# Miscellaneous Exercise on Chapter 2 Solutions

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Miscellaneous Excercise on Chapter 2 Solutions
1. Find the value of the following: {\cos^{-1} \left(\cos \dfrac{13π}{6}\right)}
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{13π}{6} ∉ [0, π]}
So, we need to simplify the argument so that it belongs to the range {[0, π]}
{\cos^{-1} \cos \dfrac{13π}{6}}
=
{\cos^{-1} \cos \left(2π + \dfrac{π}{6}\right)}
=
{\cos^{-1} \cos \dfrac{π}{6}}
=
\dfrac{π}{6}
2. Find the value of the following: {\tan^{-1} \left(\tan \dfrac{7π}{6}\right)}
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{7π}{6} ∉ \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{7π}{6}}
=
{\tan^{-1} \tan \left(π + \dfrac{π}{6}\right)}
=
{\tan^{-1} \tan \dfrac{π}{6}}
=
\dfrac{π}{6}
2. Prove that {2\sin^{-1} \dfrac{3}{5} = \tan^{-1} \dfrac{24}{7}}
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{7π}{6} ∉ \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{7π}{6}}
=
{\tan^{-1} \tan \left(π + \dfrac{π}{6}\right)}
=
{\tan^{-1} \tan \dfrac{π}{6}}
=
\dfrac{π}{6}
3. Prove that {2\sin^{-1} \dfrac{3}{5} = \tan \dfrac{24}{7}}
Let’s first express sin^{-1} interms of tan^{-1} so that we can simplify it using the identity {2\tan^{-1} x = \tan^{-1} \dfrac{2x}{1 - x^2}}
Let
{\sin^{-1} \dfrac{3}{5}}
=
θ
{⇒ \sin θ}
=
\dfrac{3}{5}
{⇒ \cos θ}
=
\sqrt{1 - \sin^2 θ}
=
\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 θ}
=
\dfrac{3}{4}
{⇒ θ}
=
{\tan^{-1} \dfrac{3}{4}}
{⇒ \sin^{-1} \dfrac{3}{5}}
=
{\tan^{-1} \dfrac{3}{4}}
Substituting, we have
L.H.S.
{= 2\sin^{-1} \dfrac{3}{5}}
{= 2\tan^{-1} \dfrac{3}{4}}
{= \tan^{-1} \left(\dfrac{2 × \dfrac{3}{4}}{1 - \left(\dfrac{3}{4}\right)^2}\right)} {\left(∵ 2\tan^{-1} x = \tan^{-1} \dfrac{2x}{1 - x^2}\right)}
{= \tan^{-1} \left(\dfrac{2 × \dfrac{3}{4}}{1 - \dfrac{9}{16}}\right)}
{= \tan^{-1} \left(\dfrac{2 × \dfrac{3}{4}}{\dfrac{16 - 9}{16}}\right)}
{= \tan^{-1} \left(\dfrac{2 × \dfrac{3}{4}}{\dfrac{7}{16}}\right)}
{= \tan^{-1} \left(2 × \dfrac{3}{4} × \dfrac{16}{7}\right)}
{= \tan^{-1} \dfrac{24}{7}}
= R.H.S.
4. Prove that {\sin^{-1} \dfrac{8}{17} + \sin^{-1} \dfrac{3}{5} = \tan \dfrac{77}{36}}
Let’s first express both the sin^{-1} functions on the L.H.S. interms of tan^{-1} so that we can simplify it using the identity {\tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}}
Let
{\sin^{-1} \dfrac{8}{17}}
=
α
{⇒ \sin α}
=
\dfrac{8}{17}
{⇒ \cos α}
=
\sqrt{1 - \sin^2 α}
=
\sqrt{1 - \left(\dfrac{8}{17}\right)^2}
=
\sqrt{1 - \dfrac{64}{289}}
=
\sqrt{\dfrac{289 - 64}{289}}
=
\sqrt{\dfrac{225}{289}}
=
\dfrac{15}{17}
{⇒ \tan α}
=
\dfrac{8}{15}
{⇒ α}
=
{\tan^{-1} \dfrac{8}{15}}
{⇒ \sin^{-1} \dfrac{8}{17}}
=
{\tan^{-1} \dfrac{8}{15}}
Similarly, let
{\sin^{-1} \dfrac{3}{5}}
=
β
{⇒ \sin β}
=
\dfrac{3}{5}
{⇒ \cos β}
=
\sqrt{1 - \sin^2 β}
=
\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 β}
=
\dfrac{3}{4}
{⇒ β}
=
{\tan^{-1} \dfrac{3}{4}}
{⇒ \sin^{-1} \dfrac{3}{5}}
=
{\tan^{-1} \dfrac{3}{4}}
Substituting, we have
L.H.S.
