Trigonometric Ratios of Complementary Angles: Class 10

Quick answer: Two angles are complementary if they add up to 90°. In a right triangle the two acute angles are complementary, and the side opposite one is adjacent to the other. So $\sin(90^\circ - \theta) = \cos\theta$, $\cos(90^\circ - \theta) = \sin\theta$, $\tan(90^\circ - \theta) = \cot\theta$, $\cot(90^\circ - \theta) = \tan\theta$, $\sec(90^\circ - \theta) = \operatorname{cosec}\theta$ and $\operatorname{cosec}(90^\circ - \theta) = \sec\theta$. These let you simplify expressions such as $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ$, which equals 1.

Syllabus note: The section "Trigonometric ratios of complementary angles" was removed from the rationalised NCERT Class 10 textbook. These results are still used in higher classes and competitive exams, so this article is useful for extra practice, but check your current board syllabus before studying it for the board exam.

Why the results hold

Take a right triangle ABC with $\angle B = 90^\circ$ and $\angle C = \theta$. Then $\angle A = 90^\circ - \theta$.

Step 1: For angle θ: $\sin\theta = \frac{AB}{AC}$ and $\cos\theta = \frac{BC}{AC}$.

Step 2: For angle $90^\circ - \theta$: $\sin(90^\circ - \theta) = \frac{BC}{AC}$ and $\cos(90^\circ - \theta) = \frac{AB}{AC}$.

Step 3: Comparing, $\sin(90^\circ - \theta) = \cos\theta$ and $\cos(90^\circ - \theta) = \sin\theta$. The other four results follow in the same way.

The six results

These hold for every acute angle θ.

Example 1: a product of tangents

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

Step 1: Pair the complementary angles: $48^\circ + 42^\circ = 90^\circ$ and $23^\circ + 67^\circ = 90^\circ$.

Step 2: Rewrite one angle in each pair: $\tan 42^\circ = \cot 48^\circ$ and $\tan 67^\circ = \cot 23^\circ$.

Step 3: Since $\tan\theta \cot\theta = 1$:

$\tan 48^\circ \cot 48^\circ \times \tan 23^\circ \cot 23^\circ = 1 \times 1 = 1$

Example 2: angles of a triangle

If A, B and C are the angles of a triangle, show that $\sin\frac{B + C}{2} = \cos\frac{A}{2}$.

Step 1: $A + B + C = 180^\circ$, so $B + C = 180^\circ - A$.

Step 2: Divide by 2: $\frac{B + C}{2} = 90^\circ - \frac{A}{2}$.

Step 3: Take the sine of both sides and use $\sin(90^\circ - \theta) = \cos\theta$:

$\sin\frac{B + C}{2} = \cos\frac{A}{2}$

Hence proved.

Example 3: finding an unknown angle

If $\sin 3A = \cos(A - 26^\circ)$, where $3A$ is acute, find A.

Step 1: Write $\sin 3A = \cos(90^\circ - 3A)$.

Step 2: So $\cos(90^\circ - 3A) = \cos(A - 26^\circ)$. Both angles are acute, so they are equal:

$90^\circ - 3A = A - 26^\circ$

Step 3: $4A = 116^\circ$.

Answer: $A = 29^\circ$. Check: $\sin 87^\circ = \cos 3^\circ$, and $87^\circ + 3^\circ = 90^\circ$.

Summary table

Exam tips

• Convert only one angle in each complementary pair; changing both brings you back to the start.
• Check that the two angles really add up to 90° before using the results.
• In triangle questions, start from the angle sum property.

Common mistakes

• Writing $\sin(180^\circ - \theta) = \cos\theta$. The complementary results use 90°, not 180°.
• Pairing angles that add up to something other than 90°.
• Equating angles without checking that both are acute.
• Forgetting that sec pairs with cosec, not with cos.

Frequently Asked Questions

What are complementary angles?

Two angles whose sum is 90°. The two acute angles of a right triangle are always complementary.

Why is sin(90° − θ) equal to cos θ?

In a right triangle, the side opposite one acute angle is the side adjacent to the other, so the sine of one angle equals the cosine of its complement.

What is the value of tan 48° tan 23° tan 42° tan 67°?

It is 1, because tan 42° = cot 48° and tan 67° = cot 23°, and tan θ × cot θ = 1.

How do you prove sin((B + C)/2) = cos(A/2)?

Since B + C = 180° − A, half of it is 90° − A/2, and the sine of that is cos(A/2).

Is this topic in the current CBSE Class 10 syllabus?

The complementary angles section was removed from the rationalised NCERT textbook, so it is mainly useful for practice and later classes.