NCERT Solutions For Class 12 Maths Chapter 2 Exercise 2.2

Go back to  'Inverse Trigonometric Functions'

Chapter 2 Ex.2.2 Question 1

Prove \(3{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right),x \in \left[ { - \frac{1}{2},\frac{1}{2}} \right]\).

Solution

Let \(x = \sin {\rm{\theta }}\)

Hence, \({\sin ^{ - 1}}\left( x \right) = {\rm{\theta }}\)

Now,

\[\begin{align}RHS &= {\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)\\&= {\sin ^{ - 1}}\left( {3\sin {\rm{\theta }} - 4{{\sin }^3}{\rm{\theta }}} \right)\\&= {\sin ^{ - 1}}\left( {\sin 3{\rm{\theta }}} \right)\\&= 3{\rm{\theta }}\\&= 3{\sin ^{ - 1}}x\\&= LHS\end{align}\]

Chapter 2 Ex.2.2 Question 2

Prove \(3{\cos ^{ - 1}}x = {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right),x \in \left[ {\frac{1}{2},1} \right]\).

Solution

Let \(x = \cos {\rm{\theta }}\)

Hence, \({\cos ^{ - 1}}\left( x \right) = {\rm{\theta }}\)

Now,

\[\begin{align}RHS &= {\cos ^{ - 1}}\left( {4{x^3} - 3x} \right)\\&= {\cos ^{ - 1}}\left( {4{{\cos }^3}{\rm{\theta }} - 3\cos {\rm{\theta }}} \right)\\&= {\cos ^{ - 1}}\left( {\cos 3{\rm{\theta }}} \right)\\&= 3{\rm{\theta }}\\&= 3{\cos ^{ - 1}}x\\&= LHS\end{align}\]

Chapter 2 Ex.2.2 Question 3

Prove \({\tan ^{ - 1}}\frac{2}{{11}} + {\tan ^{ - 1}}\frac{7}{{24}} = {\tan ^{ - 1}}\frac{1}{2}\).

Solution

Since we know that \({\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\frac{{x + y}}{{1 - xy}}\)

Now,

\[\begin{align}LHS &= {\tan ^{ - 1}}\frac{2}{{11}} + {\tan ^{ - 1}}\frac{7}{{24}}\\&= {\tan ^{ - 1}}\frac{{\frac{2}{{11}} + \frac{7}{{24}}}}{{1 - \frac{2}{{11}} \cdot \frac{7}{{24}}}}\\&= {\tan ^{ - 1}}\left( {\frac{{\frac{{48 + 77}}{{264}}}}{{\frac{{264 - 14}}{{264}}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{125}}{{250}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)\\&= RHS\end{align}\]

Chapter 2 Ex.2.2 Question 4

Prove \(2{\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{1}{7} = {\tan ^{ - 1}}\frac{{31}}{{17}}\).

Solution

Since we know that \(2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\) and \({\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\frac{{x + y}}{{1 - xy}}\)

Now,

\[\begin{align}LHS &= 2{\tan ^{ - 1}}\frac{1}{2} + {\tan ^{ - 1}}\frac{1}{7}\\&= {\tan ^{ - 1}}\frac{{2 \times \frac{1}{2}}}{{1 - {{\left( {\frac{1}{2}} \right)}^2}}} + {\tan ^{ - 1}}\frac{1}{7}\\&= {\tan ^{ - 1}}\left( {\frac{4}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\frac{4}{3} + \frac{1}{7}}}{{1 - \frac{4}{3} \cdot \frac{1}{7}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\frac{{28 + 3}}{{21}}}}{{\frac{{21 - 4}}{{21}}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{31}}{{17}}} \right)\\&= RHS\end{align}\]

Chapter 2 Ex.2.2 Question 5

Write the function in the simplest form: \({\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}}- 1}}{x},x \ne 0\)

Solution

Let \(x = \tan \theta\Rightarrow \theta= {\tan ^{ - 1}}x\)

