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sin invisible function application alpha minus cos invisible function application alpha equals 1 third not stretchy rightwards double arrow sin to the power of 4 invisible function application alpha minus cos to the power of 4 invisible function application alpha equals

  1. fraction numerator 8 over denominator square root of 17 end fraction
  2. negative fraction numerator square root of 17 over denominator 9 end fraction
  3. fraction numerator square root of 17 over denominator 9 end fraction
  4. negative fraction numerator 8 over denominator square root of 17 end fraction

The correct answer is: negative fraction numerator square root of 17 over denominator 9 end fraction

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