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Name: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 1 Subquestion (iv) Maths Textbook Solution.
Uploaded: Thu, 2021-07-08 20:32
Description: Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.6 Question 1 Subquestion (iv) Maths Textbook Solution.

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

Answers (1)

Answer: $\frac{3\pi }{4}$
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 $cot ^{-1}: R \rightarrow (0,\pi )$
First we will convert $tan\frac{3\pi }{4}$ into $\cot$
Given: $cot^{-1}\left ( \tan \frac{3\pi }{4} \right )$
Solution:
Let $y= cot^{-1}\left ( \tan \frac{3\pi }{4} \right )$                                                                                                                    $\cdot \cdot \cdot 1$
$cot\: y= \tan \frac{3\pi }{4}$
$cot\: y = cot (\frac{\pi }{2}- \frac{3\pi }{2} ) \; \; \; \; \; \; \left [ \because tan\theta = \cot \left ( \frac{\pi }{2}-\theta\right ) \right ]$
$cot\: y = cot ( \frac{-\pi }{4} )$
$cot\: y = - cot \left ( \frac{\pi }{4} \right ) \; \; \; \; [ \because cot\left ( -\theta \right ) = - cot\: \theta ]$
$cot\: y = cot \left ( \pi -\frac{\pi }{4} \right ) \; \; \; \; \; \; \; \; [ cot\left ( \pi -\theta \right ) = - cot\: \theta ]$
$cot\: y = cot \left ( \frac{3\pi }{4} \right )$
$y = \frac{3\pi }{4}$
$cot^{-1}\left ( tan\frac{3\pi }{4}\right ) = \frac{3\pi }{4}$                                                                       (From equation 1)
$\because$ The principal value branch of the function $cot^{-1}$ is $(0, \pi )$
$cot^{-1}\left ( tan\frac{3\pi }{4} \right ) =\frac{3\pi }{4}\: \epsilon \: (0, \pi )$
Hence the principal value of $cot^{-1}\left ( tan\frac{3\pi }{4} \right )$ is $\frac{3\pi }{4}$