CBSE Class 12 Mathematics Inverse Trigonometric Functions Assignment Set 02

Question. Write the value of \(\cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)\).
Answer: Given \(\cos^{-1}\left(-\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \cos^{-1}\left(\cos \frac{2\pi}{3}\right) + 2\sin^{-1}\left(\sin \frac{\pi}{6}\right) = \frac{2\pi}{3} + 2 \times \frac{\pi}{6} = \frac{2\pi}{3} + \frac{\pi}{3} = \pi\).
\([\because \text{Range of } \cos^{-1} \text{ is } [0, \pi] \text{ & of } \sin^{-1} \text{ is } [-\pi/2, \pi/2]]\)

Question. Write the principal value of \(\tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right]\). 
Answer: Here, \(\tan^{-1}\left[\sin\left(-\frac{\pi}{2}\right)\right] = \tan^{-1}(-1) = -\frac{\pi}{4}\).
This is the required principal value as it should lie in \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).

Question. Find the value of the following : \(\cot\left(\frac{\pi}{2} - 2\cot^{-1}\sqrt{3}\right)\). 
Answer: \[\cot\left(\frac{\pi}{2} - 2\cot^{-1}\sqrt{3}\right)\] \[= \cot\left(\frac{\pi}{2} - 2\cot^{-1}\left(\cot\frac{\pi}{6}\right)\right) = \cot\left(\frac{\pi}{2} - 2 \cdot \frac{\pi}{6}\right)\] \[= \cot\left(\frac{\pi}{2} - \frac{\pi}{3}\right) = \cot\frac{\pi}{6} = \sqrt{3}\]

Question. Write the principal value of \(\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right)\). 
Answer: \[\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right)\] \[= \tan^{-1}\left(\tan\frac{\pi}{4}\right) + \cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{\pi}{4} + \frac{2\pi}{3} = \frac{11\pi}{12}\]

Question. Write the value of \(\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]\).
Answer: \[\tan^{-1}\left[2\sin\left(2\cos^{-1}\frac{\sqrt{3}}{2}\right)\right]\] \[= \tan^{-1}\left[2\sin\left(2 \cdot \frac{\pi}{6}\right)\right] = \tan^{-1}\left[2\sin\frac{\pi}{3}\right]\] \[= \tan^{-1}\left[2 \cdot \frac{\sqrt{3}}{2}\right] = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\]

Question. Write the principal value of \(\left[\cos^{-1}\frac{\sqrt{3}}{2} + \cos^{-1}\left(-\frac{1}{2}\right)\right]\).
Answer: \[\cos^{-1}\frac{\sqrt{3}}{2} + \cos^{-1}\left(-\frac{1}{2}\right)\] \[= \cos^{-1}\left(\cos\frac{\pi}{6}\right) + \cos^{-1}\left(\cos\frac{2\pi}{3}\right) = \frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6}\]

Question. Write the principal value of \([\tan^{-1}(-\sqrt{3}) + \tan^{-1}(1)]\). 
Answer: \[\tan^{-1}(-\sqrt{3}) + \tan^{-1}(1)\] \[= \tan^{-1}\left(-\tan\frac{\pi}{3}\right) + \tan^{-1}\left(\tan\frac{\pi}{4}\right)\] \[= \tan^{-1}\left(\tan\left(-\frac{\pi}{3}\right)\right) + \frac{\pi}{4} = -\frac{\pi}{3} + \frac{\pi}{4} = -\frac{\pi}{12}\]

Question. Write the principal value of \(\cos^{-1}\left(\frac{1}{2}\right) - 2\sin^{-1}\left(-\frac{1}{2}\right)\). 
Answer: \[\cos^{-1}\left(\frac{1}{2}\right) - 2\sin^{-1}\left(-\frac{1}{2}\right)\] \[= \cos^{-1}\left(\cos\frac{\pi}{3}\right) - 2\sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right) = \frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{2\pi}{3}\]

Question. Using principal values, write the value of \(\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)\). 
Answer: Principal value of \(\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{2\pi}{3}\).

