Question 1: Find the principal value of each of the following:

\displaystyle \text{(i) } \tan^{-1} \Bigg(  \frac{1}{\sqrt{3}} \Bigg)

\displaystyle \text{(ii) } \tan^{-1} \Bigg(  -\frac{1}{\sqrt{3}} \Bigg)

\displaystyle \text{(iii) } \tan^{-1} \Bigg(  \cos \frac{\pi}{2} \Bigg)

\displaystyle \text{(iv) } \tan^{-1} \Bigg(  2 \cos \frac{2\pi}{3} \Bigg)

Answer:

\displaystyle \text{(i) } \tan^{-1} \Bigg(  \frac{1}{\sqrt{3}} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in \Bigg( -\frac{\pi}{2}, \frac{\pi}{2} \Bigg)  \text{ such that } \tan \theta = \frac{1}{\sqrt{3}} = \tan \Bigg( \frac{\pi}{6} \Bigg) \Bigg) = \frac{\pi}{6}

\displaystyle \text{(ii) } \tan^{-1} \Bigg(  -\frac{1}{\sqrt{3}} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in \Bigg( -\frac{\pi}{2}, \frac{\pi}{2} \Bigg)  \text{ such that } \tan \theta = -\frac{1}{\sqrt{3}} = \tan \Bigg( -\frac{\pi}{6} \Bigg) \Bigg) = -\frac{\pi}{6}

\displaystyle \text{(iii) } \tan^{-1} \Bigg(  \cos \frac{\pi}{2} \Bigg) \\ \\ = \Bigg( \text{An angle } \theta \in \Bigg( -\frac{\pi}{2}, \frac{\pi}{2} \Bigg)  \text{ such that } \tan \theta = \cos\frac{\pi}{2} = \tan (0) \Bigg) = 0

\text{(iv) } \tan^{-1} \Big(  2 \cos \frac{2\pi}{3} \Big) \\ \\ = \Big( \text{An angle } \theta \in \Big( -\frac{\pi}{2}, \frac{\pi}{2} \Big)  \text{ such that } \tan \theta = 2\cos\frac{2\pi}{3} = \tan \Big(-\frac{\pi}{4} \Big) \Big) = -\frac{\pi}{4}

\\

Question 2: Find the principal values, evaluate each of the following:

\displaystyle \text{(i) } \tan^{-1} (-1) + \cos^{-1} \Bigg(  -\frac{1}{\sqrt{2}} \Bigg)

\displaystyle \text{(ii) } \tan^{-1} \Bigg\{  2 \sin \Bigg(  4 \cos ^{-1} \frac{\sqrt{3}}{2} \Bigg)  \Bigg\}

Answer:

\displaystyle \text{(i) } \tan^{-1} (-1) + \cos^{-1} \Bigg(  -\frac{1}{\sqrt{2}} \Bigg) \\ \\ \\ = \tan^{-1} \Bigg( \tan \Bigg( -\frac{\pi}{4} \Bigg) \Bigg) + \cos^{-1} \Bigg(  \cos \frac{3\pi}{4} \Bigg) \\ \\ \\ =  -\frac{\pi}{4}+\frac{3\pi}{4} \\ \\ \\ =  \frac{\pi}{2}

\displaystyle \text{(ii) } \tan^{-1} \Bigg\{  2 \sin \Bigg(  4 \cos ^{-1} \frac{\sqrt{3}}{2} \Bigg)  \Bigg\}  \\ \\ \\ = \tan^{-1} \Bigg\{  2 \sin \Bigg(  4 \cos ^{-1} \Bigg(  \cos \frac{\pi}{6} \Bigg) \Bigg)  \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{  2 \sin \Bigg(  \frac{2\pi}{3} \Bigg) \Bigg\} \\ \\ \\ = \tan^{-1} \Bigg\{  2 \times \Bigg( \frac{\sqrt{3}}{2} \Bigg) \Bigg\} \\ \\ \\ = \tan^{-1} ( \sqrt{3}) \\ \\ \\ = \tan^{-1} \Bigg( \tan \frac{\pi}{3} \Bigg) = \frac{\pi}{3}

\\

Question 3: Evaluate each of the following:

\displaystyle \text{(i) } \tan^{-1} (-1) + \cos^{-1} \Bigg(  -\frac{1}{2} \Bigg)+ \sin^{-1} \Bigg(  -\frac{1}{2} \Bigg)

\displaystyle \text{(ii) } \tan^{-1} \Bigg(  -\frac{1}{\sqrt{3}} \Bigg) + \tan^{-1} (-\sqrt{3})+ \tan^{-1} \Bigg(  \sin \Bigg(-\frac{\pi}{2} \Bigg) \Bigg)

\displaystyle \text{(iii) } \tan^{-1} \Bigg( \tan \Bigg( \frac{5\pi}{6} \Bigg) \Bigg) + \cos^{-1} \Bigg\{ \cos \Bigg( \frac{13\pi}{6} \Bigg) \Bigg\}

Answer:

\displaystyle \text{(i) } \tan^{-1} (-1) + \cos^{-1} \Bigg(  -\frac{1}{2} \Bigg)+ \sin^{-1} \Bigg(  -\frac{1}{2} \Bigg) \\ \\ \\ = \tan^{-1} \Bigg(\tan \frac{\pi}{4}\Bigg) + \cos^{-1} \Bigg(  \cos \frac{2\pi}{3} \Bigg)+ \sin^{-1} \Bigg( \sin \frac{-\pi}{6} \Bigg) \\ \\ \\ = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi}{4}

\displaystyle \text{(ii) } \tan^{-1} \Bigg(  -\frac{1}{\sqrt{3}} \Bigg) + \tan^{-1} (-\sqrt{3})+ \tan^{-1} \Bigg(  \sin \Bigg(-\frac{\pi}{2} \Bigg) \Bigg) \\ \\ \\ =  -\tan^{-1} \Bigg(  \frac{1}{\sqrt{3}} \Bigg) - \tan^{-1} (\sqrt{3})+ \tan^{-1} (  -1 ) \\ \\ \\ =  -\tan^{-1} \Bigg(  \frac{1}{\sqrt{3}} \Bigg) - \tan^{-1} (\sqrt{3})- \tan^{-1} (  1 ) \\ \\ \\ =  -\tan^{-1} \Bigg(  \tan \frac{\pi}{6} \Bigg) - \tan^{-1} \Bigg(  \tan \frac{\pi}{3} \Bigg)- \tan^{-1} \Bigg(  \tan \frac{\pi}{4} \Bigg) \\ \\ \\ = - \frac{\pi}{6} - \frac{\pi}{3} - \frac{\pi}{4} = -\frac{3\pi}{4}  

\displaystyle \text{(iii) } \tan^{-1} \Bigg( \tan \Bigg( \frac{5\pi}{6} \Bigg) \Bigg) + \cos^{-1} \Bigg\{ \cos \Bigg( \frac{13\pi}{6} \Bigg) \Bigg\} \\ \\ \\ =  \tan^{-1} \Bigg( \tan \Bigg( \pi - \frac{5\pi}{6} \Bigg) \Bigg) + \cos^{-1} \Bigg\{ \cos \Bigg( 2\pi + \frac{\pi}{6} \Bigg) \Bigg\} \\ \\ \\ =  \tan^{-1} \Bigg( - \tan \Bigg( \frac{\pi}{6} \Bigg) \Bigg) + \cos^{-1} \Bigg\{ \cos \Bigg( \frac{\pi}{6} \Bigg) \Bigg\} \\ \\ \\ = - \frac{\pi}{6} + \frac{\pi}{6} = 0

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.