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**Equations Involving Inverse Trigonometric Functions **

**Ex.1 Solve **

**Sol. **

Given ...(1)

or cos^{â€“1} xâˆš3 = Ï€/2 â€“ cos^{â€“1} x

or cos cos^{â€“}^{1} xâˆš3 = cos (Ï€/2 â€“ cos^{â€“1}x)

or xâˆš3 = sin cos^{â€“1}x or

or xâˆš3 =

Squaring we get 3x^{2} = 1 â€“ x^{2} or 4x^{2} = 1 = x = Â± 1/2

Verification : When x = 1/2

L.H.S. of equation = cos1 ( 3 /2) + cos^{â€“1} (1/2) = Ï€/6 + Ï€/3 +Ï€/2 = R.H.S. of equation

When x = â€“1/2

L.H.S. of equation = cos^{â€“1} (â€“ 3 /2) + cos^{â€“1} (â€“1/2) = Ï€ â€“ cos^{â€“1} ( 3 /2) + Ï€ â€“ cos^{â€“1 }(1/2)

= Ï€ â€“ Ï€/6 + Ï€ â€“ Ï€/3 = 3p/2 â‰ R.H.S. of equation

âˆ´ x = 1/2 is the only solution

**Ex.2 Solve for x : (tan ^{-1} x)^{2} + (cot^{-1} x)^{2} = **

**Sol. **

tan^{â€“1} x = - Ï€/4, 3 Ï€/4 = tan^{â€“1} x = â€“ Ï€/4; x = â€“1,

**Ex.3 Determine the integral values of ' k ' for which the system , (arc tan x) ^{2} + (arc cos y)^{2} = **Ï€

**Sol.**

= 1 - 2 + 8 k â‰¥ 0 = k â‰¥ 1/2 ..(2)

From (1) and (2) k = 1

**G. Inequations involving inverse trigonometric functions **

**Ex.1 Find the interval in which cos ^{-1} x > sin^{-1} x.**

**Sol. **

We have, cos^{â€“1} x > sin^{â€“1} {for cosâ€“1 x to be real; x E [â€“1, 1]}

2 cos^{â€“1} x > Ï€/2 = cos^{â€“1} x > Ï€/4 or cos (cos^{â€“1 }x) < cos Ï€/4

**Ex.2 Find the solution set of the inequation sin ^{-1}(sin 5) > x^{2} - 4x**

**Sol. **

sin^{â€“1}(sin 5) > x^{2} â€“ 4x â‡’ sinâ€“1[sin(5 â€“ 2Ï€)] > x^{2 }â€“ 4x

â‡’ x^{2} â€“ 4x < 5 â€“ 2Ï€ â‡’ x^{2 }â€“ 4x + (2Ï€ â€“ 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 ,**

**Sol. **

= tan^{ -1} (n + 2) â€“ 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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