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Q15:
$\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$ is equal to
Solution :
Step 1: Evaluation of the First Term $\tan^{-1}(\sqrt{3})$
To evaluate $\tan^{-1}(\sqrt{3})$, let $\theta_1 = \tan^{-1}(\sqrt{3})$. This implies $\tan(\theta_1) = \sqrt{3}$.
The principal value branch of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. Since $\tan(\frac{\pi}{3}) = \sqrt{3}$, we have:
$\theta_1 = \frac{\pi}{3}$ [Per the definition of Principal Value Branches]
Step 2: Evaluation of the Second Term $\cot^{-1}(-\sqrt{3})$
To evaluate $\cot^{-1}(-\sqrt{3})$, let $\theta_2 = \cot^{-1}(-\sqrt{3})$. This implies $\cot(\theta_2) = -\sqrt{3}$.
The principal value branch of $\cot^{-1}(x)$ is $(0, \pi)$. [Per the definition of Principal Value Branches for inverse cotangent functions]
Using the identity $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$:
$\theta_2 = \pi - \cot^{-1}(\sqrt{3})$
Since $\cot(\frac{\pi}{6}) = \sqrt{3}$, it follows that $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.
$\theta_2 = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$
Step 3: Calculating the Final Difference
Now, substitute the evaluated values into the original expression:
Expression $= \tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$
Expression $= \frac{\pi}{3} - \frac{5\pi}{6}$
To perform the subtraction, find a common denominator:
Expression $= \frac{2\pi}{6} - \frac{5\pi}{6}$
Expression $= -\frac{3\pi}{6}$
Expression $= -\frac{\pi}{2}$
Visual Representation of Angles in the Unit Circle
Final Solution: -\frac{\pi}{2}
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.2
- Q1: Prove the following: $3\sin^{-1} x = \sin^{-1} (3x - 4x^3)$, $x \in [\frac{-1}{2}, \frac{1}{2}]$
- Q2: Prove the following: $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, $x \in [\frac{1}{2}, 1]$
- Q3: Write the following functions in the simplest form: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}$, $x \neq 0$
- Q4: Write the following functions in the simplest form: $\tan^{-1} (\sqrt{\frac{1-\cos x}{1+\cos x}})$, $0 < x < \pi$
- Q5: Write the following functions in the simplest form: $\tan^{-1} (\frac{\cos x - \sin x}{\cos x + \sin x})$, $\frac{-\pi}{4} < x < \frac{3\pi}{4}$
- Q6: Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}$, $|x| < a$
- Q7: Write the following functions in the simplest form: $\tan^{-1} (\frac{3a^2x - x^3}{a^3 - 3ax^2})$, $a > 0$; $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$
- Q8: Find the values of each of the following: $\tan^{-1} [2\cos(2\sin^{-1}(\frac{1}{2}))]$
- Q9: Find the values of each of the following: $\tan \frac{1}{2}[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}]$, $|x| < 1$, $y > 0$ and $xy < 1$
- Q10: Find the values of each of the expressions: $\sin^{-1}(\sin(\frac{2\pi}{3}))$
- Q11: Find the values of each of the expressions: $\tan^{-1}(\tan(\frac{3\pi}{4}))$
- Q12: Find the values of each of the expressions: $\tan(\sin^{-1}(\frac{3}{5}) + \cot^{-1}(\frac{3}{2}))$
- Q13: $\cos^{-1}(\cos(\frac{7\pi}{6}))$ is equal to
- Q14: $\sin(\frac{\pi}{3} - \sin^{-1}(\frac{-1}{2}))$ is equal to
CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions
Chapters in CBSE - Class 12 Mathematics
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