Hence,

\[\begin{align}{\tan ^{ - 1}}\frac{{\sqrt {1 + {x^2}}- 1}}{x} &= {\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {{\tan }^2}\theta }- 1}}{{\tan \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\sec \theta- 1}}{{\tan \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{2{{\sin }^2}\frac{\theta }{2}}}{{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)\\&= {\tan ^{ - 1}}\left( {\tan \frac{\theta }{2}} \right)\\&= \frac{\theta }{2}\\&= \frac{1}{2}{\tan ^{ - 1}}x\end{align}\]

Chapter 2 Ex.2.2 Question 6

Write the function in the simplest form:

\({\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }},\left| x \right| > 1\)

Solution

Let \(x = {\text{cosec}}\theta\Rightarrow \theta= {\text{cose}}{{\text{c}}^{ - 1}}x\)

Hence,\({\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }},\left| x \right| > 1\)

\[\begin{align}{\tan ^{ - 1}}\frac{1}{{\sqrt {{x^2} - 1} }} &= {\tan ^{ - 1}}\frac{1}{{\sqrt {{\rm{cose}}{{\rm{c}}^2}\theta- 1} }}\\&= {\tan ^{ - 1}}\left( {\frac{1}{{\cot \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\tan \theta } \right)\\&= \theta \\&= \cos e{c^{ - 1}}x\\&= \frac{\pi }{2} - {\sec ^{ - 1}}x\end{align}\]

Chapter 2 Ex.2.2 Question 7

Write the function in the simplest form: \({\tan ^{ - 1}}\left( {\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right),0 < x < \pi \)

Solution

Since,\(1 - \cos x = 2{\sin ^2}\frac{x}{2}\)and \(1 + \cos x = 2{\cos ^2}\frac{x}{2}\)

Hence,

\[\begin{align}{\tan ^{ - 1}}\left( {\sqrt {\frac{{1 - \cos x}}{{1 + \cos x}}} } \right) &= {\tan ^{ - 1}}\left( {\sqrt {\frac{{2{{\sin }^2}\frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} } \right)\\&= {\tan ^{ - 1}}\left( {\frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}} \right)\\&= {\tan ^{ - 1}}\left( {\tan \frac{x}{2}} \right)\\&= \frac{x}{2}\end{align}\]

Chapter 2 Ex.2.2 Question 8

Write the function in the simplest form: \({\tan ^{ - 1}}\left( {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right),0 < x < \pi \)

Solution

\[\begin{align}{\tan ^{ - 1}}\left( {\frac{{\cos x - \sin x}}{{\cos x + \sin x}}} \right) &= {\tan ^{ - 1}}\left( {\frac{{\frac{{\cos x - \sin x}}{{\cos x}}}}{{\frac{{\cos x + \sin x}}{{\cos x}}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{1 - \frac{{\sin x}}{{\cos x}}}}{{1 + \frac{{\sin x}}{{\cos x}}}}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{1 - \tan x}}{{1 + \tan x}}} \right)\\&= {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( {\tan x} \right)\\&= \frac{\pi }{4} - x\end{align}\]

Chapter 2 Ex.2.2 Question 9

Write the function in the simplest form: \({\tan ^{ - 1}}\frac{x}{{\sqrt {{a^2} - {x^2}} }},\left| x \right| < a\)

Solution

Let \(x = a\sin \theta\Rightarrow \theta= {\sin ^{ - 1}}\left( {\frac{x}{a}} \right)\)

Hence,

\[\begin{align}{\tan ^{ - 1}}\frac{x}{{\sqrt {{a^2} - {x^2}} }}&= {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{\sqrt {{a^2} - {a^2}{{\sin }^2}\theta } }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{a\sqrt {1 - {{\sin }^2}\theta } }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{a\sin \theta }}{{a\cos \theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\tan \theta } \right)\\&= \theta \\&= {\sin ^{ - 1}}\frac{x}{a}\end{align}\]

Chapter 2 Ex.2.2 Question 10

Write the function in the simplest form: \({\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right),a > 0;\frac{{ - a}}{{\sqrt 3 }} \le x \le \frac{a}{{\sqrt 3 }}\)

