Trigonometric identities

One triangle, three identities. Picture a right triangle with angle $A$, side opposite $A$ of length $a$, side adjacent to $A$ of length $b$, and hypotenuse $h$. Pythagoras' theorem gives $a^2+b^2=h^2$. Dividing every term by $h^2$, then by $b^2$, then by $a^2$, and using $\sin A = a/h$, $\cos A = b/h$, $\tan A = a/b$, $\sec A = h/b$, $\cot A = b/a$, $\csc A = h/a$, produces the three identities you must know cold:

where: $A$ is the acute angle of the triangle, and each pair of ratios ($\sin,\cos$), ($\tan,\sec$), ($\cot,\csc$) is grouped because they come from dividing the same Pythagorean relation by the same square.

Rearranged forms are the real exam skill. Since these are identities, every term can be isolated and the resulting equation is just as valid:

Almost every "prove that…" question is really asking you to spot which of these six forms (three identities, three rearrangements) turns a messy expression into something that cancels.

Two algebraic patterns do most of the work. Expanding $(\sin A + \cos A)^2$ with the ordinary identity $(x+y)^2 = x^2+2xy+y^2$ gives $\sin^2A+2\sin A\cos A+\cos^2A = 1+2\sin A\cos A$, using $\sin^2A+\cos^2A=1$. And a difference of squares, $(\sec A-\tan A)(\sec A+\tan A) = \sec^2A-\tan^2A = 1$, is the single most-tested trick in this unit — it appears whenever an expression has $\sec A \pm \tan A$ or $\csc A \pm \cot A$ in it. Recognising these two shapes on sight is what separates a fast, confident proof from a stalled one.

Prove that $\dfrac{1}{\sec A+\tan A} = \sec A - \tan A$, for $0^\circ \le A \le 90^\circ$, $A \ne 90^\circ$, and then use the result to simplify $(\sec A+\tan A)(\sec A - \tan A) + 2\tan^2A - \sec^2A$.

Step 1. Start the proof from the right-hand side, since it is the simpler expression: $\text{RHS} = \sec A - \tan A$. Multiply and divide by $(\sec A+\tan A)$ so the difference-of-squares pattern appears: $\sec A-\tan A = \dfrac{(\sec A-\tan A)(\sec A+\tan A)}{\sec A+\tan A} = \dfrac{\sec^2A-\tan^2A}{\sec A+\tan A}$.

Step 2. Apply $\sec^2A-\tan^2A=1$ (from the derivation above) to the numerator: $\dfrac{\sec^2A-\tan^2A}{\sec A+\tan A} = \dfrac{1}{\sec A+\tan A}$. This is exactly the left-hand side, so $\text{LHS}=\text{RHS}$, and the identity is proved.

Step 3. Now use it to simplify $(\sec A+\tan A)(\sec A-\tan A)+2\tan^2A-\sec^2A$. The bracketed product is again a difference of squares: $(\sec A+\tan A)(\sec A-\tan A) = \sec^2A-\tan^2A = 1$.

Step 4. Substitute this back: $1+2\tan^2A-\sec^2A$. Replace $\sec^2A$ using $\sec^2A=1+\tan^2A$: $1+2\tan^2A-(1+\tan^2A) = 1+2\tan^2A-1-\tan^2A = \tan^2A$. The proof holds, and the expression simplifies to $\tan^2A$.

Competitive stretch

If instead you were given $\sec A + \tan A = 3$, the just-proved identity lets you find $\sec A - \tan A$ instantly without knowing $A$: since $(\sec A+\tan A)(\sec A-\tan A)=1$, it follows that $\sec A - \tan A = 1/3$. Adding the two equations gives $2\sec A = 3+\tfrac13 = \tfrac{10}{3}$, so $\sec A = \tfrac{5}{3}$ and $\tan A = 3-\tfrac{5}{3}=\tfrac{4}{3}$ — a full solution using only the identity, no triangle-solving required.

• Confusing $\sin^2A$ with $\sin A^2$. $\sin^2A$ means $(\sin A)^2$ — the whole ratio is squared, not the angle.
• Trying to "prove" an identity by plugging in one angle, such as $A=30^\circ$. That checks a single case, not a general proof; full marks need the algebraic route shown above.
• Manipulating both sides at once. Pick the messier side, simplify it step by step using the identities, and show it becomes the other side — mixing operations across both sides gets confusing and error-prone.
• Sign slips when rearranging, e.g. writing $\tan^2A = 1-\sec^2A$ instead of the correct $\sec^2A-1$. Always rearrange one identity at a time and re-check the sign.
• Forgetting the domain restriction. $\tan A$ and $\sec A$ blow up at $A=90^\circ$; $\cot A$ and $\csc A$ blow up at $A=0^\circ$. ICSE questions are usually set to avoid these values, but stay alert if a range is given explicitly.

Ask the tutor: if a "prove that…" question is not clicking, ask the AI tutor for a fresh practice expression and step-by-step feedback on which identity and pattern to use.