# Exercise 2.2 Solutions

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Exercise 2.2 Solutions
1. Prove the following: {3\sin^{-1} x = \sin^{-1} (3x - 4x^3)}, {x ∈ \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]}
Let {x = \sin θ}
{\sin^{-1} x = θ}
Now, as {x ∈ \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]}
{⇒ (3x -4x^3) ∈ [-1, 1]}
Also, as {x ∈ \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]}
{⇒ θ ∈ \left[-\dfrac{π}{6}, \dfrac{π}{6}\right]}
{⇒ 3θ ∈ \left[-\dfrac{π}{2}, \dfrac{π}{2}\right]}
We have,
R.H.S.
=
{\sin^{-1} (3x - 4x^3)}
=
{\sin^{-1} (3\sin θ - 4\sin^3 θ)}
=
{\sin^{-1} (\sin 3θ)}
=
{3θ} {(∵ \sin^{-1} (\sin x) = x)}
=
{3 \sin^{-1} x}
=
L.H.S.
2. Prove the following: {3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)}, {x ∈ \left[\dfrac{1}{2}, 1\right]}
Let {x = \cos θ}
{\cos^{-1} x = θ}
Now, as {x ∈ \left[\dfrac{1}{2}, 1\right]}
{⇒ (4x^3 - 3x) ∈ [-1, 1]}
Also, as {x ∈ \left[\dfrac{1}{2}, 1\right]}
{⇒ θ ∈ \left[0, \dfrac{π}{3}\right]}
{⇒ 3θ ∈ [0, π]}
We have,
R.H.S.
=
{\cos^{-1} (4x^3 - 3x)}
=
{\cos^{-1} (4\cos^3 θ - 3\cos θ)}
=
{\cos^{-1} (\cos 3θ)}
=
{3θ} {(∵ \cos^{-1} (\cos x) = x)}
=
{3 \cos^{-1} x}
=
L.H.S.
3. Prove the following: {\tan^{-1} \dfrac{2}{11} + \tan^{-1} \dfrac{7}{24} = \tan^{-1} \dfrac{1}{2}}
L.H.S.
{= \tan^{-1} \dfrac{2}{11} + \tan^{-1} \dfrac{7}{24}}
{= \tan^{-1} \cfrac{\dfrac{2}{11} + \cfrac{7}{24}}{1 - \cfrac{2}{11} × \cfrac{7}{24}}} {(∵ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}}, when {xy \lt 1)}
{= \tan^{-1} \dfrac{\dfrac{(2 × 24) + (7 × 11)}{11 × 24}}{\dfrac{(11 × 24) - (2 × 7)}{11 × 24}}}
{= \tan^{-1} \dfrac{48 + 77}{264 - 14}}
{= \tan^{-1} \dfrac{125}{250}}
= {\tan^{-1} \dfrac{1}{2}}
= R.H.S.
4. Prove the following: {2\tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{7} = \tan^{-1} \dfrac{31}{17}}
L.H.S.
{= 2\tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{7}}
{= \tan^{-1} \cfrac{2 × \dfrac{1}{2}}{1 - \left(\cfrac{1}{2}\right)^2} + \tan^{-1} \dfrac{1}{7}} {\left(∵ 2\tan^{-1} x = \tan^{-1} \dfrac{2x}{1 - x^2}\right)}
{= \tan^{-1} \cfrac{1}{1 - \cfrac{1}{4}} + \tan^{-1} \dfrac{1}{7}}
{= \tan^{-1} \dfrac{1}{\left(\dfrac{4 - 1}{4}\right)} + \tan^{-1} \dfrac{1}{7}}
{= \tan^{-1} \dfrac{1}{\left(\dfrac{3}{4}\right)} + \tan^{-1} \dfrac{1}{7}}
{= \tan^{-1} \left(1 × \dfrac{4}{3}\right) + \tan^{-1} \dfrac{1}{7}}
{= \tan^{-1} \dfrac{4}{3} + \tan^{-1} \dfrac{1}{7}}
{= \tan^{-1} \dfrac{\dfrac{4}{3} + \dfrac{1}{7}}{1 - \dfrac{4}{3} × \dfrac{1}{7}}} ({∵ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}}, when {xy \lt 1)}
{= \tan^{-1} \dfrac{\dfrac{(4 × 7) + (3 × 1)}{3 × 7}}{\dfrac{(3 × 7) - (4 × 1)}{3 × 7}}}
{= \tan^{-1} \left(\dfrac{28 + 3}{21 - 4}\right)}
= {\tan^{-1} \dfrac{31}{17}}
= R.H.S.