{= \sin \dfrac{8}{17} + \sin^{-1} \dfrac{3}{5}}
{= \tan^{-1} \dfrac{8}{15} + \tan^{-1} \dfrac{3}{4}}
{= \tan^{-1} \left(\dfrac{\dfrac{8}{15} + \dfrac{3}{4}}{1 - \dfrac{8}{15} × \dfrac{3}{4}}\right)} {\left(∵ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}\right)}
{= \tan^{-1} \left(\dfrac{\dfrac{(8 × 4) + (15 × 3)}{15 × 4}}{\dfrac{(15 × 4) - (8 × 3)}{15 × 4}}\right)}
{= \tan^{-1} \left(\dfrac{32 + 45}{60 - 24}\right)}
{= \tan^{-1} \dfrac{77}{36}}
= R.H.S.
5. Prove that {\cos^{-1} \dfrac{4}{5} + \cos^{-1} \dfrac{12}{13} = \cos \dfrac{33}{65}}
As the right hand side is expressed interms of \cos^{-1}, let’s first find the values of the left hand side operands of the equation interms of \sin^{-1} also, so that we can use the identity
cos (A + B) = cos A cos B – sin A sin B
Let
A
=
{\cos^{-1} \dfrac{4}{5}}
⇒ cos A
=
\dfrac{4}{5}
⇒ sin A
=
\sqrt{1 - \cos^2 \text{A}}
=
\sqrt{1 - \left(\dfrac{4}{5}\right)^2}
=
\sqrt{1 - \dfrac{16}{25}}
=
\sqrt{\dfrac{25 - 16}{25}}
=
\sqrt{\dfrac{9}{25}}
=
\dfrac{3}{5}
Similarly, let
B
=
{\cos^{-1} \dfrac{12}{13}}
⇒ cos B
=
\dfrac{12}{13}
⇒ sin B
=
\sqrt{1 - \cos^2 \text{B}}
=
\sqrt{1 - \left(\dfrac{12}{13}\right)^2}
=
\sqrt{1 - \dfrac{144}{169}}
=
\sqrt{\dfrac{169 - 144}{169}}
=
\sqrt{\dfrac{25}{169}}
=
\dfrac{5}{13}
Now, let’s consider
cos (A + B)
=
cos A cos B – sin A sin B
=
{\dfrac{4}{5} × \dfrac{12}{13} - \dfrac{3}{5} × \dfrac{5}{13}}
=
{\dfrac{48}{65} - \dfrac{15}{65}}
=
{\dfrac{48 - 15}{65}}
=
{\dfrac{33}{65}}
So,
cos (A + B)
=
\dfrac{33}{65}
⇒ A + B
=
{\cos^{-1} \dfrac{33}{65}}
{\cos^{-1} \dfrac{4}{5} + \cos^{-1} \dfrac{12}{13} = \cos^{-1} \dfrac{33}{65}} (∵ as per our initial assumption {\text{A} = \cos^{-1} \dfrac{4}{5}} and {\left.\text{B} = \cos^{-1} \dfrac{12}{13}\right)}
Hence, proved.
