Referring to the AP Inter 2nd Year Maths Study Material Chapter 2 Inverse Trigonometric Functions Exercise 2a Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Inverse Trigonometric Functions Solutions Exercise 2a
I.
Question 1.
Find the principal value of sin-1\(\left(-\frac{1}{2}\right)\)
Solution:
We know that sin-1(-x) = -sin-1x
∴ \(\sin ^{-1}\left(\frac{-1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)=-\sin ^{-1}\left[\sin \left(\frac{\pi}{6}\right)\right]=-\frac{\pi}{6}\) Here \(-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
Question 2.
Find the principal value of cos-1\(\left(\frac{\sqrt{3}}{2}\right)\)
Solution:
\(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\) Here \(\frac{\pi}{6}\) ∈ [0, π]
Question 3.
Find the principal value of cosec-1 (2)
Solution:
cosec-1 (2) = cosec-1[cosec\(\left(\frac{\pi}{6}\right)\)] = \(\frac{\pi}{6}\) Here \(\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)
Question 4.
Find the principal value of tan-1 (-\(\sqrt{3}\))
Solution:
We know that tan-1(-x)= -tan-1 x
tan-1 (-\(\sqrt{3}\)) = -tan-1 (\(\sqrt{3}\)) = \(-\tan ^{-1}\left(\tan \frac{\pi}{3}\right)=-\frac{\pi}{3}\) Here \(-\frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Question 5.
Find the principal value of cos-1\(\left(-\frac{1}{2}\right)\)
Solution:
We know that cos-1(-x) = π = cos-1x
\(\cos ^{-1}\left(\frac{-1}{2}\right)=\pi-\cos ^{-1}\left(\frac{1}{2}\right)=\pi-\cos ^{-1}\left[\cos \left(\frac{\pi}{3}\right)\right]=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}\) Here \(-\frac{\pi}{3} \in[0, \pi]\)
Question 6.
Find the principal value of tan-1 (-x)
Solution:
We know that tan-1 (-x) = -tan-1 x
tan-1 (-1) = -tan-1 1 = \(-\tan ^{-1}\left[\tan \left(\frac{\pi}{4}\right)\right]=-\frac{\pi}{4}\) Here \(-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Question 7.
Find the principal value of sec-1 \(\left(\frac{2}{\sqrt{3}}\right)\)
Solution:
\(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\sec ^{-1}\left(\sec \frac{\pi}{6}\right)=\frac{\pi}{6}\) Here \(\frac{\pi}{6} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}\)
Question 8.
Find the principal value of cot-1 (\(\sqrt{3}\))
Solution:
cot-1 (\(\sqrt{3}\)) = \(\cot ^{-1}\left[\cot \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6}\) Here \(\frac{\pi}{6}\) ∈ (0, π)
Question 9.
Find the principal value of cos-1 \(\left(-\frac{1}{\sqrt{2}}\right)\)
Solution:
We know that cos-1(-x) = π – cos-1x
\(\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\pi-\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\pi-\cos ^{-1}\left[\cos \left(\frac{\pi}{4}\right)\right]=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}\) Here \(\frac{3 \pi}{4}\) ∈ [0, π]
Question 10.
Find the principal value of cosec-1 (-\(\sqrt{2}\))
Solution:
We know that cosec-1(-x) = cosec-1x
cosec-1 (-\(\sqrt{2}\)) = -cosec-1 (\(\sqrt{2}\)) = -cosec-1[cosec\(\left(\frac{\pi}{4}\right)\)] = \(-\frac{\pi}{4}\) Here \(-\frac{\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)
Question 11.
Find the value of \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)
Solution:
\(\cos ^{-1}\left[\cos \frac{\pi}{3}\right]+2 \sin ^{-1}\left[\sin \frac{\pi}{6}\right]=\frac{\pi}{3}+2\left(\frac{\pi}{6}\right)=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}\) Here \(\frac{2 \pi}{3}\) ∈ [0, π]
II.
Question 1.
Find the value of tan-1(1) + cos-1\(\left(-\frac{1}{2}\right)+\) + sin-1\(\left(-\frac{1}{2}\right)+\)
Solution:
We know that cos-1 (-x) = π – cos-1 x and sin-1(-x) = -sin-1 x
∴ \(\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)=\tan ^{-1}(1)+\left[\pi-\cos ^{-1}\left(\frac{1}{2}\right)\right]-\sin ^{-1}\left(\frac{1}{2}\right)\)
= \(\frac{\pi}{4}+\left(\pi-\frac{\pi}{3}\right)-\frac{\pi}{6}\) = 45° + 180° – 60° – 30° = 135° = \(\frac{3 \pi}{4}\)