Question

Question 21: Find the values of tan¯¹ √3 - cot-1 (-√3)is equal to (A) π (B) -π/2 (C) 0 (D) 2√3

Class 12 - Math - Inverse Trigonometric Functions

Answer 1

We know that tan¯¹ √3 is at the angle π/3.
Now,

⇒ tan¯¹ √3 - cot-1 (-√3)
⇒ π/3 - (π-π/3)
⇒ π/3 - 5π/6
⇒ -3π/6
⇒ -π/2

Hope it helps...

Answer 2

$\mathsf{tan^{-1}(\sqrt{3})-cot^{-1}(-\sqrt{3})=-\dfrac{\pi}{2}}$

Correct option: (B) $\mathsf{-\dfrac{\pi}{2}}$

To find:

The value of $\mathsf{tan^{-1}(\sqrt{3})-cot^{-1}(-\sqrt{3})}$

Concept to be used:

$\mathsf{cot^{-1}(-x)=\pi-cot^{-1}x,x\in\mathbb{R}}$, since the principal value of $\mathsf{cot^{-1}x}$ lies in $(0,\pi)$.

Values to remember:

• $\mathsf{tan^{-1}\sqrt{3}=\dfrac{\pi}{3}}$

• $\mathsf{cot^{-1}\sqrt{3}=\dfrac{\pi}{6}}$

Step-by-step explanation:

Now, $\mathsf{tan^{-1}(\sqrt{3})-cot^{-1}(-\sqrt{3})}$

$\mathsf{=\dfrac{\pi}{3}-(\pi-cot^{-1}\sqrt{3})}$

• since $\mathsf{cot^{-1}(-x)=\pi-cot^{-1}x,x}\in\mathbb{R}$

$\mathsf{=\dfrac{\pi}{3}-(\pi-\dfrac{\pi}{6})}$

$\mathsf{=\dfrac{\pi}{3}-\dfrac{6\pi-\pi}{6}}$

$\mathsf{=\dfrac{\pi}{3}-\dfrac{5\pi}{6}}$

$\mathsf{=\dfrac{2\pi-5\pi}{6}}$

$\mathsf{=-\dfrac{3\pi}{6}}$

$\mathsf{=-\dfrac{\pi}{2}}$

Final answer:

$\boxed{\mathsf{tan^{-1}(\sqrt{3})-cot^{-1}(-\sqrt{3})=-\dfrac{\pi}{2}}}$

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