5. Write the following function in the simplest form: {\tan^{-1} \dfrac{\sqrt{(1 + x^2)} - 1}{x}}, x ≠ 0
Let {x = \tan θ}
{⇒ \tan^{-1} x = θ}
Substituting {x = \tan θ} into the given function, we have
{\tan^{-1} \dfrac{\sqrt{(1 + x^2)} - 1}{x}}
{= \tan^{-1} \dfrac{\sqrt{(1 + \tan^2 θ)} - 1}{\tan θ}}
{= \tan^{-1} \dfrac{\sqrt{(\sec^2 θ)} - 1}{\tan θ}} {(∵ 1 + \tan^2 θ = \sec^2 θ)}
{= \tan^{-1} \dfrac{\sec θ - 1}{\tan θ}}
{= \tan^{-1} \dfrac{\dfrac{1}{\cos θ} - 1}{\dfrac{\sin θ}{\cos θ}}}
{= \tan^{-1} \dfrac{\dfrac{1 - \cos θ}{\cos θ}}{\dfrac{\sin θ}{\cos θ}}}
{= \tan^{-1} \dfrac{1 - \cos θ}{\sin θ}}
{= \tan^{-1} \dfrac{2\sin^2 \dfrac{θ}{2}}{2\sin \dfrac{θ}{2} \cos \dfrac{θ}{2}}} {(∵ 1 - \cos θ = 2 \sin^2 \dfrac{θ}{2}} and {\left.\sin θ = 2 \sin \dfrac{θ}{2} \cos \dfrac{θ}{2}\right)}
{= \tan^{-1} \dfrac{\sin \dfrac{θ}{2}}{\cos \dfrac{θ}{2}}}
{= \tan^{-1} \tan \dfrac{θ}{2}}
{= \tan^{-1} \tan \dfrac{θ}{2}}
{= \dfrac{θ}{2}} \text{ } {(∵ \tan^{-1} \tan x = x)}
{= \dfrac{\tan^{-1} x}{2}} \text{ } {(∵ θ = \tan^{-1} x)}
= {\dfrac{1}{2} \tan^{-1} x}
6. Write the following function in the simplest form: {\tan^{-1} \dfrac{1}{\sqrt{x^2 - 1}},} \text{ } {|x| \gt 1}
Let {x = \sec θ}
{⇒ \sec^{-1} x = θ}
Substituting {x = \sec θ} into the given function, we have
{\tan^{-1} \dfrac{1}{\sqrt{x^2 - 1}}}
{= \tan^{-1} \dfrac{1}{\sqrt{\sec^2 θ - 1}}}
{= \tan^{-1} \dfrac{1}{\sqrt{\tan^2 θ}}} {\left(∵ \sec^2 θ - 1 = \tan^2 θ\right)}
{= \tan^{-1} \dfrac{1}{\tan θ}}
{= \tan^{-1} \cot θ}
{= \tan^{-1} \tan \left(\dfrac{π}{2} - θ\right)}
{= \dfrac{π}{2} - θ} \text{ } {(∵ \tan^{-1} \tan α = α)}
= {\dfrac{π}{2} - \sec^{-1} x} {(∵ θ = \sec^{-1} x)}
7. Write the following function in the simplest form: {\tan^{-1} \left(\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right),} \text{ } {0 \lt x \lt π}
We know that {1 - \cos x = 2\sin^2 \dfrac{x}{2}} and {1 + \cos x = 2\cos^2 \dfrac{x}{2}}.
Substituting these in the given function, we have
{\tan^{-1} \left(\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right)}
{= \tan^{-1} \left(\sqrt{\cfrac{2\sin^2 \dfrac{x}{2}}{2\cos^2 \cfrac{x}{2}}}\right)}
{= \tan^{-1} \left(\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}\right)}
{= \tan^{-1} \tan \dfrac{x}{2}}
= \dfrac{x}{2}
8. Write the following function in the simplest form: {\tan^{-1} \left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right),} \text{ } {\dfrac{-π}{4} \lt x \lt \dfrac{3π}{4}}
We know that {\tan \dfrac{π}{4} = 1}. We will use this to simplify this function.
