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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 1 Subquestion (iii) Maths Textbook Solution.
Uploaded: Thu, 2021-07-08 19:52
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 1 Subquestion (iii) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 1 Subquestion (iii) Maths Textbook Solution.

Answers (1)

Answer: $\frac{2\pi }{3}$
Hint: The $cot ^{-1}$ function is defined as a function whose domain R and the principal value branch of the function $cot ^{-1}$ is $(0, \pi )$ . Thus $cot ^{-1}: R\rightarrow \left ( 0,\pi \right )$ Given: $cot^{-1}\left ( \frac{-1}{\sqrt{3}} \right )$
Solution:
Let $y = cot^{-1}\left ( \frac{-1}{\sqrt{3}} \right )$ $\cdot \cdot \cdot 1$
$cot\: y = \frac{-1}{\sqrt{3}}$
$cot\: y = - cot\frac{\pi }{3} \; \; \; \; \; \; \left [cot\frac{\pi }{3}= \frac{1}{\sqrt{3}} \right ]$
$cot\: y = cot (\pi -\frac{\pi }{3} ) \; \; \; \; \; \; \; \; \; \; \; [ \because cot\left ( \pi -\theta \right ) = - cot\theta ]$
$y = \frac{2\pi }{3}$
$cot^{-1}\left (\frac{-1}{\sqrt{3}} \right ) = \frac{2\pi }{3}$ (From equation 1)
$\because$ The principal value branch of the function $cot ^{-1}$ is $(0, \pi )$
$cot^{-1}\left ( \frac{-1}{\sqrt{3}} \right )$ $= \frac{2\pi }{3}\: \epsilon \: (0, \pi )$
Hence the principal value of $cot^{-1}\left ( \frac{-1}{\sqrt{3}} \right )$ is $\frac{2\pi }{3}$