6. Prove that {\cos^{-1} \dfrac{12}{13} + \sin^{-1} \dfrac{3}{5} = \sin^{-1} \dfrac{56}{65}}
As the right hand side is expressed interms of \sin^{-1}, let’s first find the values of the left hand side operands of the equation interms of \cos^{-1} as well as \cos^{-1}, so that we can use the identity
sin (A + B) = sin A cos B + cos A sin B
Let
A
=
{\cos^{-1} \dfrac{12}{13}}
⇒ cos A
=
\dfrac{12}{13}
⇒ sin A
=
\sqrt{1 - \cos^2 \text{A}}
=
\sqrt{1 - \left(\dfrac{12}{13}\right)^2}
=
\sqrt{1 - \dfrac{144}{169}}
=
\sqrt{\dfrac{169 - 144}{169}}
=
\sqrt{\dfrac{25}{169}}
=
\dfrac{5}{13}
Similarly, let
B
=
{\sin^{-1} \dfrac{3}{5}}
⇒ sin B
=
\dfrac{3}{5}
⇒ cos B
=
\sqrt{1 - \sin^2 \text{B}}
=
\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}
Now, let’s consider
sin (A + B)
=
sin A cos B + cos A sin B
=
{\dfrac{5}{13} × \dfrac{4}{5} + \dfrac{12}{13} × \dfrac{3}{5}}
=
{\dfrac{20}{65} + \dfrac{36}{65}}
=
{\dfrac{20 + 36}{65}}
=
{\dfrac{56}{65}}
So,
sin (A + B)
=
\dfrac{56}{65}
⇒ A + B
=
{\sin^{-1} \dfrac{56}{65}}
{\cos^{-1} \dfrac{12}{13} + \sin^{-1} \dfrac{3}{5} = \sin^{-1} \dfrac{56}{65}} (∵ as per our initial assumption {\text{A} = \cos^{-1} \dfrac{12}{13}} and {\left.\text{B} = \sin^{-1} \dfrac{3}{5}\right)}
Hence, proved.
7. Prove that {\tan^{-1} \dfrac{63}{16} = \sin^{-1} \dfrac{5}{13} + \cos^{-1} \dfrac{3}{5}}
Let’s solve this problem, by considering the right hand side of the equation. Let’s first express both the terms on the right hand side interms of \tan^{-1} so that we can use the identity
{\tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}}
Let
α
=
{\sin^{-1} \dfrac{5}{13}}
⇒ sin α
=
\dfrac{5}{13}
⇒ cos α
=
\sqrt{1 - \sin^2 α}
=
\sqrt{1 - \left(\dfrac{5}{13}\right)^2}
=
\sqrt{1 - \dfrac{25}{169}}
=
\sqrt{\dfrac{169 - 25}{169}}
=
\sqrt{\dfrac{144}{169}}
=
\dfrac{12}{13}
⇒ tan α
=
\dfrac{5}{12}
Similarly, let
β
=
{\cos^{-1} \dfrac{3}{5}}
⇒ cos β
=
\dfrac{3}{5}
⇒ sin β
=
\sqrt{1 - \cos^2 β}
=
\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 β
=
\dfrac{4}{3}
Now, let’s consider
R.H.S.
=
{\sin^{-1} \dfrac{5}{13} + \cos^{-1} \dfrac{3}{5}}
=
{\tan^{-1} \dfrac{5}{12} + \tan^{-1} \dfrac{4}{3}}
=
{\tan^{-1} \left(\dfrac{\dfrac{5}{12} + \dfrac{4}{3}}{1 - \dfrac{5}{12} × \dfrac{4}{3}}\right)}
=
{\tan^{-1} \left(\dfrac{\dfrac{(5 × 3) + (12 × 4)}{12 × 3}}{\dfrac{(12 × 3) - (5 × 4)}{12 × 3}}\right)}
=
{\tan^{-1} \left(\dfrac{15 + 48}{36 - 20}\right)}
=
{\tan^{-1} \dfrac{63}{16}}
=
L.H.S.
Hence, proved.
8. Prove that {\tan^{-1} \dfrac{1}{5} + \tan^{-1} \dfrac{1}{7} + \tan^{-1} \dfrac{1}{3} + \tan^{-1} \dfrac{1}{8} = \dfrac{π}{4}}
To solve this problem let’s start with L.H.S. and take two terms as a group so that we can use the identity
{\tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}}
L.H.S.
