What are Complementary Angles? Two angles are complementary if they add up to exactly $90^{\circ}$. In any right triangle the right angle uses $90^{\circ}$, so the two acute angles must total $90^{\circ}$ — they are always a complementary pair. If one acute angle is $\theta$, the other is $90^{\circ}-\theta$.
Why the ratios swap partners. For the angle $\theta$ the "opposite" side faces $\theta$; for the other acute angle $90^{\circ}-\theta$ that very same side becomes the "adjacent" side, and vice versa. The hypotenuse is shared. So $\sin(90^{\circ}-\theta)$, which is opposite-over-hypotenuse for the angle $90^{\circ}-\theta$, is exactly cosine of $\theta$. Tracing each ratio the same way gives the full set of co-ratio identities.
| Function of $90^{\circ}-\theta$ | Equals |
|---|
| $\sin(90^{\circ}-\theta)$ | $\cos\theta$ |
| $\cos(90^{\circ}-\theta)$ | $\sin\theta$ |
| $\tan(90^{\circ}-\theta)$ | $\cot\theta$ |
| $\cot(90^{\circ}-\theta)$ | $\tan\theta$ |
| $\sec(90^{\circ}-\theta)$ | $\operatorname{cosec}\theta$ |
| $\operatorname{cosec}(90^{\circ}-\theta)$ | $\sec\theta$ |
The pairing rule (memory hook). Think of three tag-teams: sine–cosine, tangent–cotangent, secant–cosecant. Subtracting the angle from $90^{\circ}$ hands the value to the partner of the same team. The "co-" in cosine, cotangent and cosecant literally stands for "complement."
Three classic problem types.
- Direct evaluation: ratios like $\dfrac{\sin35^{\circ}}{\cos55^{\circ}}$ collapse to $1$ because $55^{\circ}=90^{\circ}-35^{\circ}$, so $\cos55^{\circ}=\sin35^{\circ}$.
- Products that pair off: in a long product such as $\tan10^{\circ}\tan20^{\circ}\cdots\tan80^{\circ}$, group angles that add to $90^{\circ}$; each pair $\tan\theta\cdot\tan(90^{\circ}-\theta)=\tan\theta\cot\theta=1$.
- Solving for an unknown angle: equations like $\sin3A=\cos(A-26^{\circ})$ are solved by turning one side into the other ratio — write $\sin3A=\cos(90^{\circ}-3A)$, then equate angles.
Mixing co-ratios with the Pythagorean identities. Many board questions blend both ideas. For instance, $\dfrac{\cos^{2}20^{\circ}+\cos^{2}70^{\circ}}{\ \ }$ uses $\cos70^{\circ}=\sin20^{\circ}$ so the numerator becomes $\cos^{2}20^{\circ}+\sin^{2}20^{\circ}=1$. The plan is always the same: first apply $90^{\circ}-\theta$ to make the two angles equal, then collapse the sum of squares with $\sin^{2}\theta+\cos^{2}\theta=1$. Products such as $\sec70^{\circ}\sin20^{\circ}$ also vanish neatly, because $\sec70^{\circ}=\operatorname{cosec}20^{\circ}=\dfrac{1}{\sin20^{\circ}}$ leaves $\sin20^{\circ}\cdot\dfrac{1}{\sin20^{\circ}}=1$.
A quick worked check of the table. Pick $\theta=30^{\circ}$. Then $90^{\circ}-\theta=60^{\circ}$, and indeed $\sin60^{\circ}=\dfrac{\sqrt3}{2}=\cos30^{\circ}$, while $\tan60^{\circ}=\sqrt3=\cot30^{\circ}$. Substituting a standard angle like this is the fastest way to confirm you have the right co-function before committing to a long simplification.
Why the formulas only use $90^{\circ}-\theta$ in Class 10. The co-ratio identities here are stated for acute angles, where both $\theta$ and $90^{\circ}-\theta$ lie between $0^{\circ}$ and $90^{\circ}$ and every ratio is positive. You do not need to worry about negative signs or angles beyond $90^{\circ}$ at this level — that extension comes in higher classes. So whenever you rewrite an angle, keep it inside the range $0^{\circ}$ to $90^{\circ}$, which is exactly why you check that the resulting angle stays acute.
