11SURELY IT IS NOT 3Ï€4. [4] [69]
Else u wud not ave given it. [67]
What is the value of cot-1(-2)+cot-1(-3) ?
(A simple one which ppl often make mistakes at)
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10 Answers
11@ manmay, the sum turns out to be tan-1(-1). R8.
So it cannot be in 3rd quad.
39tan-1(-1/2) + tan-1(-1/3)
= tan-1(-5/6/(1-1/6))
= tan-1(-1)
= pi-pi/4, 2pi-pi/4
= 3pi/4, 7pi/4
How is 7pi/4 in the third quadrant???
62manmay your answer is correct..
Pritish what is the logic?
@Soumik: You have picked the right reason .. :D "Else u wud not ave given it."
1i got it like this bhaiya
1. cot^{-1}x=tan^{-1}\left(\frac{1}{x} \right) , x\epsilon (0,\infty )
AND
2. cot^{-1}x=\pi +tan^{-1}\left(\frac{1}{x} \right) , x\epsilon (-\infty,0 )
since here x < 0 , so we use 2 property and the given expression reduces to
2\pi +tan^{-1}\left(\frac{-1}{2} \right)+tan^{-1}\left(\frac{-1}{3} \right) ..................(3)
we know that tan^{-1}(-x) = - tan^{-1}x,for,x\epsilon R
hence (3) reduces to
2\pi -\left(tan^{-1}\left(\frac{1}{2}\right) +tan^{-1}\left(\frac{1}{3} \right) \right)
= 2\pi -\frac{\pi }{4}=\frac{7\pi }{4}
39Gosh I had to write as much as Manmay did Nishant bhaiya??? LOL aisa bolte...I thought it was understood :P
62no you are not.. basically the thing to keep in mind here is that the range of cot inverse ;is not -pi/2 to pi/2... but 0 to pi..
and for -ve values cot inverse takes values between pi/2 to pi
62@Pritish.. Yaar sorry din get the clue that you were giving from the post!