{= \left(\tan^{-1} \dfrac{1}{5} + \tan^{-1} \dfrac{1}{7}\right) + \left(\tan^{-1} \dfrac{1}{3} + \tan^{-1} \dfrac{1}{8}\right)}
{= \tan^{-1} \left(\dfrac{\dfrac{1}{5} + \dfrac{1}{7}}{1 - \dfrac{1}{5} × \dfrac{1}{7}}\right) + \tan^{-1} \left(\dfrac{\dfrac{1}{3} + \dfrac{1}{8}}{1 - \dfrac{1}{3} × \dfrac{1}{8}}\right)}
{= \tan^{-1} \left(\dfrac{\dfrac{(1 × 7) + (5 × 1)}{5 × 7}}{\dfrac{(5 × 7) - (1 × 1)}{5 × 7}}\right) + \left(\tan^{-1} \dfrac{\dfrac{(1 × 8) + (3 × 1)}{3 × 8}}{\dfrac{(3 × 8) - (1 × 1)}{3 × 8}}\right)} {\left(∵ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}\right)}
{= \tan^{-1} \left(\dfrac{7 + 5}{35 - 1}\right) + \tan^{-1} \left(\dfrac{8 + 3}{24 - 1}\right)}
{= \tan^{-1} \dfrac{12}{34} + \tan^{-1} \dfrac{11}{23}}
{= \tan^{-1} \dfrac{6}{17} + \tan^{-1} \dfrac{11}{23}}
{= \tan^{-1} \left(\dfrac{\dfrac{6}{17} + \dfrac{11}{23}}{1 - \dfrac{6}{17} × \dfrac{11}{23}}\right)} {\left(∵ \tan^{-1} x + \tan^{-1} y = \tan^{-1} \dfrac{x + y}{1 - xy}\right)}
{= \tan^{-1} \left(\dfrac{\dfrac{(6 × 23) + (17 × 11)}{17 × 23}}{\dfrac{(17 × 23) - (11 × 6)}{17 × 23}}\right)}
{= \tan^{-1} \left(\dfrac{138 + 187}{391 - 66}\right)}
{= \tan^{-1} \dfrac{325}{325}}
{= \tan^{-1} 1}
= \dfrac{π}{4}
= R.H.S.
Hence, proved.
9. Prove that {\tan^{-1} \sqrt{x} = \dfrac{1}{2} \cos^{-1} \dfrac{1 - x}{1 + x}}, {x ∈ [0, 1]}
We use the following identity to solve this problem.
{\cos^{-1} \dfrac{1 - x^2}{1 + x^2} = 2\tan^{-1} x}
R.H.S.
=
{\dfrac{1}{2}\cos^{-1} \dfrac{1 - x}{1 + x}}
=
{\dfrac{1}{2}\cos^{-1} \dfrac{1 - (\sqrt{x})^2}{1 + (\sqrt{x})^2}}
=
{\dfrac{1}{2} × 2\tan^{-1} \sqrt{x}} {(∵ \cos^{-1} \dfrac{1 - x^2}{1 + x^2} = 2\tan^{-1} x)}
=
{\tan^{-1} \sqrt{x}}
=
L.H.S.
Hence, proved.
10. Prove that {\cot^{-1} \left(\dfrac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right) = \dfrac{x}{2}}, {x ∈ \left(0, \dfrac{π}{4}\right)}
We know that {1 = \sin^2 \dfrac{x}{2} + \cos^2 \dfrac{x}{2}} and {\sin x = 2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}}
Also, {x ∈ \left(0, \dfrac{π}{4}\right)}, {\dfrac{x}{2} ∈ \left(0, \dfrac{π}{8}\right)} and in the range {\left(0, \dfrac{π}{8}\right)}, we have {\cos \dfrac{x}{2} - \sin \dfrac{x}{2} \gt 0}
Substituting these, we will have
{1 + \sin x}
=
{\cos^2 \dfrac{x}{2} + \sin^2 \dfrac{x}{2} + 2 \cos \dfrac{x}{2} \sin \dfrac{x}{2}}
=
{\left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)^2}
{∴ \sqrt{1 + \sin x}}
=
{\sqrt{\left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)^2}}
=
{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}
Similarly,
{1 - \sin x}
=
{\cos^2 \dfrac{x}{2} + \sin^2 \dfrac{x}{2} - 2 \cos \dfrac{x}{2} \sin \dfrac{x}{2}}
=
{\left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right)^2}
{∴ \sqrt{1 - \sin x}}
=
{\sqrt{\left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right)^2}}
=
{\cos \dfrac{x}{2} - \sin \dfrac{x}{2}}
Substituting these values in the L.H.S., we have
L.H.S.