Now dividing the numerator and denominator of the argument with \cos x, we get
{\tan^{-1} \left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)}
{= \tan^{-1} \left(\dfrac{\dfrac{\cos x - \sin x}{\cos x}}{\dfrac{\cos x + \sin x}{\cos x}}\right)}
{= \tan^{-1} \left(\dfrac{1 - \tan x}{1 + \tan x}\right)}
{= \tan^{-1} \left(\dfrac{\tan \dfrac{π}{4} - \tan x}{\tan \dfrac{π}{4} + \tan x}\right)} {\left(∵ 1 = \tan \dfrac{π}{4}\right)}
{= \tan^{-1} \tan \left(\dfrac{π}{4} + x\right)}
= {\dfrac{π}{4} + x} {\left(∵ \tan^{-1} \tan x = x\right)}
9. Write the following function in the simplest form: {\tan^{-1} \dfrac{x}{\sqrt{a^2 - x^2}},} \text{ } {|x| \lt a}
Let x = a \sin θ.
{⇒ \dfrac{x}{a} = \sin θ}
{⇒ θ = \sin^{-1} \dfrac{x}{a}}
Substituting these into the given function, we have
{\tan^{-1} \dfrac{x}{\sqrt{a^2 - x^2}}}
=
{\tan^{-1} \dfrac{a \sin θ}{\sqrt{a^2 - (a \sin θ)^2}}}
=
{\tan^{-1} \dfrac{a \sin θ}{\sqrt{a^2 - a^2 \sin^2 θ}}}
=
{\tan^{-1} \dfrac{a \sin θ}{\sqrt{a^2(1 - \sin^2 θ)}}}
=
{\tan^{-1} \dfrac{a \sin θ}{\sqrt{a^2 \cos^2 θ}}}
=
{\tan^{-1} \dfrac{a \sin θ}{a \cos θ}}
=
{\tan^{-1} \tan θ}
=
θ \text{ } {(∵ \tan^{-1} \tan x = x)}
=
{\sin^{-1} \dfrac{x}{a}}
10. Write the following function in the simplest form: {\tan^{-1} \left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right),} \text{ } {a \gt 0;} {\dfrac{-a}{\sqrt{3}} \lt x \lt \dfrac{a}{\sqrt{3}}}
Let x = a \tan θ.
{⇒ \dfrac{x}{a} = \tan θ}
{⇒ θ = \tan^{-1} \dfrac{x}{a}}
Substituting these into the given function, we have
{\tan^{-1} \left(\dfrac{3a^2x - x^3}{a^3 - 3ax^2}\right)}
{\tan^{-1} \left(\dfrac{3a^2.a\tan θ - (a\tan θ)^3}{a^3 - 3a(a\tan θ)^2}\right)}
{= \tan^{-1} \left(\dfrac{3a^3.\tan θ - a^3\tan^3 θ}{a^3 - 3a^3\tan^2 θ}\right)}
{= \tan^{-1} \left(\dfrac{a^3\left(3\tan θ - \tan^3 θ\right)}{a^3\left(1 - 3\tan^2 θ\right)}\right)}
{= \tan^{-1} \tan 3θ}
= 3θ
= {3\tan^{-1} \dfrac{x}{a}}
11. Find the value of each of the following: {\tan^{-1} \left[2\cos\left(2\sin^{-1} \dfrac{1}{2}\right)\right]}
{\tan^{-1} \left[2\cos\left(2\sin^{-1} \dfrac{1}{2}\right)\right]}
{= \tan^{-1} \left[2\cos(\left(2\sin^{-1} \sin \dfrac{π}{6}\right)\right]} {\left(∵ \sin \dfrac{π}{6} = \dfrac{1}{2}\right)}
{= \tan^{-1} \left[2\cos \left(2 × \dfrac{π}{6}\right)\right]} {\left(∵ \sin^{-1} \sin x = x\right)}
