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Question

If a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript horizontal ellipsis horizontal ellipsis horizontal ellipsis a subscript n end subscript are in A.P with common difference d then sind open square brackets c o s e c invisible function application a subscript 1 end subscript c o s e c invisible function application a subscript 2 end subscript plus close c o s e c invisible function application a subscript 2 end subscript c o s e c invisible function application a subscript 3 end subscript plus c o s e c invisible function application a subscript 3 end subscript c o s e c invisible function application a subscript 4 end subscript plus horizontal ellipsis horizontal ellipsis n terms] equals

  1. c o t invisible function application a subscript 1 minus c o t invisible function application a subscript n    
  2. t a n invisible function application a subscript 1 end subscript minus t a n invisible function application a subscript n plus 1 end subscript    
  3. s e c invisible function application a subscript 1 end subscript minus s e c invisible function application a subscript n plus 1 end subscript    
  4. c o s e c invisible function application a subscript 1 end subscript minus c o s e c invisible function application a subscript n plus 1 end subscript    

Hint:

Multiply sind inside the brackets and simplify.

The correct answer is: c o t invisible function application a subscript 1 minus c o t invisible function application a subscript n


    Givena subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript horizontal ellipsis horizontal ellipsis horizontal ellipsis a subscript n end subscript are in A.P with common difference d
    rightwards double arrow a subscript 2 space minus space a subscript 1 space equals d space comma space a subscript 3 space minus space a subscript 2 space equals d space comma space a subscript 4 space minus space a subscript 3 space equals d space comma...... space a subscript n space minus space a subscript n minus 1 end subscript space equals space d
    To find : sind open square brackets c o s e c invisible function application a subscript 1 end subscript c o s e c invisible function application a subscript 2 end subscript plus close c o s e c invisible function application a subscript 2 end subscript c o s e c invisible function application a subscript 3 end subscript plus c o s e c invisible function application a subscript 3 end subscript c o s e c invisible function application a subscript 4 end subscript plus horizontal ellipsis horizontal ellipsis n terms]
     We can also write the above terms as
    fraction numerator sin d over denominator sin a subscript 1 sin a subscript 2 end fraction space plus space fraction numerator sin d over denominator sin a subscript 2 sin a subscript 3 end fraction space plus fraction numerator sin d over denominator sin a subscript 3 sin a subscript 4 end fraction space...... space plus space fraction numerator sin d over denominator sin a subscript n minus 1 end subscript sin a subscript n end fraction
    Substituting the value of d in the above
    rightwards double arrow fraction numerator sin left parenthesis a subscript 2 space minus a subscript 1 right parenthesis space over denominator sin a subscript 1 sin a subscript 2 end fraction space plus space fraction numerator sin left parenthesis a subscript 3 space minus a subscript 2 right parenthesis over denominator sin a subscript 2 sin a subscript 3 end fraction space plus fraction numerator sin left parenthesis a subscript 4 space minus a subscript 3 right parenthesis over denominator sin a subscript 3 sin a subscript 4 end fraction space...... space plus space fraction numerator sin left parenthesis a subscript n space minus space a subscript n minus 1 end subscript right parenthesis over denominator sin a subscript n minus 1 end subscript sin a subscript n end fraction

rightwards double arrow fraction numerator sin a subscript 2 cos a subscript 1 space minus sin a subscript 1 cos a subscript 2 space over denominator sin a subscript 1 sin a subscript 2 end fraction space plus space fraction numerator sin a subscript 3 cos a subscript 2 space minus space sin a subscript 2 cos a subscript 3 over denominator sin a subscript 2 sin a subscript 3 end fraction space plus fraction numerator sin a subscript 4 cos a subscript 3 space minus space sin a subscript 3 cos a subscript 4 over denominator sin a subscript 3 sin a subscript 4 end fraction space...... space plus space fraction numerator sin a subscript n cos space a subscript n minus 1 end subscript space minus cos a subscript n sin space a subscript n minus 1 end subscript space over denominator sin a subscript n minus 1 end subscript sin a subscript n end fraction

S e p e r a t i n g space t h e space a b o v e space t e r m s space colon
space
rightwards double arrow fraction numerator cos a subscript 1 over denominator sin a subscript 1 end fraction space minus space fraction numerator cos a subscript 2 over denominator sin a subscript 2 end fraction space plus space fraction numerator cos a subscript 2 over denominator sin a subscript 2 end fraction space minus space fraction numerator cos a subscript 3 over denominator sin a subscript 3 end fraction space plus space fraction numerator cos a subscript 3 over denominator sin a subscript 3 end fraction space minus space fraction numerator cos a subscript 4 over denominator sin a subscript 4 end fraction plus space........ plus space fraction numerator cos space a subscript n minus 1 end subscript over denominator sin space a subscript n minus 1 end subscript end fraction space minus space fraction numerator cos a subscript n over denominator sin a subscript n end fraction

    Cancelling out like terms :
    rightwards double arrow c o t a subscript 1 space minus space c o t space a subscript n

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