Solution

Let \(x = a\tan \theta\Rightarrow \theta= {\tan ^{ - 1}}\left( {\frac{x}{a}} \right)\)

Hence,

\[\begin{align}{\tan ^{ - 1}}\left( {\frac{{3{a^2}x - {x^3}}}{{{a^3} - 3a{x^2}}}} \right) &= {\tan ^{ - 1}}\left( {\frac{{3{a^2} \cdot a\tan \theta- {a^3}{{\tan }^3}\theta }}{{{a^3} - 3a.{a^2}{{\tan }^2}\theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\frac{{3{a^3}\tan \theta- {a^3}{{\tan }^3}\theta }}{{{a^3} - 3{a^3}{{\tan }^2}\theta }}} \right)\\&= {\tan ^{ - 1}}\left( {\tan 3\theta } \right)\\&= 3\theta \\&= 3{\tan ^{ - 1}}\frac{x}{a}\end{align}\]

Chapter 2 Ex.2.2 Question 11

Write the function in the simplest form: \({\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right]\)

Solution

Let \({\sin ^{ - 1}}\frac{1}{2} = x\)

Hence,

\[\begin{align}\sin x &= \frac{1}{2}\\&= \sin \left( {\frac{\pi }{6}} \right)\\x &= \frac{\pi }{6}\\{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) &= \frac{\pi }{6}\end{align}\]

Therefore,

\[\begin{align}{\tan ^{ - 1}}\left[ {2\cos \left( {2{{\sin }^{ - 1}}\frac{1}{2}} \right)} \right] &= {\tan ^{ - 1}}\left[ {2\cos \left( {2 \times \frac{\pi }{6}} \right)} \right]\\&= {\tan ^{ - 1}}\left[ {2\cos \frac{\pi }{3}} \right]\\&= {\tan ^{ - 1}}\left[ {2 \times \frac{1}{2}} \right]\\&= {\tan ^{ - 1}}\left[ 1 \right]\\&= \frac{\pi }{4}\end{align}\]

Chapter 2 Ex.2.2 Question 12

Find the value of \(\cot \left( {{{\tan }^{ - 1}}a + {{\cot }^{ - 1}}a} \right)\)

Solution

Since \({\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \frac{\pi }{2}\)

Hence,

\[\begin{align}\cot \left( {{{\tan }^{ - 1}}a + {{\cot }^{ - 1}}a} \right) &= \cot \left( {\frac{\pi }{2}} \right)\\&= 0\end{align}\]

Chapter 2 Ex.2.2 Question 13

Find the value of \(\tan \frac{1}{2}\left( {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right),\left| x \right| < 1,y > 0\) and \(xy < 1\).

Solution

Let \(x = \tan \theta\Rightarrow \theta= {\tan ^{ - 1}}x\)

Hence,

\[\begin{align}{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} &= {\sin ^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)\\&= {\sin ^{ - 1}}\left( {\sin 2\theta } \right)\\&= 2\theta \\&= 2{\tan ^{ - 1}}x\end{align}\]

Now, let \(y = \tan \phi\Rightarrow \phi= {\tan ^{ - 1}}y\)

Hence,

\[\begin{align}{\cos ^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}} &= {\cos ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\phi }}{{1 + {{\tan }^2}\phi }}} \right)\\&= {\cos ^{ - 1}}\left( {\cos 2\phi } \right)\\&= 2\phi \\&= 2{\tan ^{ - 1}}y\end{align}\]

Therefore,

\[\begin{align}\tan \frac{1}{2}\left( {{{\sin }^{ - 1}}\frac{{2x}}{{1 + {x^2}}} + {{\cos }^{ - 1}}\frac{{1 - {y^2}}}{{1 + {y^2}}}} \right) &= \tan \frac{1}{2}\left( {2{{\tan }^{ - 1}}x + 2{{\tan }^{ - 1}}y} \right)\\&= \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y} \right)\\&= \tan \left[ {{{\tan }^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)} \right]\\&= \left( {\frac{{x + y}}{{1 - xy}}} \right)\end{align}\]

Chapter 2 Ex.2.2 Question 14

If \(\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1\), find the value of \(x\).

