Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.1 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1
Direction (1-10):
Find the principal values of the folLowing.
Question 1.
sin-1 (- \(\frac{1}{2}\))
Solution.
Let sin-1 (- \(\frac{1}{2}\)) = y
Then, sin y = – \(\frac{1}{2}\)
= – sin (\(\left(\frac{\pi}{6}\right)\))
= sin (- \(\left(\frac{\pi}{6}\right)\))
We know that the range of the principal value of sin-1 y is
[latex]-\frac{\pi}{2}, \frac{\pi}{2}[/latex] and sin[- \(\left(\frac{\pi}{6}\right)\)] = – \(\frac{1}{2}\)
Therefore, the principal value of sin-1 (- \(\frac{1}{2}\)) is – \(\left(\frac{\pi}{6}\right)\).
Question 2.
cos-1 (\(\frac{\sqrt{3}}{2}\))
Solution.
Let cos-1 (\(\frac{\sqrt{3}}{2}\)) = y.
Then, cos y = \(\frac{\sqrt{3}}{2}\) = cos (\(\left(\frac{\pi}{6}\right)\))
We know that the range of the principal value of cos-1 y is [0, π] and cos (\(\left(\frac{\pi}{6}\right)\)) = \(\frac{\sqrt{3}}{2}\).
Therefore, the principal value of cos-1 (\(\frac{\sqrt{3}}{2}\)) is \(\left(\frac{\pi}{6}\right)\).
Question 3.
cosec-1 (2)
Solution.
Let cosec-1 (2) = y. Then, cosec y = 2 = cosec(\(\left(\frac{\pi}{6}\right)\))
We know that the range of the principal value of cosec-1 y is [latex]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/latex] – {0} and cosec (\(\left(\frac{\pi}{6}\right)\)) = 2.
Therefore, the principal value of cosec-1 (2) is \(\left(\frac{\pi}{6}\right)\).
Question 4.
tan-1 (- √3)
Solution.
Let tan-1 (- √3) = y.
Then, tan y = – √3 = – tan \(\left(\frac{\pi}{3}\right)\) = tan (- \(\left(\frac{\pi}{3}\right)\))
We know that the range of the principal value of tan-1 y is [latex]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/latex] and tan(- \(\left(\frac{\pi}{2}\right)\)) is – √3
Therefore, the principal value of tan-1 (- √3) is – \(\frac{\pi}{3}\).
Question 5.
cos-1 (- \(\frac{1}{2}\))
Solution.
Let cos-1 (- \(\frac{1}{2}\)) = y. Then,
cos y = – \(\frac{1}{2}\) = – cos (\(\left(\frac{\pi}{3}\right)\))
= cos (π – \(\frac{\pi}{3}\)) = cos \(\left(\frac{2 \pi}{3}\right)\)
We know that the range of the principal value of cos-1 y is [0, π] and cos \(\left(\frac{2 \pi}{3}\right)\) = – \(\frac{1}{2}\).
Therefore, the principal value of cos-1 (- \(\frac{1}{2}\)) is \(\left(\frac{2 \pi}{3}\right)\).
Question 6.
tan-1 (- 1)
Solution.
Let tan-1 (- 1) = y.
Then, tan y = – 1 = – tan (\(\frac{\pi}{4}\)) = tan (- \(\frac{\pi}{4}\))
We know that the range of the principal value of tan-1 y is (\(-\frac{\pi}{2}, \frac{\pi}{2}\)) and tan (- \(\frac{\pi}{4}\)) = – 1.
Therefore, the principal value of tan-1 (- 1) is (- \(\frac{\pi}{4}\))
Question 7.
sec-1 (\(\frac{2}{\sqrt{3}}\))
Solution.
Let sec-1 (\(\frac{2}{\sqrt{3}}\)) = y.
Then, sec y = \(\frac{2}{\sqrt{3}}\) = sec (\(\frac{\pi}{6}\))
We know that the range of the principal value of sec-1 y is [0, π] – {\(\frac{\pi}{2}\)} and sec (\(\frac{\pi}{6}\)) = \(\frac{2}{\sqrt{3}}\).
Therefore, the principal value of sec-1 (\(\frac{2}{\sqrt{3}}\)) is \(\frac{\pi}{6}\).
Question 8.
cot-1 (√3)
Solution.
Let cot-1 (√3) = y. Then, cot y = √3 = cot (\(\frac{\pi}{6}\))
We know that the range of the principal value of cot-1 y is (0, π) and cot (\(\frac{\pi}{6}\)) = √3
Therefore, the principal value of cot-1 (√3) is \(\frac{\pi}{6}\).
Question 9.
cos-1 (- \(\frac{1}{\sqrt{2}}\))
Solution.
