Lesson notes

This lesson finishes the remaining proofs from Exercise 8.3. In each one we start from the left-hand side (LHS), simplify it using the basic trigonometric ratios and the Pythagorean identities, and show that it equals the right-hand side (RHS).

Proof 1: $\dfrac{1+\sec A}{\sec A} = \dfrac{\sin^2 A}{1-\cos A}$

We show both sides reduce to $1+\cos A$.

Left side. Use $\sec A = \dfrac{1}{\cos A}$:

$\frac{1+\sec A}{\sec A} = \frac{1+\tfrac{1}{\cos A}}{\tfrac{1}{\cos A}} = \frac{\tfrac{\cos A + 1}{\cos A}}{\tfrac{1}{\cos A}} = (\cos A + 1)\cdot\frac{\cos A}{\cos A} = 1+\cos A.$

Right side. Use $\sin^2 A = 1-\cos^2 A$ and factor as a difference of squares:

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

Both sides equal $1+\cos A$, so LHS $=$ RHS. Hence proved.

Proof 2: $\dfrac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$

This is a longer, important proof. We use the identity $\csc^2 A = 1 + \cot^2 A$.

Divide top and bottom by $\sin A$ to bring in $\cot A$ and $\csc A$:

$\text{LHS} = \frac{\tfrac{\cos A}{\sin A} - \tfrac{\sin A}{\sin A} + \tfrac{1}{\sin A}}{\tfrac{\cos A}{\sin A} + \tfrac{\sin A}{\sin A} - \tfrac{1}{\sin A}} = \frac{\cot A - 1 + \csc A}{\cot A + 1 - \csc A}.$

Rearranging the terms:

$\text{LHS} = \frac{(\cot A + \csc A) - 1}{(\cot A - \csc A) + 1}.$

Multiply top and bottom by $\cot A - \csc A$ to set up a square relation:

$\text{LHS} = \frac{\big[(\cot A + \csc A) - 1\big](\cot A - \csc A)}{\big[(\cot A - \csc A) + 1\big](\cot A - \csc A)}.$

Expand the numerator. Note $(\cot A + \csc A)(\cot A - \csc A) = \cot^2 A - \csc^2 A = -1$, since $\csc^2 A - \cot^2 A = 1$:

$\text{numerator} = (\cot^2 A - \csc^2 A) - (\cot A - \csc A) = -1 - (\cot A - \csc A).$

So

$\text{LHS} = \frac{-1 - \cot A + \csc A}{(\cot A - \csc A + 1)(\cot A - \csc A)} = \frac{-(1 + \cot A - \csc A)}{(1 + \cot A - \csc A)(\cot A - \csc A)}.$

The factor $1 + \cot A - \csc A$ cancels:

$\text{LHS} = \frac{-1}{\cot A - \csc A} = \frac{1}{\csc A - \cot A}.$

Multiply top and bottom by the conjugate $\csc A + \cot A$:

$\text{LHS} = \frac{\csc A + \cot A}{(\csc A - \cot A)(\csc A + \cot A)} = \frac{\csc A + \cot A}{\csc^2 A - \cot^2 A} = \frac{\csc A + \cot A}{1} = \csc A + \cot A.$

This is the RHS. Hence proved.

Proof 3: $(\csc A - \sin A)(\sec A - \cos A) = \dfrac{1}{\tan A + \cot A}$

We simplify each side to $\sin A \cos A$.

Left side. Write each bracket over a common denominator:

$\csc A - \sin A = \frac{1}{\sin A} - \sin A = \frac{1-\sin^2 A}{\sin A} = \frac{\cos^2 A}{\sin A},$

$\sec A - \cos A = \frac{1}{\cos A} - \cos A = \frac{1-\cos^2 A}{\cos A} = \frac{\sin^2 A}{\cos A}.$

Multiplying and cancelling:

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

Right side. Combine $\tan A + \cot A$:

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

So

$\text{RHS} = \frac{1}{\tan A + \cot A} = \sin A \cos A.$

Both sides equal $\sin A \cos A$, so LHS $=$ RHS. Hence proved.

Key takeaways

• To prove an identity, simplify one side (or each side) using $\sec A = \tfrac{1}{\cos A}$, $\csc A = \tfrac{1}{\sin A}$, $\cot A = \tfrac{\cos A}{\sin A}$ until it matches the other.
• The Pythagorean identities $\sin^2 A + \cos^2 A = 1$ and $\csc^2 A - \cot^2 A = 1$ are the main tools, often combined with the difference of squares.
• For a tough fraction, multiplying top and bottom by a conjugate turns the denominator into one of these square identities and clears it.