Common mistakes. Applying the rule without checking that the angles really add to $90^{\circ}$; forgetting that the function name changes to its co-function (e.g. writing $\sin(90^{\circ}-\theta)=\sin\theta$ instead of $\cos\theta$); and, when solving for an angle, equating two different ratios directly (you must first make both sides the same ratio). Always verify your answer keeps the stated angle acute.
Evaluate $\dfrac{\sin18^{\circ}}{\cos72^{\circ}}$.
Solution- Step 1: Check the angles: $18^{\circ}+72^{\circ}=90^{\circ}$, so they are complementary.
- Step 2: Write $\sin18^{\circ}=\sin(90^{\circ}-72^{\circ})=\cos72^{\circ}$.
- Step 3: The expression becomes $\dfrac{\cos72^{\circ}}{\cos72^{\circ}}$.
- Step 4: $=1$.
Answer: $1$.
Express $\cos40^{\circ}$ in terms of a sine, and $\tan65^{\circ}$ in terms of a cotangent.
Solution- Step 1: $\cos40^{\circ}=\cos(90^{\circ}-50^{\circ})$.
- Step 2: Using $\cos(90^{\circ}-\theta)=\sin\theta$, this is $\sin50^{\circ}$.
- Step 3: $\tan65^{\circ}=\tan(90^{\circ}-25^{\circ})$.
- Step 4: Using $\tan(90^{\circ}-\theta)=\cot\theta$, this is $\cot25^{\circ}$.
Answer: $\cos40^{\circ}=\sin50^{\circ}$ and $\tan65^{\circ}=\cot25^{\circ}$.
Evaluate $\cos48^{\circ}-\sin42^{\circ}$.
Solution- Step 1: Note $42^{\circ}=90^{\circ}-48^{\circ}$, so $\sin42^{\circ}=\sin(90^{\circ}-48^{\circ})=\cos48^{\circ}$.
- Step 2: Substitute: $\cos48^{\circ}-\cos48^{\circ}$.
- Step 3: $=0$.
Answer: $0$.
Evaluate $\tan48^{\circ}\,\tan23^{\circ}\,\tan42^{\circ}\,\tan67^{\circ}$.
Solution- Step 1: Group complementary pairs: $48^{\circ}+42^{\circ}=90^{\circ}$ and $23^{\circ}+67^{\circ}=90^{\circ}$.
- Step 2: $\tan42^{\circ}=\tan(90^{\circ}-48^{\circ})=\cot48^{\circ}$ and $\tan67^{\circ}=\tan(90^{\circ}-23^{\circ})=\cot23^{\circ}$.
- Step 3: The product becomes $(\tan48^{\circ}\cot48^{\circ})(\tan23^{\circ}\cot23^{\circ})$.
- Step 4: Each bracket is $1$ because $\tan\theta\cot\theta=1$, so the product $=1\times1=1$.
Answer: $1$.
Evaluate $\dfrac{\sin35^{\circ}}{\cos55^{\circ}}+\dfrac{\cos15^{\circ}}{\sin75^{\circ}}$.
Solution- Step 1: $\cos55^{\circ}=\cos(90^{\circ}-35^{\circ})=\sin35^{\circ}$, so the first term is $\dfrac{\sin35^{\circ}}{\sin35^{\circ}}=1$.
- Step 2: $\sin75^{\circ}=\sin(90^{\circ}-15^{\circ})=\cos15^{\circ}$, so the second term is $\dfrac{\cos15^{\circ}}{\cos15^{\circ}}=1$.
- Step 3: Add: $1+1=2$.
Answer: $2$.
Evaluate $\sec70^{\circ}\sin20^{\circ}+\cos20^{\circ}\operatorname{cosec}70^{\circ}$.
Solution- Step 1: $\sec70^{\circ}=\sec(90^{\circ}-20^{\circ})=\operatorname{cosec}20^{\circ}$, so $\sec70^{\circ}\sin20^{\circ}=\operatorname{cosec}20^{\circ}\sin20^{\circ}=1$.
- Step 2: $\operatorname{cosec}70^{\circ}=\operatorname{cosec}(90^{\circ}-20^{\circ})=\sec20^{\circ}$, so $\cos20^{\circ}\operatorname{cosec}70^{\circ}=\cos20^{\circ}\sec20^{\circ}=1$.