Solution

It is given that \(\sin \left( {{{\sin }^{ - 1}}\frac{1}{5} + {{\cos }^{ - 1}}x} \right) = 1\)

Since we know that \(\sin \left( {x + y} \right) = \sin x\cos y + \cos x\sin y\)

Therefore,

\[\begin{align}&\sin \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\cos \left( {{{\cos }^{ - 1}}x} \right) + \cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\sin \left( {{{\cos }^{ - 1}}x} \right) = 1\\&\left( {\frac{1}{5}} \right) \times \left( x \right) + \cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\sin \left( {{{\cos }^{ - 1}}x} \right) = 1\\&\frac{x}{5} + \cos \left( {{{\sin }^{ - 1}}\frac{1}{5}} \right)\sin \left( {{{\cos }^{ - 1}}x} \right) = 1\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

Now, let \({\sin ^{ - 1}}\frac{1}{5} = y \Rightarrow \sin y = \frac{1}{5}\)

Then,

\[\begin{align}\cos y &= \sqrt {1 - {{\left( {\frac{1}{5}} \right)}^2}} \\&= \frac{{2\sqrt 6 }}{5}\\y &= {\cos ^{ - 1}}\left( {\frac{{2\sqrt 6 }}{5}} \right)\end{align}\]

Therefore,

\({\sin ^{ - 1}}\frac{1}{5} = {\cos ^{ - 1}}\left( {\frac{{2\sqrt 6 }}{5}} \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\)

Now, let \({\cos ^{ - 1}}x = z \Rightarrow \cos z = x\)

Then,

\[\begin{align}&\sin z = \sqrt {1 - {x^2}} \\&z = {\sin ^{ - 1}}\sqrt {1 - {x^2}}\end{align}\]

Therefore,

\({\cos ^{ - 1}}x = {\sin ^{ - 1}}\sqrt {1 - {x^2}} \;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\)

From \(\left( 1 \right),\left( 2 \right)\) and \(\left( 3 \right)\), we have

\[\begin{align}&\Rightarrow \frac{x}{5} + \cos \left( {{{\cos }^{ - 1}}\frac{{2\sqrt 6 }}{5}} \right)\sin \left( {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } \right) = 1\\&\Rightarrow \frac{x}{5} + \frac{{2\sqrt 6 }}{5}\sqrt {1 - {x^2}}= 1\\&\Rightarrow x + 2\sqrt 6 \sqrt {1 - {x^2}}= 5\\&\Rightarrow 5 - x = 2\sqrt 6 \sqrt {1 - {x^2}}\end{align}\]

On squaring both the sides

\[\begin{align}25 + {x^2} - 10x &= 24 - 24{x^2}\\25{x^2} - 10x + 1& = 0\\{\left( {5x - 1} \right)^2} &= 0\\\left( {5x - 1} \right) &= 0\\x &= \frac{1}{5}\end{align}\]

Chapter 2 Ex.2.2 Question 15

If \({\tan ^{ - 1}}\frac{{x - 1}}{{x - 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x - 1}} = \frac{\pi }{4}\), find the value of \(x\).

Solution

It is given that \({\tan ^{ - 1}}\frac{{x - 1}}{{x - 2}} + {\tan ^{ - 1}}\frac{{x + 1}}{{x - 1}} = \frac{\pi }{4}\)

Since \({\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x + y}}{{1 - xy}}} \right)\)

