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The value of taninvisible function application open square brackets s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses plus t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 over denominator 3 end fraction close parentheses close square brackets is

  1. fraction numerator 6 over denominator 17 end fraction    
  2. fraction numerator 7 over denominator 16 end fraction    
  3. fraction numerator 5 over denominator 7 end fraction    
  4. fraction numerator 17 over denominator 6 end fraction    

The correct answer is: fraction numerator 7 over denominator 16 end fraction


    for given expression taninvisible function application open square brackets s i n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 3 over denominator 5 end fraction close parentheses plus t a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 2 over denominator 3 end fraction close parentheses close square brackets  we need to convert sin-1 into tan-1
    sin-1(3/5) = tan-1(3/4)

    tan[sin-1(3/5)+tan-1(2/3c)] = tan[tan-1(3/4)+tan-1(2/3)]
    tan-1(x) + tan-1(y) = tan-1[(x + y) / (1-xy)]
    tan[tan-1(3/4)+tan-1(2/3)]=tan[tan-1((3/4+2/3) / (1-3/4*2/3))]
    =tan[tan-1(17/6)]
    =17/6
    ans is 17/6
    tan-1(x) + tan-1(y) =tan-1 [(x + y) / (1-xy)]   (1 > x y)

    tan[tan-1(x)] = x

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