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Prove the trigonometric identity: ()/(1-) = +

Mathematics

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Answer

cos⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ\frac{\cos\theta}{1-\sin\theta} = \sec\theta + \tan\theta

Step-by-step solution

7 steps

  1. Start with the left-hand side (LHS) of the identity.

    The LHS is cos⁡θ1−sin⁡θ\frac{\cos\theta}{1-\sin\theta}.

  2. Multiply the numerator and denominator by the conjugate of the denominator, which is 1+sin⁡θ1+\sin\theta.

    cos⁡θ1−sin⁡θ=cos⁡θ1−sin⁡θ×1+sin⁡θ1+sin⁡θ\frac{\cos\theta}{1-\sin\theta} = \frac{\cos\theta}{1-\sin\theta} \times \frac{1+\sin\theta}{1+\sin\theta} =cos⁡θ(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)= \frac{\cos\theta(1+\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}

  3. Expand the numerator and use the difference of squares formula (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2-b^2 in the denominator.

    =cos⁡θ+cos⁡θsin⁡θ12−sin⁡2θ= \frac{\cos\theta + \cos\theta\sin\theta}{1^2 - \sin^2\theta} =cos⁡θ+cos⁡θsin⁡θ1−sin⁡2θ= \frac{\cos\theta + \cos\theta\sin\theta}{1 - \sin^2\theta}

  4. Apply the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, which means 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta.

    =cos⁡θ+cos⁡θsin⁡θcos⁡2θ= \frac{\cos\theta + \cos\theta\sin\theta}{\cos^2\theta}

  5. Separate the fraction into two terms.

    =cos⁡θcos⁡2θ+cos⁡θsin⁡θcos⁡2θ= \frac{\cos\theta}{\cos^2\theta} + \frac{\cos\theta\sin\theta}{\cos^2\theta}

  6. Simplify each term.

    =1cos⁡θ+sin⁡θcos⁡θ= \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta}

  7. Express the terms using sec⁡θ\sec\theta and tan⁡θ\tan\theta.

    We know that 1cos⁡θ=sec⁡θ\frac{1}{\cos\theta} = \sec\theta and sin⁡θcos⁡θ=tan⁡θ\frac{\sin\theta}{\cos\theta} = \tan\theta.

    =sec⁡θ+tan⁡θ= \sec\theta + \tan\theta

    This is equal to the right-hand side (RHS) of the identity.

    Thus, cos⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ\boxed{\frac{\cos\theta}{1-\sin\theta} = \sec\theta + \tan\theta} is proven.

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