Two Important Trigonometry Proofs

This lesson works through two important trigonometric identity proofs. Both use a Pythagorean identity to rewrite the expression, then cancel and rationalise to reach the required side.

Proof 1: $\dfrac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \sec A + \tan A$

We start from the left-hand side and use the identity $\sec^2 A - \tan^2 A = 1$, which rearranges to $\tan^2 A - \sec^2 A = -1$.

$\text{LHS} = \frac{\tan A + \sec A - 1}{\tan A - \sec A + 1}$

Multiply the numerator and denominator by $(\tan A - \sec A)$:

$\text{LHS} = \frac{(\tan A + \sec A - 1)(\tan A - \sec A)}{(\tan A - \sec A + 1)(\tan A - \sec A)}$

Numerator. Group the first two terms so they form $(\tan A + \sec A)(\tan A - \sec A)$:

$(\tan A + \sec A)(\tan A - \sec A) - 1\cdot(\tan A - \sec A) = (\tan^2 A - \sec^2 A) - (\tan A - \sec A)$

Since $\tan^2 A - \sec^2 A = -1$, this becomes

$-1 - \tan A + \sec A = -\big(\tan A - \sec A + 1\big).$

So the numerator is the negative of $(\tan A - \sec A + 1)$, which also appears in the denominator.

$\text{LHS} = \frac{-(\tan A - \sec A + 1)}{(\tan A - \sec A + 1)(\tan A - \sec A)} = \frac{-1}{\tan A - \sec A} = \frac{1}{\sec A - \tan A}.$

Rationalise. Multiply the numerator and denominator by the conjugate $(\sec A + \tan A)$:

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

This equals the right-hand side, so the identity is proved.

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

Here we use the identity $\operatorname{cosec}^2 A = 1 + \cot^2 A$, so that $\cot^2 A - \operatorname{cosec}^2 A = -1$.

$\text{LHS} = \frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$

Divide every term in the numerator and denominator by $\sin A$. Using $\tfrac{\cos A}{\sin A} = \cot A$ and $\tfrac{1}{\sin A} = \operatorname{cosec} A$:

$\text{LHS} = \frac{\cot A - 1 + \operatorname{cosec} A}{\cot A + 1 - \operatorname{cosec} A} = \frac{\cot A + \operatorname{cosec} A - 1}{\cot A - \operatorname{cosec} A + 1}.$

This now has exactly the same form as Proof 1. Multiply the numerator and denominator by $(\cot A - \operatorname{cosec} A)$.

Numerator. Group the first two terms as $(\cot A + \operatorname{cosec} A)(\cot A - \operatorname{cosec} A)$:

$(\cot^2 A - \operatorname{cosec}^2 A) - (\cot A - \operatorname{cosec} A) = -1 - \cot A + \operatorname{cosec} A = -\big(\cot A - \operatorname{cosec} A + 1\big).$

The denominator becomes $(\cot A - \operatorname{cosec} A + 1)(\cot A - \operatorname{cosec} A)$, so the common factor cancels:

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

Rationalise. Multiply by the conjugate $(\operatorname{cosec} A + \cot A)$:

$\text{LHS} = \frac{\operatorname{cosec} A + \cot A}{\operatorname{cosec}^2 A - \cot^2 A} = \frac{\operatorname{cosec} A + \cot A}{1} = \operatorname{cosec} A + \cot A.$

This equals the right-hand side, so the identity is proved.

Key takeaways

• Multiplying the numerator and denominator by a well-chosen factor lets a Pythagorean identity collapse the expression.
• $\sec^2 A - \tan^2 A = 1$ and $\operatorname{cosec}^2 A - \cot^2 A = 1$ are the engines of both proofs.
• Dividing a sine and cosine expression through by $\sin A$ converts it into cotangent and cosecant, turning the second proof into the same shape as the first.
• A final conjugate multiplication rationalises the denominator and delivers the required side.