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`        This question is related to trigonometry.pls solve as fast as possible and send solution .`
one year ago

```							ok so the question you have asked is a wonderful ques no doubt. let beta=b. now, i am sure you are familiar with the identity cotx – tanx= 2cot2xso the general term of the summation becomes 2^r*tan2^rb = 2^r*(cot2^rb – 2cot2^(r+1)b)= 2^rcot2^rb – 2^(r+1)cot2^(r+1)b which is TELESCOPICso ∑r=0 to n – 1 2^rcot2^rb – 2^(r+1)cot2^(r+1)b = cotb – 2^ncot2^nbor tanb+2tan2b+...+2^n-1tan2^n-1b+ 2^ncot2^nb= cotbso now all we gotta do is to find the value of b.given that cos(a)+cos(a+b)+cos(a+b+y)=0and sin(a)+sin(a+b)+sin(a+b+y)=0multiply second eqn by iota and add the first eqn to it to gete^ia+e^i(a+b)+e^i(a+b+y)=0divide by e^ia to get1+e^ib+e^i(b+y)=0similarly, multiply second eqn by minus iota and add the first eqn to it to gete^(-ia)+e^–i(a+b)+e^–i(a+b+y)=0multiply by e^ia to get1+e^–ib+e^–i(b+y)=0let e^ib = x and e^i(b+y) = zthen x+z+1=0 and 1+1/x+1/z=0eliminating z from these eqns, we havex^2+x+1=0so x= omega, omega square=  e^i(2pi/3) or  e^i(4pi/3)but since b belongs to (0, pi)so x= e^i(2pi/3) so e^ib=e^i(2pi/3) or b=2pi/3so that sum= cotb= cot2pi/3= – 1/√3
```
one year ago
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