{= \tan^{-1} \left[2\cos \left(\dfrac{π}{3}\right)\right]}
{= \tan^{-1} \left(2 × \dfrac{1}{2}\right)} {\left(∵ \cos \dfrac{π}{3} = \dfrac{1}{2}\right)}
{= \tan^{-1} 1}
{= \tan^{-1} \tan \dfrac{π}{4}} {\left(∵ 1 = \tan \dfrac{π}{4}\right)}
= \dfrac{π}{4} {\left(∵ \tan^{-1} \tan x = x\right)}
12. Find the value of each of the following: {\cot\text{ }(\tan^{-1} a + \cot^{-1} a)}
{\cot\text{ }(\tan^{-1} a + \cot^{-1} a)}
{= \cot \dfrac{π}{2}} {\left(∵ \tan^{-1} x + \cot^{-1} x = \dfrac{π}{2}\right)}
= 0
13. Find the value of each of the following: {\tan \dfrac{1}{2}\left[\sin^{-1} \dfrac{2x}{1 + x^2} + \cos^{-1} \dfrac{1 - y^2}{1 + y^2}\right]}, {|x| \lt 1}, {y \gt 0} and {xy \lt 1}
Let x = \tan α and y = \tan β
{⇒ α = \tan x} and {β = \tan y}
Substituting these into the given function, we have
{\tan \dfrac{1}{2}\left[\sin^{-1} \dfrac{2x}{1 + x^2} + \cos^{-1} \dfrac{1 - y^2}{1 + y^2}\right]}
{= \tan \dfrac{1}{2}\left[\sin^{-1} \dfrac{2\tan α}{1 + \tan^2 α} + \cos^{-1} \dfrac{1 - \tan^2 β}{1 + \tan^2 β}\right]} (substituting {x = \tan α} and {y = \tan β})
{= \tan \dfrac{1}{2}\left(\sin^{-1} \sin 2α + \cos^{-1} \cos 2β\right)} {\left(∵ \dfrac{2\tan α}{1 + \tan^2 α} = \sin 2α\right.} and {\left.\dfrac{1 - \tan^2 β}{1 + \tan^2 β} = \cos 2β\right)}
{= \tan \dfrac{1}{2}(2α + 2β)} {\left(∵ \sin^{-1} \sin θ = θ\right.} and {\left.\cos^{-1} \cos θ = θ\right)}
{= \tan \left[\dfrac{1}{2}× 2 × (α + β)\right]}
{= \tan (α + β)}
{= \dfrac{\tan α + \tan β}{1 - \tan α\tan β}}
= \dfrac{x + y}{1 - xy} (substituting {\tan α = x} and {\tan β = y})
14. If {\sin \left(\sin^{-1} \dfrac{1}{5} + \cos^{-1} x\right) = 1}, then find the value of x.
We know that {\cos^{-1} x + \sin^{-1} x = \dfrac{π}{2}}.
We’ll use this identity to sold this equation.
The given equation is
{\sin \left(\sin^{-1} \dfrac{1}{5} + \cos^{-1} x\right) = 1}
{⇒ \sin \left(\sin^{-1} \dfrac{1}{5} + \cos^{-1} x\right) = \sin \dfrac{π}{2}}
{⇒ \sin^{-1} \dfrac{1}{5} + \cos^{-1} x = \dfrac{π}{2}}
{⇒ \cos^{-1} x = \dfrac{π}{2} - \sin^{-1} \dfrac{1}{5}}
{⇒ \cos^{-1} x = \cos^{-1} \dfrac{1}{5}}
{x =} {\dfrac{1}{5}}
15. If {\tan^{-1} \dfrac{x - 1}{x - 2} + \tan^{-1} \dfrac{x + 1}{x + 2} = \dfrac{π}{4}}, then find the value of x
We use {\tan x + \tan y = \tan \dfrac{x + y}{1 - xy}}, to solve this equation.