Therefore,

\[\begin{align}&\Rightarrow {\tan ^{ - 1}}\left( {\frac{{\frac{{x - 1}}{{x - 2}} + \frac{{x + 1}}{{x + 2}}}}{{1 - \left( {\frac{{x - 1}}{{x - 2}}} \right)\left( {\frac{{x + 1}}{{x + 2}}} \right)}}} \right) = \frac{\pi }{4}\\&\Rightarrow {\tan ^{ - 1}}\left[ {\frac{{\left( {x - 1} \right)\left( {x + 2} \right) + \left( {x + 1} \right)\left( {x - 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right) - \left( {x - 1} \right)\left( {x + 1} \right)}}} \right] = \frac{\pi }{4}\\&\Rightarrow {\tan ^{ - 1}}\left[ {\frac{{{x^2} + x - 2 + {x^2} - x - 2}}{{{x^2} - 4 - {x^2} + 1}}} \right] = \frac{\pi }{4}\\&\Rightarrow {\tan ^{ - 1}}\left[ {\frac{{2{x^2} - 4}}{{ - 3}}} \right] = \frac{\pi }{4}\\&\Rightarrow \tan \left[ {{{\tan }^{ - 1}}\frac{{4 - 2{x^2}}}{3}} \right] = \tan \frac{\pi }{4}\\&\Rightarrow \frac{{4 - 2{x^2}}}{3} = 1\\&\Rightarrow 4 - 2{x^2} = 3\\&\Rightarrow 2{x^2} = 1\\&\Rightarrow {x^2} = \frac{1}{2}\\&\Rightarrow x =\pm \frac{1}{{\sqrt 2 }}\end{align}\]

Chapter 2 Ex.2.2 Question 16

Find the value of \({\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right)\).

Solution

Since, \(\sin {\rm{\theta }} = \sin \left( {\pi- {\rm{\theta }}} \right)\)

Therefore,

\[\begin{align}{\sin ^{ - 1}}\left( {\sin \frac{{2\pi }}{3}} \right)& = {\sin ^{ - 1}}\left[ {\sin \left( {\pi- \frac{{2\pi }}{3}} \right)} \right]\\&= {\sin ^{ - 1}}\left( {\sin \frac{\pi }{3}} \right)\\&= \frac{\pi }{3}\end{align}\]

Chapter 2 Ex.2.2 Question 17

Find the value of \({\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right)\).

Solution

Since, \(\tan {\rm{\theta }} = \tan \left( {\pi- {\rm{\theta }}} \right)\)

Therefore,

\[\begin{align}{\tan ^{ - 1}}\left( {\tan \frac{{3\pi }}{4}} \right) &= {\tan ^{ - 1}}\left[ { - \tan \left( { - \frac{{3\pi }}{4}} \right)} \right]\\&= {\tan ^{ - 1}}\left[ { - \tan \left( {\pi- \frac{\pi }{4}} \right)} \right]\\&= {\tan ^{ - 1}}\left[ { - \tan \left( {\frac{\pi }{4}} \right)} \right]\\&= {\tan ^{ - 1}}\left[ {\tan \left( { - \frac{\pi }{4}} \right)} \right]\\&= \left( { - \frac{\pi }{4}} \right)\end{align}\]

Chapter 2 Ex.2.2 Question 18

Find the value of \(\tan \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cot }^{ - 1}}\frac{3}{2}} \right)\).

Solution

Let \({\sin ^{ - 1}}\frac{3}{5} = x \Rightarrow \sin x = \frac{3}{5}\)

Then,

\[\begin{align}&\Rightarrow \cos x = \sqrt {1 - {{\sin }^2}x}= \frac{4}{5}\\&\Rightarrow \sec x = \frac{5}{4}\end{align}\]

Therefore,

\[\begin{align}\tan x &= \sqrt {{{\sec }^2}x - 1} \\&= \sqrt {\frac{{25}}{{16}} - 1} \\&= \frac{3}{4}\\x &= {\tan ^{ - 1}}\frac{3}{4}\\{\sin ^{ - 1}}\frac{3}{5} &= {\tan ^{ - 1}}\frac{3}{4}\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

Now,

\[{\cot ^{ - 1}}\frac{3}{2} = {\tan ^{ - 1}}\frac{2}{3}\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\]

