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# Trigo equations watch

1. The value of (tan1)(tan2)+(tan2)(tan3)...(tan 88)(tan89) is (cot^2x)-n (all values in degrees) , where n is a 2 digit number ab. Then find x+b-a. Please show complete solution. Rgds...
2. (Original post by Spandy)
The value of (tan1)(tan2)+(tan2)(tan3)...(tan 88)(tan89) is (cot^2x)-n (all values in degrees) , where n is a 2 digit number ab. Then find x+b-a. Please show complete solution. Rgds...
I don't know the solution to this, but regardless we don't give out full solutions on TSR anyway!

What thoughts have you had on this? Are there any trig formulae that you know that might be useful here?

Where does this question come from btw?
3. (Original post by davros)
I don't know the solution to this, but regardless we don't give out full solutions on TSR anyway!

What thoughts have you had on this? Are there any trig formulae that you know that might be useful here?

Where does this question come from btw?
It looks like a heap of nonsense to me.
4. (Original post by Mr M)
It looks like a heap of nonsense to me.
I was sort of giving the benefit of the doubt on the assumption that it might be possible to extract something meaningful from it if it had been formatted better, but you may be right
5. (Original post by Spandy)
The value of (tan1)(tan2)+(tan2)(tan3)...(tan 88)(tan89) is (cot^2x)-n (all values in degrees) , where n is a 2 digit number ab. Then find x+b-a. Please show complete solution. Rgds...
note that tanx = 1/tan(90-x)

eg tan10 = 1/tan80
6. This Q was in the maths section of AITS-3, conducted by the institute IITian's pace, anyway, thanks! I will post the solution over here when its declared in the pace website.
7. PS- the Q was- summation from x=1 to x=88, of [tan(x)][tan(x+1)]
8. Hello, I succeeded in solving it by myself, note that 1+[tan(x)][tan(x+1)] can be written as [tan(x+1)-tan(x)]/(tan1), a simple telescoping series. Thanks again.

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Updated: January 14, 2015
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