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Prove that ( A)/(1 - A) + (1 - A)/( A) = 2cosec A.

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( A)/(1 - A) + (1 - A)/( A) = 2cosec A

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We need to prove that sin⁡A1−cos⁡A+1−cos⁡Asin⁡A=2cosec⁡A\frac{\sin A}{1 - \cos A} + \frac{1 - \cos A}{\sin A} = 2\operatorname{cosec} A.

  1. Start with the Left Hand Side (LHS) of the equation.

    LHS=sin⁡A1−cos⁡A+1−cos⁡Asin⁡A\text{LHS} = \frac{\sin A}{1 - \cos A} + \frac{1 - \cos A}{\sin A}

  2. Find a common denominator for the two fractions, which is (1−cos⁡A)sin⁡A(1 - \cos A)\sin A.

    LHS=sin⁡A⋅sin⁡A+(1−cos⁡A)(1−cos⁡A)(1−cos⁡A)sin⁡A\text{LHS} = \frac{\sin A \cdot \sin A + (1 - \cos A)(1 - \cos A)}{(1 - \cos A)\sin A} LHS=sin⁡2A+(1−cos⁡A)2(1−cos⁡A)sin⁡A\text{LHS} = \frac{\sin^2 A + (1 - \cos A)^2}{(1 - \cos A)\sin A}

  3. Expand the term (1−cos⁡A)2(1 - \cos A)^2 in the numerator.

    Recall the algebraic identity (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2.
    So, (1−cos⁡A)2=12−2(1)(cos⁡A)+cos⁡2A=1−2cos⁡A+cos⁡2A(1 - \cos A)^2 = 1^2 - 2(1)(\cos A) + \cos^2 A = 1 - 2\cos A + \cos^2 A.
    Substitute this back into the expression:

    LHS=sin⁡2A+1−2cos⁡A+cos⁡2A(1−cos⁡A)sin⁡A\text{LHS} = \frac{\sin^2 A + 1 - 2\cos A + \cos^2 A}{(1 - \cos A)\sin A}

  4. Rearrange the terms in the numerator and use the trigonometric identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1.

    LHS=(sin⁡2A+cos⁡2A)+1−2cos⁡A(1−cos⁡A)sin⁡A\text{LHS} = \frac{(\sin^2 A + \cos^2 A) + 1 - 2\cos A}{(1 - \cos A)\sin A} LHS=1+1−2cos⁡A(1−cos⁡A)sin⁡A\text{LHS} = \frac{1 + 1 - 2\cos A}{(1 - \cos A)\sin A} LHS=2−2cos⁡A(1−cos⁡A)sin⁡A\text{LHS} = \frac{2 - 2\cos A}{(1 - \cos A)\sin A}

  5. Factor out 2 from the numerator.

    LHS=2(1−cos⁡A)(1−cos⁡A)sin⁡A\text{LHS} = \frac{2(1 - \cos A)}{(1 - \cos A)\sin A}

  6. Cancel out the common term (1−cos⁡A)(1 - \cos A) from the numerator and the denominator.

    LHS=2sin⁡A\text{LHS} = \frac{2}{\sin A}

  7. Use the reciprocal identity cosec⁡A=1sin⁡A\operatorname{cosec} A = \frac{1}{\sin A}.

    LHS=2⋅1sin⁡A=2cosec⁡A\text{LHS} = 2 \cdot \frac{1}{\sin A} = 2\operatorname{cosec} A

    This is equal to the Right Hand Side (RHS).

    Therefore, the identity is proven:

    sin⁡A1−cos⁡A+1−cos⁡Asin⁡A=2cosec⁡A\boxed{\frac{\sin A}{1 - \cos A} + \frac{1 - \cos A}{\sin A} = 2\operatorname{cosec} A}
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