{= \cot^{-1} \left(\dfrac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}\right)}
{= \cot^{-1} \left(\dfrac{\left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right) + \left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right)}{\left(\cos \dfrac{x}{2} + \sin \dfrac{x}{2}\right) - \left(\cos \dfrac{x}{2} - \sin \dfrac{x}{2}\right)}\right)}
{= \cot^{-1} \left(\dfrac{\cos \dfrac{x}{2} + \sin \dfrac{x}{2} + \cos \dfrac{x}{2} - \sin \dfrac{x}{2}}{\cos \dfrac{x}{2} + \sin \dfrac{x}{2} - \cos \dfrac{x}{2} + \sin \dfrac{x}{2}}\right)}
{= \cot^{-1} \left(\dfrac{2\cos \dfrac{x}{2}}{2\sin \dfrac{x}{2}}\right)}
{= \cot^{-1} \cot \dfrac{x}{2}}
= \dfrac{x}{2}
= R.H.S.
Hence, proved.
11. Prove that {\tan^{-1} \left(\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}\right) = \dfrac{π}{4} - \dfrac{1}{2}\cos^{-1} x}, {-\dfrac{1}{\sqrt{2}} \le x \le 1} [Hint: Put {x = \cos 2θ}]
Let {x = \cos 2θ}
{⇒ \cos^{-1} x = 2θ}
{⇒ θ = \dfrac{1}{2}\cos^{-1} x}
Substituting into the terms on the L.H.S., we have
{1 + x}
=
{1 + \cos 2θ}
=
{2\cos^2 θ}
{⇒ \sqrt{(1 + x)}}
=
{\sqrt{2\cos^2 θ}}
=
{\sqrt{2}\cos θ}
{1 - x}
=
{1 - \cos 2θ}
=
{2\sin^2 θ}
{⇒ \sqrt{(1 - x)}}
=
{\sqrt{2\sin^2 θ}}
=
{\sqrt{2}\sin θ}
Substituting these values on the L.H.S. of the given equation, we have
L.H.S.
{= \tan^{-1} \left(\dfrac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}\right)}
{= \tan^{-1} \left(\dfrac{\sqrt{2}\cos θ - \sqrt{2}\sin θ}{\sqrt{2}\cos θ + \sqrt{2}\sin θ}\right)}
Dividing each term in the numerator and denominator with {\sqrt{2}\cos θ}), we have
L.H.S.
{= \tan^{-1} \left(\dfrac{\dfrac{\sqrt{2}\cos θ}{\sqrt{2}\cos θ} - \dfrac{\sqrt{2}\sin θ}{\sqrt{2}\cos θ}}{\dfrac{\sqrt{2}\cos θ}{\sqrt{2}\cos θ} + \dfrac{\sqrt{2}\sin θ}{\sqrt{2}\cos θ}}\right)}
{= \tan^{-1} \left(\dfrac{1 - \tan θ}{1 + \tan θ}\right)}
{= \tan^{-1} \left(\dfrac{\tan \dfrac{π}{4} - \tan θ}{\tan \dfrac{π}{4} + \tan θ}\right)} {\left(∵ 1 = \tan \dfrac{π}{4}\right)}
{= \tan^{-1} \tan \left(\dfrac{π}{4} - θ\right)} {\left(∵ \dfrac{\tan x - \tan y}{1 - \tan x \tan y} = \tan (x - y)\right)}
{= \dfrac{π}{4} - θ}
= {\dfrac{π}{4} - \dfrac{1}{2}\cos^{-1} x} (∵ as per our initial assumption {θ = \dfrac{1}{2}\cos^{-1} x)}
= R.H.S.