Let cos-1 (- \(\frac{1}{\sqrt{2}}\)) = y. Then,
cos y = – \(\frac{1}{\sqrt{2}}\) = – cos (\(\frac{\pi}{4}\))
= cos(\(\pi-\frac{\pi}{4}\)) = cos(\(\frac{3 \pi}{4}\))
We know that the range of the principal value of cos-1 y is [0, π] and cos (\(\frac{3 \pi}{4}\)) = – \(\frac{1}{\sqrt{2}}\)
Therefore, the principal value of cos-1 (- \(\frac{1}{\sqrt{2}}\)) is \(\frac{3 \pi}{4}\)
Question 10.
cosec-1 (√2)
Solution.
Let cosec-1 (√2) = y.
Then, cosec y = – √2 = – cosec (\(\frac{\pi}{4}\)) = cosec (- \(\frac{\pi}{4}\))
We know that the range of the principal value of cosec-1 y is
[latex]-\frac{\pi}{2}, \frac{\pi}{2}[/latex] – {0} and cosec (- \(\frac{\pi}{4}\)) = – √2.
Therefore, the principal value of cosec-1 (- √2) is – \(\frac{\pi}{4}\).
DirectIon (11 – 14): Find the value of the following.
Question 11.
tan-1 (1) + cos-1 (- \(\frac{1}{2}\)) + sin-1 (- \(\frac{1}{2}\))
Solution.
Let tan-1 (1) = x. Then, tan x = 1 = tan \(\frac{\pi}{4}\)
∴ tan-1 (1) = \(\frac{\pi}{4}\)
Let cos-1 (- \(\frac{1}{2}\)) = y.
Then, cos y = – \(\frac{1}{2}\)
= – cos (\(\frac{\pi}{3}\))
= cos (π – \(\frac{\pi}{3}\))
= cos \(\frac{2 \pi}{3}\)
∴ cos-1 (- \(\frac{1}{2}\)) = \(\frac{2 \pi}{3}\)
Let sin-1 (- \(\frac{1}{2}\)) = z.
Then, sin z = – \(\frac{1}{2}\)
= – sin (\(\frac{\pi}{6}\))
= sin (- \(\frac{\pi}{6}\))
∴ sin-1 (- \(\frac{1}{2}\)) = – \(\frac{\pi}{6}\)
∴ tan-1 (1) + cos-1 (- \(\frac{1}{2}\)) + sin-1 (- \(\frac{1}{2}\)) = \(\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}\)
= \(\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\).
Question 12.
cos-1 (\(\frac{1}{2}\)) + 2 sin-1 \(\frac{1}{2}\)
Solution.
Let cos-1 (\(\frac{1}{2}\)) = x.
Then, cos x = \(\frac{1}{2}\) = cos (\(\frac{\pi}{3}\)).
∴ cos-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{3}\)
Let sin-1 (\(\frac{1}{2}\)) = y.
Then, sin y = \(\frac{1}{2}\) = sin (\(\frac{\pi}{6}\))
∴ sin-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{6}\)
∴ cos-1 (\(\frac{1}{2}\)) + 2 sin-1 (\(\frac{1}{2}\)) = \(\frac{\pi}{3}+\frac{2 \pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}\)
Question 13.
If sin-1 x = y, then
(A) 0 ≤ y ≤ K
(B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
(C) 0 < y < π
(D) \(-\frac{\pi}{2}<y<\frac{\pi}{2}\)
Solution.
It is given that sin-1 x = y.
We know that the range of the principal value branch of sin-1 is [latex]-\frac{\pi}{2}, \frac{\pi}{2}[/latex]
Therefore, \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).
Hence, the correct option is (B).
Question 14.
tan-1 √3 – sec-1 (- 2) is equal to
(A) π
(B) – \(\frac{\pi}{3}\)
(C) \(\frac{\pi}{3}\)
(D) \(\frac{2 \pi}{6}\)
Solution.
Let tan-1 √3 = x.
Then, tan x = √3 = tan \(\frac{\pi}{3}\)
We know that the range of the principal value of tan-1 x is (\(-\frac{\pi}{2}, \frac{\pi}{2}\))
∴ tan-1 √3 = \(\frac{\pi}{3}\)
Let sec-1 (- 2) = y.
Then, sec y = – 2 = – sec (\(\frac{\pi}{3}\))
= sec (π – \(\frac{\pi}{3}\)) = sec(\(\frac{2 \pi}{3}\))
Now, tan-1 (√3) – sec-1 (- 2) = \(\frac{\pi}{3}\) – \(\frac{2 \pi}{3}\)
= – \(\frac{\pi}{3}\)
Hence, correct option is (B).