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Q11:
Find the values of the following: $\tan^{-1}(1) + \cos^{-1}(\frac{-1}{2}) + \sin^{-1}(\frac{-1}{2})$
Solution :
Step 1: Evaluation of the Principal Value of $\tan^{-1}(1)$
Let $x = \tan^{-1}(1)$. By definition of the inverse tangent function, $\tan(x) = 1$. The principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. [Per the definition of Inverse Trigonometric Functions].
Since $\tan(\frac{\pi}{4}) = 1$, and $\frac{\pi}{4} \in (-\frac{\pi}{2}, \frac{\pi}{2})$:
$\tan^{-1}(1) = \frac{\pi}{4}$
Step 2: Evaluation of the Principal Value of $\cos^{-1}(-\frac{1}{2})$
Let $y = \cos^{-1}(-\frac{1}{2})$. By definition, $\cos(y) = -\frac{1}{2}$. The principal value branch of $\cos^{-1}(x)$ is $[0, \pi]$.
We know that $\cos(\frac{\pi}{3}) = \frac{1}{2}$. Since the cosine function is negative in the second quadrant, we use the identity $\cos(\pi - \theta) = -\cos(\theta)$ [Per the Trigonometric Quadrant Rules]:
$\cos(\pi - \frac{\pi}{3}) = \cos(\frac{2\pi}{3}) = -\frac{1}{2}$
Since $\frac{2\pi}{3} \in [0, \pi]$:
$\cos^{-1}(-\frac{1}{2}) = \frac{2\pi}{3}$
Step 3: Evaluation of the Principal Value of $\sin^{-1}(-\frac{1}{2})$
Let $z = \sin^{-1}(-\frac{1}{2})$. By definition, $\sin(z) = -\frac{1}{2}$. The principal value branch of $\sin^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
We know that $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Using the property $\sin(-\theta) = -\sin(\theta)$ [Per the Odd Function property of Sine]:
$\sin(-\frac{\pi}{6}) = -\frac{1}{2}$
Since $-\frac{\pi}{6} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$:
$\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$
Step 4: Summation of the Evaluated Values
Substitute the obtained values into the original expression:
Expression $= \frac{\pi}{4} + \frac{2\pi}{3} + (-\frac{\pi}{6})$
To perform the addition, find the Least Common Multiple (LCM) of the denominators $4, 3,$ and $6$, which is $12$:
$= \frac{3\pi}{12} + \frac{8\pi}{12} - \frac{2\pi}{12}$
$= \frac{3\pi + 8\pi - 2\pi}{12}$
$= \frac{9\pi}{12}$
Reducing the fraction by dividing the numerator and denominator by $3$:
$= \frac{3\pi}{4}$
Final Solution: \frac{3\pi}{4}
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.1
- Q1: Find the principal values of the following: $\sin^{-1} (\frac{-1}{2})$
- Q2: Find the principal values of the following: $\cos^{-1} (\frac{\sqrt{3}}{2})$
- Q3: Find the principal values of the following: $\csc^{-1} (2)$
- Q4: Find the principal values of the following: $\tan^{-1} (-\sqrt{3})$
- Q5: Find the principal values of the following: $\cos^{-1} (\frac{-1}{2})$
- Q6: Find the principal values of the following: $\tan^{-1} (-1)$
- Q7: Find the principal values of the following: $\sec^{-1} (\frac{2}{\sqrt{3}})$
- Q8: Find the principal values of the following: $\cot^{-1} (\sqrt{3})$
- Q9: Find the principal values of the following: $\cos^{-1} (\frac{-1}{\sqrt{2}})$
- Q10: Find the principal values of the following: $\cosec^{-1} (-\sqrt{2})$
- Q12: Find the values of the following: $\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2})$
- Q13: If $\sin^{-1} x = y$, then
- Q14: $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$ is equal to
CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions
Chapters in CBSE - Class 12 Mathematics
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