Question. Evaluate : \(\sin\left[\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right]\).
Answer: \[\sin\left[\frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right)\right] = \sin\left[\frac{\pi}{3} - \sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right)\right]\] \[= \sin\left[\frac{\pi}{3} + \frac{\pi}{6}\right] = \sin\frac{\pi}{2} = 1\]

Question. Write the principal value of \(\sin^{-1}\left(-\frac{1}{2}\right)\).
Answer: Let \(\sin^{-1}\left(-\frac{1}{2}\right) = \theta\).
Then, \(\sin\theta = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)\), where \(-\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Hence, the principal value of \(\sin^{-1}\left(-\frac{1}{2}\right)\) is \(-\frac{\pi}{6}\).

Question. Using principal values, write the value of \(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\). 
Answer: The principal value of \(\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)
\[= \sin^{-1}\left(\sin\left(-\frac{\pi}{3}\right)\right) = -\frac{\pi}{3}\], where \(-\frac{\pi}{3} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Question. Find the principal value of \(\sin^{-1}\left(-\frac{1}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right)\).
Answer: \[\sin^{-1}\left(-\frac{1}{2}\right) + \cos^{-1}\left(-\frac{1}{2}\right)\] \[= \sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right) + \cos^{-1}\left(\cos\frac{2\pi}{3}\right) = -\frac{\pi}{6} + \frac{2\pi}{3} = \frac{\pi}{2}\]

Question. Find the principal value of \(\sec^{-1}(-2)\).
Answer: Let \(\sec^{-1}(-2) = y\). Then, \(\sec y = -2\).
\(\sec y = -2 = -\sec\frac{\pi}{3} = \sec\left(\pi - \frac{\pi}{3}\right) = \sec\frac{2\pi}{3}\).
We know that the range of principal value branch of \(\sec^{-1}\) is \([0, \pi] - \left\{\frac{\pi}{2}\right\}\) and \(\sec\frac{2\pi}{3} = -2\).
Hence, principal value of \(\sec^{-1}(-2) = \frac{2\pi}{3}\).

Question. Using the principal values, evaluate the following : \(\tan^{-1}(1) + \sin^{-1}\left(-\frac{1}{2}\right)\). 
Answer: \[\tan^{-1}(1) + \sin^{-1}\left(-\frac{1}{2}\right)\] \[= \tan^{-1}(1) + \sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right) = \frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}\]
\(\therefore\) Required principal value is \(\frac{\pi}{12}\).

Question. Find the principal value of \(\tan^{-1}(-1)\). 
Answer: Let \(\tan^{-1}(-1) = x \Rightarrow -1 = \tan x\).
We know that the range of principal value branch of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
Then, \(-1 = \tan\left(-\frac{\pi}{4}\right)\), where \(-\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).
Hence, the principal value of \(\tan^{-1}(-1)\) is \(-\frac{\pi}{4}\).

Question. Find the principal value of \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\).
Answer: Principal value of \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \cos^{-1}\left(\cos\frac{\pi}{6}\right) = \frac{\pi}{6}\), where \(\frac{\pi}{6} \in [0, \pi]\).

Question. Find the value of the following: \(\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)\).
Answer: Here, \(\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)\)
\[= \tan^{-1}\left(\tan\frac{\pi}{4}\right) + \cos^{-1}\left(\cos\frac{2\pi}{3}\right) + \sin^{-1}\left(\sin\left(-\frac{\pi}{6}\right)\right)\] \[= \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{4}\]

SA (4 marks)

