Prove that the L.H.S is equal to R.H.D ( in the same form)
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Let x = theta, so that I can do less typing
sin^3 x / cos^2 x - sec^2 x = tan^2 x . (sinx - 1) - 1
<=> sin^3 x / cos^2 x - 1/cos^2 x = sin^2 x / cos^2 x . (sinx - 1) - 1
<=> sin^3 x - 1 = sin^2 x (sinx - 1) - cos^2 x
<=> sin^3 x - 1 = sin^2 x (sinx - 1) - 1 + sin^2 x
<=> sin^3 x = sin^2 x (sinx - 1) + sin^2 x
<=> sin^3 x = sin^3 x - sin^2 x + sin^2 x
<=> 0 = 0
<=> Means if and only if, and so we can say that LHS = RHS if and only if 0=0.
So 0=0 implies LHS = RHS
Since 0=0 is obviously true, LHS = RHS
sin^3 x / cos^2 x - sec^2 x = tan^2 x . (sinx - 1) - 1
<=> sin^3 x / cos^2 x - 1/cos^2 x = sin^2 x / cos^2 x . (sinx - 1) - 1
<=> sin^3 x - 1 = sin^2 x (sinx - 1) - cos^2 x
<=> sin^3 x - 1 = sin^2 x (sinx - 1) - 1 + sin^2 x
<=> sin^3 x = sin^2 x (sinx - 1) + sin^2 x
<=> sin^3 x = sin^3 x - sin^2 x + sin^2 x
<=> 0 = 0
<=> Means if and only if, and so we can say that LHS = RHS if and only if 0=0.
So 0=0 implies LHS = RHS
Since 0=0 is obviously true, LHS = RHS
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keywords: Trigonometry,Trigonometry?