Trigonometric identities

Take a right triangle with acute angle $\theta$, side opposite $a$, side adjacent $b$, and hypotenuse $c$. The Pythagoras theorem states

where: $a$ is the side opposite $\theta$, $b$ is the side adjacent to $\theta$, and $c$ is the hypotenuse. Dividing every term by $c^2$ turns each ratio into a trigonometric ratio, since $a/c=\sin\theta$ and $b/c=\cos\theta$:

Dividing instead by $b^2$ (valid only when $\cos\theta\ne0$, i.e. $\theta\ne90^\circ$) gives

and dividing by $a^2$ (valid only when $\sin\theta\ne0$, i.e. $\theta\ne0^\circ$) gives

where: in each identity, $\tan\theta=\sin\theta/\cos\theta$, $\sec\theta=1/\cos\theta$, $\cot\theta=\cos\theta/\sin\theta$, $\csc\theta=1/\sin\theta$. These are called identities, not just equations, because they hold for every acute $\theta$ — unlike $\sin\theta=\tfrac12$, which is true only at $\theta=30^\circ$. Each identity rearranges into a companion form worth memorising alongside it: $\sin^2\theta=1-\cos^2\theta$; $\sec^2\theta-\tan^2\theta=1$; $\csc^2\theta-\cot^2\theta=1$. A problem will often hand you a hint (a $\sec\theta-\tan\theta$ pairing, say) pointing at the exact identity to reach for. The single most useful problem-solving habit here: whenever an expression mixes $\tan\theta$, $\sec\theta$, $\cot\theta$ or $\csc\theta$ with fractions or square roots and looks stuck, rewrite everything in $\sin\theta$ and $\cos\theta$ first using the ratio definitions above. Very often the expression collapses using Identity 1 alone; only if it does not do you reach for Identity 2 or Identity 3. It also helps to notice the family resemblance between the three identities: each pairs a "$1$" with the square of one ratio to produce the square of a related ratio, and in every case the two ratios on the squared side are reciprocal-adjacent — $\tan\theta$ sits next to $\sec\theta$ because both are built from $\cos\theta$ in the denominator, and $\cot\theta$ sits next to $\csc\theta$ because both are built from $\sin\theta$ in the denominator. Spotting which pair a question is nudging you toward — by looking at which ratios appear together — is often faster than blindly converting everything to $\sin\theta$ and $\cos\theta$, though that fallback always works. A CBSE-style question will typically hand you an expression like $\dfrac{\cos\theta}{1-\sin\theta}+\dfrac{\cos\theta}{1+\sin\theta}$ and ask you to simplify or prove it equals something specific, say $2\sec\theta$. Combine the fractions over a common denominator $(1-\sin\theta)(1+\sin\theta)=1-\sin^2\theta$, which Identity 1 immediately tells you equals $\cos^2\theta$. The numerator becomes $\cos\theta(1+\sin\theta)+\cos\theta(1-\sin\theta)=2\cos\theta$, so the whole expression collapses to $\dfrac{2\cos\theta}{\cos^2\theta}=\dfrac{2}{\cos\theta}=2\sec\theta$ — exactly the target, and the entire simplification used nothing beyond Identity 1 and basic algebra.

Problem. Verify $1+\tan^2\theta=\sec^2\theta$ numerically at $\theta=30^\circ$, then use it to prove that $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=1$ for every acute $\theta$.

Step 1. Use the standard values $\tan30^\circ=\dfrac{1}{\sqrt3}$ and $\sec30^\circ=\dfrac{2}{\sqrt3}$.

Step 2. Compute the left side: $\text{LHS}=1+\left(\dfrac{1}{\sqrt3}\right)^2=1+\dfrac13=\dfrac43$.

Step 3. Compute the right side: $\text{RHS}=\left(\dfrac{2}{\sqrt3}\right)^2=\dfrac{4}{3}$. Since $\text{LHS}=\text{RHS}=\dfrac43$, the identity checks out at $\theta=30^\circ$, as it must for every acute angle.

Step 4. Now prove the second claim using algebra, not numbers. The left side has the form $(x-y)(x+y)=x^2-y^2$ with $x=\sec\theta$, $y=\tan\theta$, so $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)=\sec^2\theta-\tan^2\theta$.

Step 5. Rearrange the identity just derived, $1+\tan^2\theta=\sec^2\theta$, into $\sec^2\theta-\tan^2\theta=1$, and substitute directly. The left side simplifies to exactly $1$, proving the identity for every acute $\theta$ (excluding $90^\circ$, where $\tan\theta$ and $\sec\theta$ are undefined).

Competitive stretch

Because $\sec^2\theta-\tan^2\theta=1$ always factors as $(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)$, the two bracketed expressions are always reciprocals of each other. So if a question states $\sec\theta+\tan\theta=k$, you can write $\sec\theta-\tan\theta=\dfrac1k$ immediately, without any further algebra — a shortcut that shows up repeatedly in CBSE "prove that" problems built on Identity 2.

• Reading $\sin^2\theta$ as $\sin(\theta^2)$, squaring the angle — it always means $(\sin\theta)^2$, squaring the ratio, and the same convention applies to $\cos^2\theta$, $\tan^2\theta$ and every other squared ratio you will meet.
• Treating $\sin^2\theta+\cos^2\theta=1$ as true only for special angles like $30^\circ$ — it is an identity, valid for every acute angle, not a fact to verify once and move on. Checking it numerically at one angle is a good sanity check on your algebra, never a substitute for the general proof.
• Manipulating both sides of a "prove that" statement simultaneously as if it were already known to be true — the correct method is to simplify one side, usually by converting to $\sin\theta$ and $\cos\theta$, until it matches the other, or to reduce both sides independently to the same expression.
• Dividing by $\cos\theta$ or $\sin\theta$ without checking they are nonzero — this is exactly why Identity 2 fails at $\theta=90^\circ$ and Identity 3 fails at $\theta=0^\circ$. Most CBSE problems restrict $\theta$ to acute angles inside a triangle precisely so this issue never arises, but it is worth knowing why the restriction exists.
• Forgetting the rearranged forms exist and re-deriving them from scratch every time — knowing $\sec^2\theta-\tan^2\theta=1$ and $\csc^2\theta-\cot^2\theta=1$ by heart saves a line of algebra in almost every proving question.

Ask the tutor: Stuck partway through a "prove that" identity question, or want more worked examples solved step by step? Ask the AI tutor to walk through additional problems with you.