Question. Prove that \(\cos^{-1}\frac{12}{13} + \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{56}{33}\). 
Answer: Let \(x = \cos^{-1}\frac{4}{5}\) and \(y = \cos^{-1}\frac{12}{13}\)
\(\Rightarrow \cos x = \frac{4}{5}\) and \(\cos y = \frac{12}{13}\)
Now, \(\sin x = \sqrt{1 - \cos^2 x}\) and \(\sin y = \sqrt{1 - \cos^2 y}\)
\(\Rightarrow \sin x = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}\) and \(\sin y = \sqrt{1 - \frac{144}{169}} = \frac{5}{13}\)
We know that, \(\cos(x + y) = \cos x \cos y - \sin x \sin y\)
\[= \frac{4}{5} \times \frac{12}{13} - \frac{3}{5} \times \frac{5}{13}\]
\(\Rightarrow \cos(x+y) = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}\)
\(\Rightarrow x+y = \cos^{-1}\frac{33}{65}\)
\(\therefore \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}\)
Now, \(\cos^{-1} x = \tan^{-1}\frac{\sqrt{1-x^2}}{x}\)
\(\therefore \cos^{-1}\frac{33}{65} = \tan^{-1}\left(\frac{\sqrt{1-\left(\frac{33}{65}\right)^2}}{\frac{33}{65}}\right) = \tan^{-1}\left(\frac{\frac{56}{65}}{\frac{33}{65}}\right) = \tan^{-1}\left(\frac{56}{33}\right)\)
\(\therefore \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \tan^{-1}\frac{56}{33}\)

Question. Prove that : \(\cos^{-1}\frac{12}{13} + \sin^{-1}\frac{3}{5} = \sin^{-1}\frac{56}{65}\).
Answer: Let \(x = \cos^{-1}\frac{12}{13}\) and \(y = \sin^{-1}\frac{3}{5}\)
or \(\cos x = \frac{12}{13}\) and \(\sin y = \frac{3}{5}\)
Now, \(\sin x = \sqrt{1 - \cos^2 x}\) and \(\cos y = \sqrt{1 - \sin^2 y}\)
\(\Rightarrow \sin x = \sqrt{1 - \frac{144}{169}} = \frac{5}{13}\) and \(\cos y = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}\)
We know that,
\(\sin(x+y) = \sin x \cos y + \cos x \sin y\)
\[= \frac{5}{13} \times \frac{4}{5} + \frac{12}{13} \times \frac{3}{5} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65}\]
\(\Rightarrow x+y = \sin^{-1}\frac{56}{65}\)
or, \(\cos^{-1}\left(\frac{12}{13}\right) + \sin^{-1}\left(\frac{3}{5}\right) = \sin^{-1}\left(\frac{56}{65}\right)\)

Question. Prove that : \(\sin^{-1}\frac{12}{13} + \cos^{-1}\frac{4}{5} + \tan^{-1}\frac{63}{16} = \pi\).
Answer: Let \(\sin^{-1}\left(\frac{12}{13}\right) = x\), \(\cos^{-1}\left(\frac{4}{5}\right) = y\), \(\tan^{-1}\left(\frac{63}{16}\right) = z\).
Then, \(\sin x = \frac{12}{13}\), \(\cos y = \frac{4}{5}\), \(\tan z = \frac{63}{16}\).
Therefore, \(\cos x = \frac{5}{13}\), \(\sin y = \frac{3}{5}\), \(\tan x = \frac{12}{5}\) and \(\tan y = \frac{3}{4}\).
We have, \(\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}\)
\[= \frac{\frac{12}{5} + \frac{3}{4}}{1 - \frac{12}{5} \times \frac{3}{4}} = -\frac{63}{16}\]
Hence, \(\tan(x + y) = -\tan z\)
i.e., \(\tan(x+y) = \tan(-z)\) or \(\tan(x+y) = \tan(\pi - z)\)
Therefore, \(x + y = -z\) or \(x + y = \pi - z\)
Since, \(x, y\) and \(z\) are positive, \(x+y \neq -z\)
Hence, \(x + y + z = \pi\)
or, \(\sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) = \pi\).

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