The given equation is
{\tan^{-1} \dfrac{x - 1}{x - 2} + \tan^{-1} \dfrac{x + 1}{x + 2} = \dfrac{π}{4}}
{⇒ \tan^{-1} \dfrac{\dfrac{x - 1}{x - 2} + \dfrac{x + 1}{x + 2}}{1 - \dfrac{x - 1}{x - 2} × \dfrac{x + 1}{x + 2}} = \dfrac{π}{4}}
{⇒ \tan^{-1} \dfrac{\dfrac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2)}}{\dfrac{(x - 2)(x + 2) - (x - 1)(x + 1)}{(x - 2)(x + 2)}} = \dfrac{π}{4}}
{⇒ \tan^{-1} \dfrac{x^2 + 2x - x - 2 + x^2 - 2x + x - 2}{(x^2 - 4) - (x^2 - 1)} = \dfrac{π}{4}}
{⇒ \tan^{-1} \dfrac{2x^2 - 4}{-3} = \dfrac{π}{4}}
{⇒ \dfrac{2x^2 - 4}{-3} = \tan \dfrac{π}{4}}
{⇒ \dfrac{2x^2 - 4}{-3} = 1} {\left(∵ \tan \dfrac{π}{4} = 1\right)}
{⇒ 2x^2 - 4 = -3}
{⇒ 2x^2 = 4 - 3}
{⇒ 2x^2 = 1}
{⇒ x^2 = \dfrac{1}{2}}
{⇒ x = \sqrt\dfrac{1}{2}}
{x = ±\dfrac{1}{\sqrt{2}}}
16. Find the values of each of the expressions in Exercises 16 to 18. {\sin^{-1} \sin \dfrac{2π}{3}}
We know that {\sin^{-1} \sin x = x}, when the range of principal value branch of sin-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]}.
However, we can’t apply this to the given function as the argument {\dfrac{2π}{3} ∉ \left[-\dfrac{π}{2}, \dfrac{π}{2}\right]}
So, we need to simplify the argument so that it belongs to the range {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]}
{\sin^{-1} \sin \dfrac{2π}{3}}
=
{\sin^{-1} \sin \left(π - \dfrac{π}{3}\right)}
=
{\sin^{-1} \sin \dfrac{π}{3}}
=
\dfrac{π}{3}
17. Find the values of each of the expressions in Exercises 16 to 18. {\tan^{-1} \tan \dfrac{3π}{4}}
We know that {\tan^{-1} \tan x = x}, when the range of principal value branch of tan-1 is {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)}.
However, we can’t apply this to the given function as the argument {\dfrac{3π}{4} ∉ \left(-\dfrac{π}{2}, \dfrac{π}{2}\right)}
So, we need to simplify the argument so that it belongs to the range {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)}
{\tan^{-1} \tan \dfrac{3π}{4}}
=
{\tan^{-1} \tan \left(π - \dfrac{π}{4}\right)}
=
{\tan^{-1} \left(-\tan \dfrac{π}{4}\right)}
=
{-\tan^{-1} \tan \dfrac{π}{4}} {\left(∵ \tan^{-1} (-x) = - \tan^{-1} x\right)}
=
-\dfrac{π}{4}
18. Find the values of each of the expressions in Exercises 16 to 18. {\tan \left(\sin^{-1} \dfrac{3}{5} + \cot^{-1} \dfrac{3}{2}\right)}
Let
{\sin^{-1} \dfrac{3}{5}}
=
α
{⇒ \sin α}
=
\dfrac{3}{5}
{⇒ \cos α}
=
{\sqrt{1 - \sin^2 α}}
=
{\sqrt{1 - \left(\dfrac{3}{5}\right)^2}}
=
{\sqrt{1 - \dfrac{9}{25}}}
=
{\sqrt{\dfrac{25 - 9}{25}}}
=
{\sqrt{\dfrac{16}{25}}}
=
{\dfrac{4}{5}}
{⇒ \tan α}
=
{\dfrac{\sin α}{\cos α}}
=
{\dfrac{3/5}{4/5}}
=
{\dfrac{3}{4}}
{⇒ α}
=
{\tan^{-1} \dfrac{3}{4}}
{⇒ \sin^{-1} \dfrac{3}{5}}
=
{\tan^{-1} \dfrac{3}{4}}
We also have,
{\cot^{-1} \dfrac{3}{2}}
=
{\tan^{-1} \dfrac{2}{3}} {\left(∵ \cot^{-1} x = \tan^{-1} \dfrac{1}{x}\right)}
Substituting these equivalents into the given function, we have
{\tan \left(\sin^{-1} \dfrac{3}{5} + \cot^{-1} \dfrac{3}{2}\right)}
{= \tan \left(\tan^{-1} \dfrac{3}{4} + \tan^{-1} \dfrac{2}{3}\right)}
{= \tan \tan^{-1} \dfrac{\dfrac{3}{4} + \dfrac{2}{3}}{1 - \dfrac{3}{4} × \dfrac{2}{3}}} {\left(∵ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}\right)}
{= \tan \tan^{-1} \dfrac{\dfrac{(3 × 3) + (4 × 2)}{4 × 3}}{\dfrac{(4 × 3) - (3 × 2)}{4 × 3}}}
{= \tan \tan^{-1} \dfrac{9 + 8}{12 - 6}}
{= \tan \tan^{-1} \dfrac{17}{6}}
= {\dfrac{17}{6}} {\left(∵ \tan \tan^{-1} x = x\right)}
19. {\cos^{-1}\left(\cos \dfrac{7π}{6}\right)} is equal to
(A) \dfrac{7π}{6}
(B) \dfrac{5π}{6}
(C) \dfrac{π}{3}
(D) \dfrac{π}{6}

We know that {\cos^{-1} \cos x = x}, when the range of principal value branch of cos-1 is {[0, π]}.
