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Need solution for RD Sharma Maths Class 12 Chapter 3 Inverse Trigonometric Functions Excercise Fill in the Blanks Question 29

Answers (1)

Answer:

$7$

Hints:

You must know the rules of inverse function.

Given:

$\cot\left ( \frac{\pi}{4} -2\cot^{-1}3\right )$

Solution:

$\cot\left ( \frac{\pi}{4} -2\cot^{-1}3\right )$

Let $\cot^{-1}3=a$

$\Rightarrow \cot a=3$

$\cot \cot\left ( \frac{\pi}{4}-2\left ( \cot a \right ) \right )=\cot \cot\left ( \frac{\pi}{4} -2a\right )$

$=\frac{1}{\tan\left ( \frac{\pi}{4}-2a \right )}$

$=\frac{1+\tan 2a}{1-\tan 2a}$

Now $\cot a=\frac{1}{3}$

$\tan a=\frac{1}{3}$

$\tan 2a=\frac{2\tan a}{1-\tan^{2}a}$

$=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}$

$=\frac{3}{4}$

So, $\cot\left ( \frac{\pi}{4}-2\cot^{-1} 3\right )$

$=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$

$=\frac{4+3}{4}\times\frac{4}{4-3}$

$=7$

Posted by

infoexpert27

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