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tan(2tan inverseX + cot inverseX) at x=2. I couldn`t understand the X=2 thing

tan(2tan inverseX + cot inverseX) at x=2. I couldn`t understand the X=2 thing

Grade:12

1 Answers

Harish
82 Points
4 years ago
Hello Aashi !
 
The question is that we have to find the value of tan( 2tan-1x + cot-1x ) at x = 2.
 
So, we can rewrite the question as :
          tan( tan-1x + tan-1x + cot-1x )
Since we know that tan-1x + cot-1x =  \frac{\pi }{2}  
 
Therefore, the question now becomes :
         tan(  \frac{\pi }{2}   + tan-1x ) 
 
If you see this looks like tan( a + b ) formula , where a = \frac{\pi }{2} and b = tan-1x
Therefore, solving using the formula above we will get,
 
\frac{tan(\frac{\pi }{2}) + tan(tan^{-1}x)}{1- tan(\frac{\pi }{2})tan(tan^{-1}x)}
 
\frac{\infty + x }{1 - (\infty )(x)}  
 
= 0 
 
This is your answer.
 
Hope this will help you ! 
 
If I am wrong somewhere , then please let me know !
 
 
 
 

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