12. Prove that {\dfrac{9π}{8} - \dfrac{9}{4}\sin^{-1} \dfrac{1}{3} = \dfrac{9}{4} \sin^{-1} \dfrac{2\sqrt{2}}{3}}
We know that {\sin^{-1} x + \cos^{-1} x = \dfrac{π}{2}}
So, we'll use this in the problem as {\dfrac{π}{2} = \sin^{-1} \dfrac{1}{3} + \cos^{-1} \dfrac{1}{3}}
Now,
L.H.S.
=
{\dfrac{9π}{8} - \dfrac{9}{4}\sin^{-1} \dfrac{1}{3}}
=
{\dfrac{9}{4}\left(\dfrac{π}{2} - \sin^{-1} \dfrac{1}{3}\right)}
=
{\dfrac{9}{4}\cos^{-1} \dfrac{1}{3}} \left(∵ \sin^{-1} x + \cos^{-1} x = \dfrac{π}{2}\right)
=
{\dfrac{9}{4}\sin^{-1} \sqrt{1 - \left(\dfrac{1}{3}\right)^2}} {\left(∵ \cos^{-1} x = \sin^{-1} \sqrt{\left(1 - x^2\right)}\right)}
=
{\dfrac{9}{4}\sin^{-1} \sqrt{1 - \dfrac{1}{9}}}
=
{\dfrac{9}{4}\sin^{-1} \sqrt{\dfrac{9 - 1}{9}}}
=
{\dfrac{9}{4}\sin^{-1} \sqrt{\dfrac{8}{9}}}
=
{\dfrac{9}{4}\sin^{-1} \dfrac{2\sqrt{2}}{3}}
=
R.H.S.
13. Solve the following equations: {2\tan^{-1} (\cos x) = \tan^{-1} (2 \cosec x)}
The given equation is
{2\tan^{-1} (\cos x) = \tan^{-1} (2 \cosec x)}
{⇒ \tan^{-1} \dfrac{2\cos x}{1 - \cos^2 x} = \tan^{-1} 2 × \dfrac{1}{\sin x}} {\left(∵ 2\tan^{-1} x = \tan^{-1} \dfrac{2x}{1 - x^2}\right)}
{⇒ \dfrac{2\cos x}{\sin^2 x} = 2 × \dfrac{1}{\sin x}}
{⇒ \dfrac{\cos x}{\sin x} = 1}
{⇒ \cot x = 1}
{x = \dfrac{π}{4}} {\left(∵ \cot \dfrac{π}{4} = 1\right)}
14. Solve the following equations: {\tan^{-1} \dfrac{1 - x}{1 + x} = \dfrac{1}{2}\tan^{-1} x}
We know that {\tan^{-1} \dfrac{x - y}{x + y} = \tan^{-1} x - \tan^{-1} y}
We'll apply this into the given equation.
The given equation is
{\tan^{-1} \dfrac{1 - x}{1 + x} = \dfrac{1}{2}\tan^{-1} x}
{⇒ \tan^{-1} 1 - \tan^{-1} x = \dfrac{1}{2}\tan^{-1} x} {\left(∵ \tan^{-1} \dfrac{x + y}{x - y} = \tan^{-1} x - \tan^{-1} x\right)}
{⇒ \dfrac{π}{4} - \tan^{-1} x = \dfrac{1}{2}\tan^{-1} x} {\left(∵ \tan^{-1} 1 = \dfrac{π}{4}\right)}
{⇒ \dfrac{π}{4} = \tan^{-1} x + \dfrac{1}{2}\tan^{-1} x}
{⇒ \dfrac{π}{4} = \dfrac{3}{2}\tan^{-1} x}
{⇒ \dfrac{π}{4} × \dfrac{2}{3} = \tan^{-1} x}
{⇒ \tan^{-1} x = \dfrac{π}{4} × \dfrac{2}{3}}
{⇒ \tan^{-1} x = \dfrac{π}{6}}
{⇒ x = \tan \dfrac{π}{6}}
{x = \dfrac{1}{\sqrt{3}}}
15. {\sin \left(\tan^{-1} x\right)}, {|x| \lt 1} is equal to
(A) {\dfrac{x}{\sqrt{1 - x^2}}}
(B) {\dfrac{1}{\sqrt{1 - x^2}}}
(C) {\dfrac{1}{\sqrt{1 + x^2}}}
(D) {\dfrac{x}{\sqrt{1 + x^2}}}

We know that
{\tan^{-1} x = \sin^{-1} \dfrac{x}{\sqrt{1 + x^2}}}
Substituting, we get
{\sin \left(\tan^{-1} x\right)}
=
{\sin \left(\sin^{-1} \dfrac{x}{\sqrt{1 + x^2}}\right)}
=
{\dfrac{x}{\sqrt{1 + x^2}}} {\left(∵ \sin^{-1} \sin x = x\right)}
So, option D is the correct answer.