Thus, by using \(\left( 1 \right)\) and \(\left( 2 \right)\)

\[\begin{align}\tan \left( {{{\sin }^{ - 1}}\frac{3}{5} + {{\cot }^{ - 1}}\frac{2}{3}} \right) &= \tan \left( {{{\tan }^{ - 1}}\frac{3}{4} + {{\tan }^{ - 1}}\frac{2}{3}} \right)\\&= \tan \left[ {{{\tan }^{ - 1}}\frac{{\frac{3}{4} + \frac{2}{3}}}{{1 - \frac{3}{4} \cdot \frac{2}{3}}}} \right]\\&= \tan \left( {{{\tan }^{ - 1}}\frac{{17}}{6}} \right)\\&= \frac{{17}}{6}\end{align}\]

Chapter 2 Ex.2.2 Question 19

\({\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right)\) is equal to

(A) \(\frac{{7\pi }}{6}\)

(B) \(\frac{{5\pi }}{6}\)

(C) \(\frac{\pi }{3}\)

(D) \(\frac{\pi }{6}\)

Solution

\[\begin{align}{\cos ^{ - 1}}\left( {\cos \frac{{7\pi }}{6}} \right) &= {\cos ^{ - 1}}\left( {\cos \frac{{ - 7\pi }}{6}} \right)\\&= {\cos ^{ - 1}}\left[ {\cos \left( {2\pi- \frac{{7\pi }}{6}} \right)} \right]\\&= {\cos ^{ - 1}}\left[ {\cos \left( {\frac{{5\pi }}{6}} \right)} \right]\\&= \frac{{5\pi }}{6}\end{align}\]

Thus, the correct option is B.

Chapter 2 Ex.2.2 Question 20

\(\sin \left( {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right)\) is equal to

(A) \(\frac{1}{2}\)

(B) \(\frac{1}{3}\)

(C) \(\frac{1}{4}\)

(D) \(1\)

Solution

Let \({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) = x\)

Hence,

\[\begin{align}\sin x &=- \frac{1}{2}\\&=- \sin \frac{\pi }{6}\\&= \sin \left( { - \frac{\pi }{6}} \right)\\x &=- \frac{\pi }{6}\end{align}\]

Since, Range of principal value of\({\sin ^{ - 1}}\left( x \right) = \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]\).

Therefore,

\({\sin ^{ - 1}}\left( { - \frac{1}{2}} \right) =- \frac{\pi }{6}\)

Then,

\[\begin{align}\sin \left( {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right) &= \sin \left( {\frac{\pi }{3} + \frac{\pi }{6}} \right)\\&= \sin \left( {\frac{\pi }{2}} \right)\\&= 1\end{align}\]

Thus, the correct option is D.

Chapter 2 Ex.2.2 Question 21

Find the values of \({\tan ^{ - 1}}\sqrt 3- {\cot ^{ - 1}}\left( { - \sqrt 3 } \right)\) is equal to

(A) \(\pi \)

(B) \( - \frac{\pi }{2}\)

(C) \(0\)

(D) \(2\sqrt 3 \)

Solution

Let \({\tan ^{ - 1}}\sqrt 3= x\)

Hence,

\(\tan x = \sqrt 3= \tan \frac{\pi }{3}\), where \(\frac{\pi }{3} \in \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)\)

Therefore, \({\tan ^{ - 1}}\sqrt 3= \frac{\pi }{3}\)

Now, let \({\cot ^{ - 1}}\left( { - \sqrt 3 } \right) = y\)

Hence,

\[\begin{align}\cot y &= \left( { - \sqrt 3 } \right)\\&=- \cot \left( {\frac{\pi }{6}} \right)\\&= \cot \left( {\pi- \frac{\pi }{6}} \right)\\&= \cot \left( {\frac{{5\pi }}{6}} \right)\end{align}\]

Since, Range of principal value of\({\cot ^{ - 1}}x = \left( {0,\pi } \right)\)

Therefore,

\[{\cot ^{ - 1}}\left( { - \sqrt 3 } \right) = \frac{{5\pi }}{6}\]

Then,

\[\begin{align}{\tan ^{ - 1}}\sqrt 3- {\cot ^{ - 1}}\left( { - \sqrt 3 } \right) &= \frac{\pi }{3} - \frac{{5\pi }}{6}\\&=- \frac{\pi }{2}\end{align}\]

Thus, the correct option is B.

  
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