for which the system , (arc tan x)2 + (arc cos y)2 = π2 k and tan -1 x + cos -1 y = π /2 posses solution and find all the solutions.
Sol.
= 1 - 2 + 8 k ≥ 0 = k ≥ 1/2 ..(2)
From (1) and (2) k = 1
Inequations involving inverse trigonometric functions
Ex.1 Find the interval in which cos-1 x > sin-1 x.
Sol.
We have, cos–1 x > sin–1 {for cos–1 x to be real; x E [–1, 1]}
2 cos–1 x > π/2 = cos–1 x > π/4 or cos (cos–1 x) < cos π/4
Ex.2 Find the solution set of the inequation sin-1(sin 5) > x2 - 4x
Sol.
sin–1(sin 5) > x2 – 4x ⇒ sin–1[sin(5 – 2π)] > x2 – 4x
⇒ x2 – 4x < 5 – 2π ⇒ x2 – 4x + (2π – 5) < 0
Summation of Series
Ex.1 Sum the series ,
Sol.
= tan -1 (n + 1) (n + 2) - tan -1 n (n + 1)
Put n = 1 , 2 , 3 , ........ , n and add, we get Sn = tan -1 (n + 1) (n + 2) - tan -1 2
Ex.2 Sum the series to ' n ' terms , + ...... to ' n ' terms. Also show that , S∞ = tan -1 3 .
Sol.
= tan -1 (n + 2) – tan -1 (n)
Hence, Sn = tan -1 (n + 2) + tan -1 (n + 1) - (tan -1 1 + tan -1 2)
Ex.3 If the sum , find the value of k.
Sol.
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