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  4. If the mapping f ( x )= ax + b , a >0 maps [-1,1] onto [0,2], then cot [ cot -1 7+ cot -1 8+ cot -1 18] is equal to

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Q. If the mapping $f ( x )= ax + b , a >0$ maps $[-1,1]$ onto $[0,2]$, then $\cot \left[\cot ^{-1} 7+\cot ^{-1} 8+\cot ^{-1} 18\right]$ is equal to

Inverse Trigonometric Functions

Solution:

$f(x)=a x+b $
$f'(x)=a>0$
$f ( x )$ is an increasing function.
$\Rightarrow f (-1)=0$ and $f (1)=2 $
or $ - a + b =0 $
and $ a + b =2 $
Then, $ a = b =1 $
$ \Rightarrow f ( x )= x +1$
Now, $\cot \left[\cot ^{-1} 7+\cot ^{-1} 8+\cot ^{-1} 18\right]$
$=\cot \left\{\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7} \cdot \frac{1}{8}}\right)+\tan ^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{\tan ^{-1}\left(\frac{15}{55}\right)+\tan ^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{\tan ^{-1}\left(\frac{3}{11}\right)+\tan ^{-1}\left(\frac{1}{18}\right)\right\}$
$=\cot \left\{\tan ^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{1.1} \cdot \frac{1}{18}}\right)\right\}$
$=\cot \left\{\tan ^{-1}\left(\frac{65}{195}\right)\right\}$
$=\cot \left\{\tan ^{-1}\left(\frac{1}{3}\right)\right\}$
$=\cot \left(\cot ^{-1} 3\right)=3=1+2= f (2)$