Trigonometric Ratios of Complementary Angles for CBSE Class 10

When two acute angles add up to $90^\circ$, they share an intimate geometric partnership: the side that serves as the Perpendicular for one angle becomes the Base for the other. This simple spatial symmetry produces one of the most practical and elegant sets of algebraic relationships in trigonometry: Trigonometric Ratios of Complementary Angles.

In CBSE Class 10 Mathematics, Chapter 8 (Introduction to Trigonometry), complementary angle relations allow students to simplify seemingly impossible fractions (like $\frac{\sin 18^\circ}{\cos 72^\circ}$) without consulting four-figure trigonometric tables, evaluate telescoping products, and prove geometric riders.

What You Will Learn

• Definition of complementary angles ($A + B = 90^\circ$)
• Geometric derivation of the complementary angle formulas in a right-angled triangle
• The six core formulas: $\sin(90^\circ - \theta) = \cos \theta$, $\tan(90^\circ - \theta) = \cot \theta$, etc.
• The Golden Strategic Rule: Convert ONLY ONE of the two terms in a complementary pair
• Evaluating telescoping products of tangents
• Proving conditional triangle identity riders: \sin\left(\frac{B+C}{2}\right) = \cos\left(\frac{A}{2} ight)
• Board exam tips and common pitfalls

1. What Are Complementary Angles?

Two angles are said to be complementary if their sum equals $90^\circ$. If one angle is $\theta$, its complementary angle is $(90^\circ - \theta)$.

Geometric Derivation in a Right-Angled Triangle:

Consider a right-angled triangle $\Delta ABC$, right-angled at $B$ ($\angle B = 90^\circ$): $\angle A + \angle C = 180^\circ - 90^\circ = 90^\circ \implies \angle C = 90^\circ - \angle A$

                           A
                           |                           |                         AB |  \ AC
                           |                              +----+
                           B    C (90° - A)
                             BC

Now, write the trigonometric ratios for both acute angles:

1. For angle $A$: $\sin A = \frac{BC}{AC}, \quad \cos A = \frac{AB}{AC}, \quad \tan A = \frac{BC}{AB}$
2. For angle $C = (90^\circ - A)$: Here, the side opposite to $\angle C$ is $AB$, and adjacent side is $BC$: $\sin(90^\circ - A) = \frac{AB}{AC}, \quad \cos(90^\circ - A) = \frac{BC}{AC}, \quad \tan(90^\circ - A) = \frac{AB}{BC}$

Comparing the ratios side by side, notice the remarkable crossover: $\mathbf{\sin(90^\circ - A) = \cos A} \quad \text{and} \quad \mathbf{\cos(90^\circ - A) = \sin A}$

2. The Six Complementary Angle Formulas

For any acute angle $\theta$ ($0^\circ \le \theta < 90^\circ$):

$\mathbf{\sin(90^\circ - \theta) = \cos \theta} \quad \Longleftrightarrow \quad \mathbf{\cos(90^\circ - \theta) = \sin \theta}$ $\mathbf{\tan(90^\circ - \theta) = \cot \theta} \quad \Longleftrightarrow \quad \mathbf{\cot(90^\circ - \theta) = \tan \theta}$ $\mathbf{\sec(90^\circ - \theta) = \csc \theta} \quad \Longleftrightarrow \quad \mathbf{\csc(90^\circ - \theta) = \sec \theta}$

Important: <u>Notice the naming pattern: "Cosine" literally means the "Complement's Sine". "Cotangent" means the "Complement's Tangent". "Cosecant" means the "Complement's Secant"!</u>

3. The Golden Rule of Evaluation

When simplifying complementary angle problems in board exams, students often make the mistake of converting both terms, which simply recreates the original expression in reverse.

The Golden Strategic Rule: <u>In any complementary pair $( heta ext{ and } 90^\circ - heta)$, change ONLY ONE ratio! Leave the other ratio completely untouched.</u>

4. Solved CBSE Board Examination Problems

Solved Example 1: Direct Quotient Evaluation

Problem: Evaluate $\frac{\sin 18^\circ}{\cos 72^\circ}$.

Solution:

1. Check if the angles are complementary: $18^\circ + 72^\circ = 90^\circ$
2. Convert only the numerator using $\sin \theta = \cos(90^\circ - \theta)$: $\sin 18^\circ = \cos(90^\circ - 18^\circ) = \cos 72^\circ$
3. Substitute into the fraction: $\frac{\sin 18^\circ}{\cos 72^\circ} = \frac{\cos 72^\circ}{\cos 72^\circ} = 1$
4. Therefore, <u>the value is $1$</u>.

Solved Example 2: Telescoping Tangent Products (CBSE PYQ)

Problem: Show that $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1$.

Solution:

1. Group the complementary angle pairs together:
   • $48^\circ + 42^\circ = 90^\circ$
   • $23^\circ + 67^\circ = 90^\circ$ $\text{LHS} = (\tan 48^\circ \tan 42^\circ) \times (\tan 23^\circ \tan 67^\circ)$
2. Convert only one term in each complementary pair:
   • $\tan 48^\circ = \cot(90^\circ - 48^\circ) = \cot 42^\circ$
   • $\tan 23^\circ = \cot(90^\circ - 23^\circ) = \cot 67^\circ$
3. Substitute back into the expression: $\text{LHS} = (\cot 42^\circ \tan 42^\circ) \times (\cot 67^\circ \tan 67^\circ)$
4. Use the reciprocal relation $\cot \theta \times \tan \theta = 1$: $\text{LHS} = (1) \times (1) = 1 = \text{RHS}$
5. Hence, proved.

Solved Example 3: Triangle Interior Angles Rider (Board Classic)

Problem: If $A, B,$ and $C$ are interior angles of a triangle $ABC$, then show that $\sin\left(\frac{B + C}{2}\right) = \cos\left(\frac{A}{2}\right)$.

Solution:

1. In $\Delta ABC$, the sum of interior angles is $180^\circ$: $A + B + C = 180^\circ$
2. Isolate $(B + C)$: $B + C = 180^\circ - A$
3. Divide both sides by 2: $\frac{B + C}{2} = \frac{180^\circ - A}{2} = 90^\circ - \frac{A}{2}$
4. Take the sine on both sides: $\sin\left(\frac{B + C}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right)$
5. Using the complementary angle formula $\sin(90^\circ - \theta) = \cos \theta$: $\mathbf{\sin\left(\frac{B + C}{2}\right) = \cos\left(\frac{A}{2}\right)}$
6. Hence, proved.

5. Summary and Examination Tips

Exam Tip: In questions of the type $\tan 2A = \cot(A - 18^\circ)$, convert $\tan 2A$ into $\cot(90^\circ - 2A)$ so that both sides have the identical $\cot$ function, allowing you to equate the angles directly: $90^\circ - 2A = A - 18^\circ \implies 3A = 108^\circ \implies A = 36^\circ$!

Common Mistake: Converting both numerator and denominator in a fraction. In $\frac{\sin 18^\circ}{\cos 72^\circ}$, converting both gives $\frac{\cos 72^\circ}{\sin 18^\circ}$, which leaves you right back where you started!