Solution of question asked by Mr. Ravi Raj Anand - III

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Ques: If \alpha \in \left( { - \frac{\pi }{2},0} \right) then find the value of {\tan ^{ - 1}}(\cot \alpha ) - {\cot ^{ - 1}}(\tan \alpha ) ?

Graphical Approach:

Let us take 'x' in place of \alpha .

Graph of {\tan ^{ - 1}}(\cot x):

it is a periodic function, because cot x is periodic and the period is \pi .

So, we have to draw graph for a period \pi  and repeat for entire values of x.

Range of {\tan ^{ - 1}}x = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)
so Range of {\tan ^{ - 1}}(\cot x) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right)

We know that, {\tan ^{ - 1}}(\tan x) = x , if  - \frac{\pi }{2} < x < \frac{\pi }{2}

since, {\tan ^{ - 1}}(\cot x) = {\tan ^{ - 1}}\tan (\frac{\pi }{2} - x)

= \frac{\pi }{2} - x, if  - \frac{\pi }{2} < \frac{\pi }{2} - x < \frac{\pi }{2} or 0 < x < \pi

so, first draw the line y = \frac{\pi }{2} - x for 0 < x < \pi  and repeat it for entire x.


Similarly, draw the graph of {\cot ^{ - 1}}(\tan x)

{\cot ^{ - 1}}(\tan x) = {\cot ^{ - 1}}\cot (\frac{\pi }{2} - x) = \frac{\pi }{2} - x , if 0 < \frac{\pi }{2} - x < \pi  or  - \frac{\pi }{2} < x < \frac{\pi }{2}

and range of {\cot ^{ - 1}}(\tan x) = (0,\pi )



Draw both the graph together to get the difference,



Clearly, in interval \left( { - \frac{\pi }{2},0} \right) the difference is pi, both the graph are represented by lines in this interval which are parallel. Thus, throughout the interval difference will be the same i.e. pi .

i.e.  {\tan ^{ - 1}}(\cot x) - {\cot ^{ - 1}}(\tan x) = - \pi  ( since graph of arccotx is above arctanx, therefore in difference put minus sign)


and second observation,

in interval, (0,\frac{\pi }{2})
{\tan ^{ - 1}}(\cot x) - {\cot ^{ - 1}}(\tan x) = 0

Traditional approach,
since {\tan ^{ - 1}}(\tan \alpha ) = \alpha , - \frac{\pi }{2} < \alpha < \frac{\pi }{2}
and, {\cot ^{ - 1}}(\cot \alpha ) = \alpha ,0 < \alpha < \pi

therefore first we convert the interval to (0,pi/2)


 If \alpha \in ( - \frac{\pi }{2},0)

 - \alpha \in (0,\frac{\pi }{2})
and  - \alpha  is acute angle and lies in first quadrant,

 since {\tan ^{ - 1}}(\tan \alpha ) = \alpha , - \frac{\pi }{2} < \alpha < \frac{\pi }{2}
and, {\cot ^{ - 1}}(\cot \alpha ) = \alpha ,0 < \alpha < \pi


therefore, {\tan ^{ - 1}}(\cot \alpha ) - {\cot ^{ - 1}}(\tan \alpha ) = {\tan ^{ - 1}}[ - \cot ( - \alpha )] - {\cot ^{ - 1}}[ - \tan ( - \alpha )]

= - {\tan ^{ - 1}}\cot ( - \alpha ) - [\pi - {\cot ^{ - 1}}\tan ( - \alpha )]

= - {\tan ^{ - 1}}\tan [\frac{\pi }{2} - ( - \alpha )] - \pi + {\cot ^{ - 1}}\cot [\frac{\pi }{2} - ( - \alpha )]

= - (\frac{\pi }{2} + \alpha ) - \pi + (\frac{\pi }{2} + \alpha )

= - \pi





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