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Show that cot θ + cosec θ / sin θ + tan θ = cosec θ cot θ

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cot θ + cosec θ / sin θ + tan θ = cosec θ cot θ

Step-by-step solution

7 steps

  1. Rewrite the left-hand side (LHS) of the equation in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta.

    Recall the identities:
    cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}
    csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}
    tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

    Substitute these into the LHS:

    LHS=cot⁡θ+csc⁡θsin⁡θ+tan⁡θ=cos⁡θsin⁡θ+1sin⁡θsin⁡θ+sin⁡θcos⁡θ\text{LHS} = \frac{\cot \theta + \csc \theta}{\sin \theta + \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}{\sin \theta + \frac{\sin \theta}{\cos \theta}}

  2. Simplify the numerator and the denominator separately.

    Numerator:

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

    Denominator:

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

    Factor out sin⁡θ\sin \theta from the denominator:

    sin⁡θ(cos⁡θ+1)cos⁡θ\frac{\sin \theta (\cos \theta + 1)}{\cos \theta}

  3. Substitute the simplified numerator and denominator back into the LHS expression.

    LHS=cos⁡θ+1sin⁡θsin⁡θ(cos⁡θ+1)cos⁡θ\text{LHS} = \frac{\frac{\cos \theta + 1}{\sin \theta}}{\frac{\sin \theta (\cos \theta + 1)}{\cos \theta}}

  4. Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator.

    LHS=cos⁡θ+1sin⁡θ×cos⁡θsin⁡θ(cos⁡θ+1)\text{LHS} = \frac{\cos \theta + 1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta (\cos \theta + 1)}

  5. Cancel out the common term (cos⁡θ+1)(\cos \theta + 1) from the numerator and denominator.

    LHS=1sin⁡θ×cos⁡θsin⁡θ\text{LHS} = \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta}

  6. Multiply the remaining terms.

    LHS=cos⁡θsin⁡2θ\text{LHS} = \frac{\cos \theta}{\sin^2 \theta}

  7. Rewrite the expression in terms of csc⁡θ\csc \theta and cot⁡θ\cot \theta.

    LHS=1sin⁡θ×cos⁡θsin⁡θ\text{LHS} = \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta}

    Using the identities 1sin⁡θ=csc⁡θ\frac{1}{\sin \theta} = \csc \theta and cos⁡θsin⁡θ=cot⁡θ\frac{\cos \theta}{\sin \theta} = \cot \theta:

    LHS=csc⁡θcot⁡θ\text{LHS} = \csc \theta \cot \theta

    This matches the right-hand side (RHS) of the original equation.

    Therefore, the identity is proven.

    cot⁡θ+csc⁡θsin⁡θ+tan⁡θ=csc⁡θcot⁡θ\boxed{\frac{\cot \theta + \csc \theta}{\sin \theta + \tan \theta} = \csc \theta \cot \theta}
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