16. {\sin^{-1} (1 - x) - 2 \sin^{-1} x = \dfrac{π}{2}}, then x is equal to
(A) 0, \dfrac{1}{2}
(B) 1, {\dfrac{1}{2}}
(C) 0 ✔
(D) \dfrac{1}{2}

Let {\sin^{-1} x = θ}
{⇒ x = \sin θ}
Now {\cos 2θ}
=
{1 - 2\sin^2 θ}
=
{1 - 2x^2}
⇒ 2θ
=
{\cos^{-1} (1 - 2x^2)}
{⇒ 2\sin^{-1} x}
=
{\cos^{-1} (1 - 2x^2)}
We'll apply this into the given equation to solve it.
Now {\cos 2θ = 1 - 2\sin^2 θ}
The given equation is
{\sin^{-1} (1 - x) - 2 \sin^{-1} x = \dfrac{π}{2}}
{⇒ \sin^{-1} (1 - x) = \dfrac{π}{2} + 2 \sin^{-1} x}
{⇒ 1 - x = \sin \left(\dfrac{π}{2} + 2 \sin^{-1} x\right)}
{⇒ 1 - x = \cos \left(2 \sin^{-1} x\right)}
{⇒ 1 - x = \cos \cos^{-1} \left(1 - 2x^2\right)} {\left(∵ 2\sin^{-1} x = \cos^{-1} \left(1 - 2x^2\right)\right)}
{⇒ 1 - x = 1 - 2x^2}
{⇒ 2x^2 - x = 0}
{⇒ x(2x - 1) = 0}
{⇒ x = 0 \text{ or } 2x - 1 = 0}
{⇒ x = 0 \text{ or } x = \dfrac{1}{2}}
However, when {x = \dfrac{1}{2}}, the given equation is not satisfied.
So, {x = 0}
So, option C is the corect answer
17. {\tan^{-1} \dfrac{x}{y} - \tan^{-1} \dfrac{x - y}{x + y}} is equal to
(A) \dfrac{π}{2}
(B) \dfrac{π}{3}
(C) \dfrac{π}{4}
(D) \dfrac{3π}{4}

Dividing the numerator and denominator of the second term's argument in the given function with {y,} we have
{\tan^{-1} \dfrac{x}{y} - \tan^{-1} \dfrac{x - y}{x + y}}
{= \tan^{-1} \dfrac{x}{y} - \tan^{-1} \dfrac{\dfrac{x}{y} - \dfrac{y}{y}}{\dfrac{x}{y} + \dfrac{y}{y}}}
{= \tan^{-1} \dfrac{x}{y} - \tan^{-1} \dfrac{\dfrac{x}{y} - 1}{\dfrac{x}{y} + 1}}
{= \tan^{-1} \dfrac{x}{y} - \tan^{-1} \dfrac{\dfrac{x}{y} - 1}{1 + 1 × \dfrac{x}{y}}}
{= \tan^{-1} \dfrac{x}{y} - \left(\tan^{-1} \dfrac{x}{y} - \tan^{-1} 1\right)} {\left(∵ \tan^{-1} \dfrac{x - y}{1 + xy} = \tan^{-1} x - \tan^{-1} y\right)}
{= \tan^{-1} \dfrac{x}{y} - \tan^{-1} \dfrac{x}{y} + \tan^{-1} 1}
{= \tan^{-1} 1}
= {\dfrac{π}{4}} {\left(∵ 1 = \tan \dfrac{π}{4}\right)}
So, option C is the correct answer.