Trigonometric Identity Proofs (Exercise 8.3)

This lesson works through six proofs from Class 10 trigonometry, Exercise 8.3. For each identity we simplify one or both sides, rewriting $\sec$, $\csc$ and $\cot$ in terms of $\sin$ and $\cos$, until the two sides agree.

A quick note on strategy: you can start from the left side and reach the right, start from the right side and reach the left, or simplify each side separately to the same result. Pick one route and write hence proved at the end. Do not mix all three at once.

Proof 1

Prove that

$\sqrt{\frac{1+\sin A}{1-\sin A}} = \sec A + \tan A.$

Start from the left side and rationalise: multiply numerator and denominator inside the root by the conjugate $1+\sin A$.

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

Since $1-\sin^2 A = \cos^2 A$,

$= \sqrt{\frac{(1+\sin A)^2}{\cos^2 A}} = \frac{1+\sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}.$

Proof 2

Prove that

$(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A.$

Expand each square using $(a+b)^2 = a^2 + 2ab + b^2$.

Left side

$(\sin A + \csc A)^2 = \sin^2 A + \csc^2 A + 2\sin A\,\csc A,$ $(\cos A + \sec A)^2 = \cos^2 A + \sec^2 A + 2\cos A\,\sec A.$

Now $\sin A\,\csc A = 1$ and $\cos A\,\sec A = 1$, so both cross terms give $2(1) = 2$.

$\text{LHS} = (\sin^2 A + \cos^2 A) + \csc^2 A + \sec^2 A + 2 + 2.$

Use $\sin^2 A + \cos^2 A = 1$, and the identities $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$:

$= 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4 = 7 + \tan^2 A + \cot^2 A = \text{RHS}.$

Proof 3

Prove that

$\frac{1+\tan^2 A}{1+\cot^2 A} = \left(\frac{1-\tan A}{1-\cot A}\right)^2 = \tan^2 A.$

Prove the two equalities in turn.

First equality

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

Second equality

$\left(\frac{1-\tan A}{1-\cot A}\right)^2 = \left(\frac{1-\tan A}{1-\tfrac{1}{\tan A}}\right)^2 = \left(\frac{1-\tan A}{\tfrac{\tan A - 1}{\tan A}}\right)^2 = \left(\frac{\tan A\,(1-\tan A)}{\tan A - 1}\right)^2.$

Since $1-\tan A = -(\tan A - 1)$, the bracket equals $-\tan A$, and squaring gives

$= \tan^2 A.$

Both sides equal $\tan^2 A$, so the identity holds.

Proof 4

Prove that

$\frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\,\csc\theta.$

Write everything in $\sin\theta$ and $\cos\theta$.

$\text{LHS} = \frac{\tfrac{\sin\theta}{\cos\theta}}{1-\tfrac{\cos\theta}{\sin\theta}} + \frac{\tfrac{\cos\theta}{\sin\theta}}{1-\tfrac{\sin\theta}{\cos\theta}} = \frac{\sin^2\theta}{\cos\theta\,(\sin\theta-\cos\theta)} + \frac{\cos^2\theta}{\sin\theta\,(\cos\theta-\sin\theta)}.$

Factor $-1$ from the second denominator so both share $(\sin\theta-\cos\theta)$:

$= \frac{1}{\sin\theta-\cos\theta}\left(\frac{\sin^2\theta}{\cos\theta} - \frac{\cos^2\theta}{\sin\theta}\right) = \frac{1}{\sin\theta-\cos\theta}\cdot\frac{\sin^3\theta-\cos^3\theta}{\sin\theta\cos\theta}.$

Use $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ with $a=\sin\theta$, $b=\cos\theta$, and cancel the common $(\sin\theta-\cos\theta)$:

$= \frac{\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta}.$

Splitting the fraction,

$= \frac{1}{\sin\theta\cos\theta} + 1 = \sec\theta\,\csc\theta + 1 = 1 + \sec\theta\,\csc\theta = \text{RHS}.$

Proof 5

Prove that

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

Simplify each side to the same result.

Left side

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

Right side

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

Both sides equal $1 + \cos A$, so the identity holds.

Proof 6

Prove that

$(\csc A - \sin A)(\sec A - \cos A) = \frac{1}{\tan A + \cot A}.$

Left side

$(\csc A - \sin A)(\sec A - \cos A) = \left(\frac{1}{\sin A} - \sin A\right)\left(\frac{1}{\cos A} - \cos A\right) = \frac{1-\sin^2 A}{\sin A}\cdot\frac{1-\cos^2 A}{\cos A}.$

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

Right side

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

Both sides equal $\sin A\cos A$, so the identity holds.

Key takeaways

• Rewriting $\sec$, $\csc$ and $\cot$ in terms of $\sin$ and $\cos$ turns most identities into ordinary fraction algebra.
• The Pythagorean identity $\sin^2 A + \cos^2 A = 1$, and its forms $1-\sin^2 A = \cos^2 A$ and $1-\cos^2 A = \sin^2 A$, close out almost every proof.
• When one side is hard to reach, simplify both sides separately to a common expression instead.