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

Question

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

Asked on March 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-26T14:17:11.332Z

Step 1: Rewrite the left-hand side (LHS) of the equation in terms of $\sin \theta$ and $\cos \theta$. Recall the identities: $\cot \theta = \frac{\cos \theta}{\sin \theta}$ $\csc \theta = \frac{1}{\sin \theta}$ $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Substitute these into the LHS: $$ \text{LHS} = \frac{\cot \theta + \csc \theta}{\sin \theta + \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}{\sin \theta + \frac{\sin \theta}{\cos \theta}} $$ Step 2: Simplify the numerator and the denominator separately. Numerator: $$ \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} = \frac{\cos \theta + 1}{\sin \theta} $$ Denominator: $$ \sin \theta + \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta \cos \theta}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta \cos \theta + \sin \theta}{\cos \theta} $$ Factor out $\sin \theta$ from the denominator: $$ \frac{\sin \theta (\cos \theta + 1)}{\cos \theta} $$ Step 3: Substitute the simplified numerator and denominator back into the LHS expression. $$ \text{LHS} = \frac{\frac{\cos \theta + 1}{\sin \theta}}{\frac{\sin \theta (\cos \theta + 1)}{\cos \theta}} $$ Step 4: Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator. $$ \text{LHS} = \frac{\cos \theta + 1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta (\cos \theta + 1)} $$ Step 5: Cancel out the common term $(\cos \theta + 1)$ from the numerator and denominator. $$ \text{LHS} = \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta} $$ Step 6: Multiply the remaining terms. $$ \text{LHS} = \frac{\cos \theta}{\sin^2 \theta} $$ Step 7: Rewrite the expression in terms of $\csc \theta$ and $\cot \theta$. $$ \text{LHS} = \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta} $$ Using the identities $\frac{1}{\sin \theta} = \csc \theta$ and $\frac{\cos \theta}{\sin \theta} = \cot \theta$: $$ \text{LHS} = \csc \theta \cot \theta $$ This matches the right-hand side (RHS) of the original equation. Therefore, the identity is proven. $$\boxed{\frac{\cot \theta + \csc \theta}{\sin \theta + \tan \theta} = \csc \theta \cot \theta}$$

Accepted Answer · 2026-03-26T14:17:11.332Z

Step 1: Rewrite the left-hand side (LHS) of the equation in terms of $\sin \theta$ and $\cos \theta$. Recall the identities: $\cot \theta = \frac{\cos \theta}{\sin \theta}$ $\csc \theta = \frac{1}{\sin \theta}$ $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Substitute these into the LHS: $$ \text{LHS} = \frac{\cot \theta + \csc \theta}{\sin \theta + \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}{\sin \theta + \frac{\sin \theta}{\cos \theta}} $$ Step 2: Simplify the numerator and the denominator separately. Numerator: $$ \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} = \frac{\cos \theta + 1}{\sin \theta} $$ Denominator: $$ \sin \theta + \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta \cos \theta}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta \cos \theta + \sin \theta}{\cos \theta} $$ Factor out $\sin \theta$ from the denominator: $$ \frac{\sin \theta (\cos \theta + 1)}{\cos \theta} $$ Step 3: Substitute the simplified numerator and denominator back into the LHS expression. $$ \text{LHS} = \frac{\frac{\cos \theta + 1}{\sin \theta}}{\frac{\sin \theta (\cos \theta + 1)}{\cos \theta}} $$ Step 4: Simplify the complex fraction by multiplying the numerator by the reciprocal of the denominator. $$ \text{LHS} = \frac{\cos \theta + 1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta (\cos \theta + 1)} $$ Step 5: Cancel out the common term $(\cos \theta + 1)$ from the numerator and denominator. $$ \text{LHS} = \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta} $$ Step 6: Multiply the remaining terms. $$ \text{LHS} = \frac{\cos \theta}{\sin^2 \theta} $$ Step 7: Rewrite the expression in terms of $\csc \theta$ and $\cot \theta$. $$ \text{LHS} = \frac{1}{\sin \theta} \times \frac{\cos \theta}{\sin \theta} $$ Using the identities $\frac{1}{\sin \theta} = \csc \theta$ and $\frac{\cos \theta}{\sin \theta} = \cot \theta$: $$ \text{LHS} = \csc \theta \cot \theta $$ This matches the right-hand side (RHS) of the original equation. Therefore, the identity is proven. $$\boxed{\frac{\cot \theta + \csc \theta}{\sin \theta + \tan \theta} = \csc \theta \cot \theta}$$