However, we can’t apply this to the given function as the argument {\dfrac{7π}{6} ∉ [0, π]}
So, we need to simplify the argument so that it belongs to the range {[0, π]}
{\cos^{-1} \cos \dfrac{7π}{6}}
=
{\cos^{-1} \cos \left(π + \dfrac{π}{6}\right)}
=
{\cos^{-1} \left(-\cos \dfrac{π}{6}\right)}
=
{π - \cos^{-1} \cos \dfrac{π}{6}} {\left(∵ \cos^{-1} (-x) = π - \cos^{-1} x\right)}
=
{π - \dfrac{π}{6}} {\left(∵ \cos \cos^{-1} x = x\right)}
=
\dfrac{5π}{6}
So, option B is the correct answer.
20. {\sin \left(\dfrac{π}{3} - \sin^{-1} \left(-\dfrac{1}{2}\right)\right)} is equal to
(A) \dfrac{1}{2}
(B) \dfrac{1}{3}
(C) \dfrac{1}{4}
(D) 1 ✔

Let {\sin^{-1}\left(-\dfrac{1}{2}\right) = y}. Then,
{\sin y}
=
-\dfrac{1}{2}
=
{\sin\left(-\dfrac{π}{6}\right)}
We know that the range of principal value branch of sin-1 is {\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]} and {\sin\left(-\dfrac{π}{6}\right) = -\dfrac{1}{2}}.
∴ The principal value of {\sin^{-1}\left(-\dfrac{1}{2}\right)} is {-\dfrac{π}{6}}
Substituting this into the given function, we have
{\sin \left(\dfrac{π}{3} - \sin^{-1} \left(-\dfrac{1}{2}\right)\right)}
=
{\sin \left(\dfrac{π}{3} - \left(-\dfrac{π}{6}\right)\right)}
=
{\sin \left(\dfrac{π}{3} + \dfrac{π}{6}\right)}
=
{\sin \dfrac{π}{2}}
=
1
So, option D is the correct answer.
21. {\tan^{-1} \sqrt{3} - \cot^{-1}\left(-\sqrt{3}\right)} is equal to
(A) π
(B) -\dfrac{π}{2}
(C) \dfrac{1}{4}
(D) 1

Let {\tan^{-1} \sqrt{3} = x}. Then,
{\tan x}
=
\sqrt{3}
=
{\tan \left(\dfrac{π}{3}\right)}
We know that the range of principal value branch of tan-1 is {\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)} and {\tan \left(\dfrac{π}{3}\right) = \sqrt{3}}.
∴ The principal value of {\tan^{-1} \left(\sqrt{3}\right)} is {\dfrac{π}{3}}
Now let {\cot^{-1} \left(-\sqrt{3}\right) = y}. Then,
{\cot y}
=
-\sqrt{3}
=
{\cot \left(\dfrac{5π}{6}\right)}
We know that the range of principal value branch of cot-1 is (0, π) and {\cot \left(\dfrac{5π}{6}\right) = -\sqrt{3}}.
∴ The principal value of {\cot^{-1} \left(-\sqrt{3}\right)} is {\dfrac{5π}{6}}
Substituting, we have
{\tan^{-1} \sqrt{3} - \cot^{-1}\left(-\sqrt{3}\right)}
=
{\dfrac{π}{3} - \dfrac{5π}{6}}
=
{\dfrac{2π - 5π}{6}}
=
{\dfrac{-3π}{6}}
=
{-\dfrac{π}{2}}
So, option B is the correct answer.