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^{–1}x)
or x√3 = sin cos^{–1}x or
or x√3 =
Squaring we get:
3x^{2} = 1 – x^{2} or 4x^{2} = 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} = π^{2} 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) > x^{2}  4x
Sol.
sin^{–1}(sin 5) > x^{2} – 4x ⇒ sin–1[sin(5 – 2π)] > x^{2 }– 4x
⇒ x^{2} – 4x < 5 – 2π ⇒ x^{2 }– 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 S_{n} _{ }= 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, S_{n} = 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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