Maths-
General
Easy

Question

open parentheses 1 plus c o t to the power of 2 end exponent invisible function application theta close parentheses left parenthesis 1 minus c o s invisible function application theta right parenthesis left parenthesis 1 plus c o s invisible function application theta right parenthesis equals

  1. 0    
  2. 1    
  3. tan2theta    
  4. sec2theta    

The correct answer is: 1


    To find value ofopen parentheses 1 plus c o t squared theta close parentheses left parenthesis 1 minus c o s theta right parenthesis left parenthesis 1 plus c o s theta right parenthesis
    G i v e n space e x p r e s s i o n
open parentheses 1 plus c o t squared theta close parentheses left parenthesis 1 minus c o s theta right parenthesis left parenthesis 1 plus c o s theta right parenthesis
left parenthesis 1 minus c o s theta right parenthesis left parenthesis 1 plus c o s theta right parenthesis equals 1 minus cos squared theta equals sin squared theta
open parentheses 1 plus c o t squared theta close parentheses equals 1 plus fraction numerator cos squared theta over denominator sin squared theta end fraction equals fraction numerator 1 over denominator sin squared theta end fraction
equals greater than fraction numerator 1 over denominator sin squared theta end fraction cross times sin squared theta equals 1

    Hence option 2 is correct

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