GSEB Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1

Find the principal values of the following:

Question 1.
sin^-1(-$\frac { 1 }{ 2 }$)
Solution:
Let sin^-1(-$\frac { 1 }{ 2 }$) = y
∴ sin y = $\frac { 1 }{ 2 }$ = – sin$\frac { π }{ 6 }$ = sin(-$\frac { π }{ 6 }$)
The range of principal value branch of sin^-1 is [-$\frac { π }{ 2 }$, $\frac { π }{ 2 }$].
Hence, principal value of sin^-1(-$\frac { 1 }{ 2 }$) is – $\frac { π }{ 6 }$.

Question 2.
cos^-1$\left(-\frac{\sqrt{3}}{2}\right)$
Solution:
Let cos^-1$\left(\frac{\sqrt{3}}{2}\right)$ = y
∴ cos y = $\left(\frac{\sqrt{3}}{2}\right)$ = – cos$\frac { π }{ 6 }$
The range of principal value branch is (0, π)
∴ principal value of cos^-1$\left(\frac{\sqrt{3}}{2}\right)$ is $\frac { π }{ 6 }$.

Question 3.
cosec^-1(2)
Solution:
Let cosec^-1(2) = y
∴ cosec y = 2 = cosec$\frac { π }{ 6 }$
The range of principal value branch of cosec^-1 is [-$\frac { π }{ 2 }$, $\frac { π }{ 2 }$] – [0]
∴ principal value of cosec^-1(2) is $\frac { π }{ 6 }$.

Question 4.
tan^-1(-$\sqrt{3}$)
Solution:
Let tan^-1(-$\sqrt{3}$) = y
∴ tan y = 2 = – $\sqrt{3}$ = – tan$\frac { π }{ 3 }$ = tan(-$\frac { π }{ 3 }$)
The range of principal value branch of tan^-1 is [-$\frac { π }{ 2 }$, $\frac { π }{ 2 }$]
∴ principal value of
tan^-1(-$\sqrt{3}$) is – $\frac { π }{ 3 }$.

Question 5.
cos^-1(-$\frac { 1 }{ 2 }$)
Solution:
Let cos^-1(-$\frac { 1 }{ 2 }$) = y
∴ cos y = –$\frac { 1 }{ 2 }$ = – cos$\frac { π }{ 3 }$ = cos(π – $\frac { π }{ 3 }$
= cos $\frac { 2π }{ 3 }$
The range of principal value branch is (0, π)
∴ principal value of cos^-1(-$\frac { 1 }{ 2 }$) is $\frac { 2π }{ 3 }$.

Question 6.
tan^-1(-1)
Solution:
Let tan^-1(-1) = y.
∴ tan y = – 1 = tan$\frac { π }{ 4 }$) = tan($\frac { -π }{ 4 }$)
The range of principal value branch of tan^-1 is [-$\frac { π }{ 2 }$, $\frac { π }{ 2 }$]
∴ principal value of tan^-1(-1) is $\frac { π }{4 }$.

Question 7.
sec^-1$\left(-\frac{2}{\sqrt{3}}\right)$
Solution:
Let sec^-1$\left(\frac{2}{\sqrt{3}}\right)$ = y
∴ sec y = $\left(\frac{2}{\sqrt{3}}\right)$ = sec($\frac { π }{ 6 }$)
The range of principal value branch of sec^-1 is [0, π], ($\frac { π }{ 2 }$)
∴ principal value of sec^-1$\left(\frac{2}{\sqrt{3}}\right)$ is $\frac { π }{6 }$.

Question 8.
cot^-1($\sqrt{3}$)
Solution:
Let cot^-1($\sqrt{3}$) = y
∴ cot y = $\sqrt{3}$ = cot$\frac { π }{ 6 }$
The range of principal value branch of cot^-1 is [0, π]
∴ principal value of cot^-1$\sqrt{3}$ is $\frac { π }{ 6 }$.

Question 9.
cos^-1$\left(-\frac{1}{\sqrt{2}}\right)$
Solution:
Let cos^-1$\left(-\frac{1}{\sqrt{2}}\right)$ = y
∴ cos y = $\left(-\frac{1}{\sqrt{2}}\right)$ = cos($\frac { π }{ 4 }$) = cos(π – $\frac { π }{ 4 }$)
The range of principal value branch of cos^-1 is [0, π].
∴ principal value of cos^-1$\left(-\frac{1}{\sqrt{2}}\right)$ is $\frac { 3π }{4 }$.

Question 10.
cosec^-1 (-$\sqrt{2}$)
Solution:
Let cosec^-1(-$\sqrt{2}$) = y
∴ cosec y = –$\sqrt{2}$ = – cosec$\frac { π }{ 4 }$ = cosec (- $\frac { π }{ 4 }$).
The range of principal value branch of cosec^-1 is (- $\frac { π }{ 2 }$, $\frac { π }{ 2 }$) – {0}
∴ principal value of cosec^-1(-$\sqrt{2}$) is – $\frac { π }{ 4 }$.

Question 11.
tan^-1(1) + cos^-1(-$\frac { 1 }{ 2 }$) + sin^-1($\frac { -1 }{ 2 }$)
Solution:
tan^-1(1) + cos^-1(-$\frac { 1 }{ 2 }$) + sin^-1($\frac { -1 }{ 2 }$)
Now, tan^-1 1 = $\frac { π }{ 4 }$,
Since range of principal value branch of cos^-1 is [0, π]. So,

Question 12.
cos^-1($\frac { 1 }{ 2 }$) + 2 sin^-1($\frac { 1 }{ 2 }$)
Solution:
cos^-1($\frac { 1 }{ 2 }$) + 2 sin^-1($\frac { 1 }{ 2 }$)
Now, cos^-1($\frac { 1 }{ 2 }$) = $\frac { π }{ 3 }$
and sin^-1($\frac { 1 }{ 2 }$) = $\frac { π }{ 6 }$
∴ cos^-1($\frac { 1 }{ 2 }$) + 2 sin^-1($\frac { 1 }{ 2 }$)
= $\frac { π }{ 3 }$ + 2 x $\frac { π }{ 6 }$ = $\frac { 2π }{ 3 }$.

Question 13.
If sin^-1 = y, then
(A) 0 ≤ y ≤ π
(B) – $\frac { π }{ 2 }$ ≤ y ≤ $\frac { π }{ 2 }$
(C) 0 < y < π
(D) – $\frac { π }{ 2 }$ < y $\frac { π }{ 2 }$
Solution:
The range of principal value branch of sin^-1 is [-$\frac { π }{ 2 }$, $\frac { π }{ 2 }$]
∴ if sin^-1 x = y, then – $\frac { π }{ 2 }$ ≤ y ≤ $\frac { π }{ 2 }$
∴ Part (B) is correct.

Question 14.
tan^-1$\sqrt{3}$ – sec^-1(-2) is equal to
(A) π
(B) – $\frac { π }{ 3 }$
(C) $\frac { π }{ 3 }$
(D) $\frac { 2π }{ 3 }$
Solution:
tan^-1$\sqrt{3}$ = $\frac { π }{ 3 }$
and sec^-1(-2) = π – $\frac { π }{ 3 }$ = $\frac { 2π }{ 3 }$,
since principal value branch of sec^-1 is [0, π] – {$\frac { π }{ 2 }$}
∴ tan^-1$\sqrt{3}$ – sec^-1(-2) = $\frac { π }{ 3 }$ – $\frac { 2π }{ 3 }$
= – $\frac { π }{ 3 }$
∴ Part (B) is correct.