Trigonometric identities and proofs

Reciprocal identities. These follow directly from how $\csc$, $\sec$ and $\cot$ are defined relative to $\sin$, $\cos$ and $\tan$:

where: each equality holds wherever the denominator on the right is nonzero — $\csc x$ and $\cot x$ are undefined when $\sin x = 0$, and $\sec x$ is undefined when $\cos x = 0$.

Quotient identities. $\tan x$ and $\cot x$ are ratios of the two basic functions:

The three Pythagorean identities. The first comes straight from $x^2+y^2=1$ on the unit circle; the other two are obtained by dividing every term through by $\cos^2x$ and by $\sin^2x$ respectively:

where: $x$ is any angle for which the functions involved are defined; the three identities are not independent facts to memorise separately but three views of one relation, $\sin^2x+\cos^2x=1$, seen through different lenses.

Two patterns worth spotting on sight. A denominator like $1-\sin x$ or $1+\cos x$ almost always invites multiplying top and bottom by its conjugate ($1+\sin x$ or $1-\cos x$), because the product collapses via a Pythagorean identity: $(1-\sin x)(1+\sin x) = 1-\sin^2x = \cos^2x$. Separately, an expression mixing $\sec x$ and $\tan x$ (or $\csc x$ and $\cot x$) often simplifies the moment you recall the rearranged Pythagorean forms $\sec^2x-\tan^2x=1$ and $\csc^2x-\cot^2x=1$. Recognising these two patterns before starting the algebra is what turns a long, messy proof into a short one.

Why bother proving identities at all? Two expressions that look completely different — say $\dfrac{1}{\sec x-\tan x}$ and $\sec x+\tan x$ — often turn out to be the same function written two ways, and knowing this lets you swap freely between whichever form is easier to work with in a later step, whether that is integrating, simplifying a bigger expression, or solving an equation in Class XII. The proof itself is also excellent algebra practice: it forces disciplined, reversible steps rather than guesswork, which is exactly the habit examiners are testing when they set an "identity" question rather than a "solve" question.

Problem. Prove that $\dfrac{\cos x}{1-\sin x} = \sec x + \tan x$.

Step 1. Multiply numerator and denominator of the LHS by the conjugate of $1-\sin x$, which is $1+\sin x$, so the denominator becomes a difference of squares:

Step 2. Apply the Pythagorean identity $1-\sin^2x=\cos^2x$ to the denominator:

Step 3. Cancel one factor of $\cos x$ from top and bottom — legitimate since $\cos x\ne0$ is already required for $\sec x$ and $\tan x$ on the RHS to exist:

The identity is proved. $\dfrac{\cos x}{1-\sin x} = \sec x + \tan x$, valid wherever $\cos x\ne0$.

Competitive stretch

The derivation's result $\dfrac{1}{\sec x-\tan x}=\sec x+\tan x$ can be reused rather than re-proved. Squaring both sides directly gives $\dfrac{1}{(\sec x-\tan x)^2}=(\sec x+\tan x)^2$ for free, with no fresh conjugate step needed. This is the kind of shortcut examiners reward: once an identity is proved, later questions that are algebraic consequences of it (squaring, taking reciprocals) follow in one line instead of restarting the whole proof from scratch.

• Treating "prove the identity" as "solve for $x$" — an identity must hold for every valid $x$, so the task is algebraic transformation, never finding a specific angle.
• Cross-multiplying across the equals sign or working both sides at once assuming they are already equal — this can "prove" a false statement; always reduce one side (or both sides independently) to a common form first.
• Forgetting the implicit domain restriction — in the worked example, $\cos x\ne0$ and $\sin x\ne1$ must hold at every step for $\sec x$ and $\tan x$ to even be defined.
• Mixing up which Pythagorean identity pairs with which function: $1+\tan^2x=\sec^2x$ goes with $\tan$–$\sec$, while $1+\cot^2x=\csc^2x$ goes with $\cot$–$\csc$; swapping them is a very common slip.
• Stopping the simplification one step early and declaring the proof done — the final simplified expression must match the RHS exactly (same functions, same form), not merely "look close"; $\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}$ is only finished once it is rewritten as $\sec x+\tan x$ to match the target.

Ask the tutor: Stuck partway through proving an identity — ask the AI tutor to check your algebra line by line and suggest which Pythagorean identity or conjugate move applies next.