Question

1. \frac{\cos 2a + 1}{\sin 2a} = \cot a \quad \text{hint:on the left, make double-angle identity}

Explanation:

Identify the given identity

We need to prove the following trigonometric identity:

$$ \frac{\cos 2A + 1}{\sin 2A} = \cot A $$

Apply double-angle formulas

We use the double-angle identities for cosine and sine:

$$ \cos 2A = 2\cos^2 A - 1 $$

$$ \sin 2A = 2\sin A \cos A $$

Substitute into the left-hand side

Substitute these expressions into the numerator and denominator:

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

Simplify the expression

Simplify the numerator and cancel common terms:

$$ \text{LHS} = \frac{2\cos^2 A}{2\sin A \cos A} = \frac{\cos A}{\sin A} $$

Relate to the right-hand side

Using the definition of the cotangent function:

$$ \frac{\cos A}{\sin A} = \cot A = \text{RHS} $$

Answer:

To prove the identity \(\frac{\cos 2A + 1}{\sin 2A} = \cot A\):

1. Substitute the double-angle identities \(\cos 2A = 2\cos^2 A - 1\) and \(\sin 2A = 2\sin A \cos A\) into the left-hand side:

$$ \frac{\cos 2A + 1}{\sin 2A} = \frac{(2\cos^2 A - 1) + 1}{2\sin A \cos A} $$

1. Simplify the numerator:

$$ \frac{2\cos^2 A}{2\sin A \cos A} $$

1. Cancel the common factors \(2\) and \(\cos A\):

$$ \frac{\cos A}{\sin A} = \cot A $$

Thus, the identity is proven.