Question

Hint:

Always convert all trigonometric functions to sine and cosine for simplification-based proofs.

Answer 1

Step 1: Write the LHS.
\[ \text{LHS} = (\csc A - \sin A)(\sec A - \cos A) \]
Step 2: Express in terms of sine and cosine.
\[ \text{LHS} = \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right) \]
Step 3: Simplify each bracket.
\[ \frac{1 - \sin^2 A}{\sin A} \times \frac{1 - \cos^2 A}{\cos A} \]
Step 4: Use trigonometric identity.
\[ 1 - \sin^2 A = \cos^2 A \quad \text{and} \quad 1 - \cos^2 A = \sin^2 A \] \[ \Rightarrow \text{LHS} = \frac{\cos^2 A}{\sin A} \times \frac{\sin^2 A}{\cos A} \]
Step 5: Simplify the expression.
\[ \text{LHS} = \frac{\sin A \cos A}{1} \] Step 6: Write the RHS and simplify.
\[ \text{RHS} = \frac{1}{\tan A + \cot A} \] \[ = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} \] \[ \text{RHS} = \frac{\sin A \cos A}{1} = \sin A \cos A \] Step 7: Conclusion.
\[ \text{LHS} = \text{RHS} = \sin A \cos A \] Hence proved that \[ (\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A} \]