- Step 3: Add: $1+1=2$.
Answer: $2$.
If $\sin3A=\cos(A-26^{\circ})$, where $3A$ is acute, find $A$.
Solution- Step 1: Convert the LHS to a cosine: $\sin3A=\cos(90^{\circ}-3A)$.
- Step 2: So $\cos(90^{\circ}-3A)=\cos(A-26^{\circ})$, giving $90^{\circ}-3A=A-26^{\circ}$.
- Step 3: $90^{\circ}+26^{\circ}=A+3A\Rightarrow116^{\circ}=4A$.
- Step 4: $A=29^{\circ}$.
Answer: $A=29^{\circ}$.
If $\sec5A=\operatorname{cosec}(A+36^{\circ})$, where $5A$ is acute, find $A$.
Solution- Step 1: Convert: $\sec5A=\operatorname{cosec}(90^{\circ}-5A)$.
- Step 2: So $\operatorname{cosec}(90^{\circ}-5A)=\operatorname{cosec}(A+36^{\circ})$, giving $90^{\circ}-5A=A+36^{\circ}$.
- Step 3: $90^{\circ}-36^{\circ}=A+5A\Rightarrow54^{\circ}=6A$.
- Step 4: $A=9^{\circ}$.
Answer: $A=9^{\circ}$.
Show that $\sin(50^{\circ}+\theta)-\cos(40^{\circ}-\theta)=0$.
Solution- Step 1: Look at the two angles: $(50^{\circ}+\theta)+(40^{\circ}-\theta)=90^{\circ}$, so they are complementary.
- Step 2: Hence $\cos(40^{\circ}-\theta)=\cos\bigl(90^{\circ}-(50^{\circ}+\theta)\bigr)=\sin(50^{\circ}+\theta)$.
- Step 3: Substitute: $\sin(50^{\circ}+\theta)-\sin(50^{\circ}+\theta)=0$.
Answer: $0$ (proved).
Evaluate $\dfrac{\cos^{2}20^{\circ}+\cos^{2}70^{\circ}}{\sin^{2}59^{\circ}+\sin^{2}31^{\circ}}$.
Solution- Step 1: $\cos70^{\circ}=\cos(90^{\circ}-20^{\circ})=\sin20^{\circ}$, so the numerator $=\cos^{2}20^{\circ}+\sin^{2}20^{\circ}=1$.
- Step 2: $\sin31^{\circ}=\sin(90^{\circ}-59^{\circ})=\cos59^{\circ}$, so the denominator $=\sin^{2}59^{\circ}+\cos^{2}59^{\circ}=1$.
- Step 3: The fraction $=\dfrac{1}{1}=1$.
Answer: $1$.
Find $\tan10^{\circ}\tan15^{\circ}\tan75^{\circ}\tan80^{\circ}$.
Solution- Step 1: Pair complements: $10^{\circ}+80^{\circ}=90^{\circ}$ and $15^{\circ}+75^{\circ}=90^{\circ}$.
- Step 2: $\tan80^{\circ}=\cot10^{\circ}$ and $\tan75^{\circ}=\cot15^{\circ}$.
- Step 3: Product $=(\tan10^{\circ}\cot10^{\circ})(\tan15^{\circ}\cot15^{\circ})=1\times1$.
- Step 4: $=1$.
Answer: $1$.
Without tables, evaluate $\dfrac{\tan65^{\circ}}{\cot25^{\circ}}+\dfrac{\sin18^{\circ}}{\cos72^{\circ}}-3\tan^{2}30^{\circ}$.
Solution- Step 1: $\cot25^{\circ}=\cot(90^{\circ}-65^{\circ})=\tan65^{\circ}$, so the first term $=\dfrac{\tan65^{\circ}}{\tan65^{\circ}}=1$.
- Step 2: $\cos72^{\circ}=\cos(90^{\circ}-18^{\circ})=\sin18^{\circ}$, so the second term $=\dfrac{\sin18^{\circ}}{\sin18^{\circ}}=1$.
- Step 3: $\tan30^{\circ}=\dfrac{1}{\sqrt3}$, so $3\tan^{2}30^{\circ}=3\cdot\dfrac13=1$.
- Step 4: Combine: $1+1-1=1$.
Answer: $1$.