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Trigonometry - Trigonometric Ratios of Complementary Angles

Grade 9ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Complementary angles are two angles whose sum is 90∘90^{\circ}. In a right-angled triangle, if one acute angle is θ\theta, the other must be 90∘−θ90^{\circ} - \theta because the sum of all angles is 180∘180^{\circ}.

Right-angled triangle showing complementary angles theta and 90-theta.
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The sine of an angle is equal to the cosine of its complement. For example, sin⁡30∘=cos⁡(90∘−30∘)=cos⁡60∘\sin 30^{\circ} = \cos (90^{\circ} - 30^{\circ}) = \cos 60^{\circ}.

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The tangent of an angle is the reciprocal of the tangent of its complement, which is the cotangent of that angle. Thus, tan⁡θ=cot⁡(90∘−θ)\tan \theta = \cot(90^{\circ} - \theta).

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Secant and Cosecant share a similar complementary relationship: sec⁡θ=csc⁡(90∘−θ)\sec \theta = \csc(90^{\circ} - \theta).

📐Formulae

sin⁡(90∘−θ)=cos⁡θ\sin (90^{\circ} - \theta) = \cos \theta

cos⁡(90∘−θ)=sin⁡θ\cos (90^{\circ} - \theta) = \sin \theta

tan⁡(90∘−θ)=cot⁡θ\tan (90^{\circ} - \theta) = \cot \theta

cot⁡(90∘−θ)=tan⁡θ\cot (90^{\circ} - \theta) = \tan \theta

sec⁡(90∘−θ)=csc⁡θ\sec (90^{\circ} - \theta) = \csc \theta

csc⁡(90∘−θ)=sec⁡θ\csc (90^{\circ} - \theta) = \sec \theta

💡Examples

Problem 1:

Evaluate: sin⁡36∘cos⁡54∘−tan⁡70∘cot⁡20∘\frac{\sin 36^{\circ}}{\cos 54^{\circ}} - \frac{\tan 70^{\circ}}{\cot 20^{\circ}}

Solution:

Step 1: Observe the angles. In the first term, 36∘+54∘=90∘36^{\circ} + 54^{\circ} = 90^{\circ}. In the second term, 70∘+20∘=90∘70^{\circ} + 20^{\circ} = 90^{\circ}. Step 2: Apply the complementary ratio formula to the numerators: sin⁡36∘=sin⁡(90∘−54∘)=cos⁡54∘\sin 36^{\circ} = \sin(90^{\circ} - 54^{\circ}) = \cos 54^{\circ} tan⁡70∘=tan⁡(90∘−20∘)=cot⁡20∘\tan 70^{\circ} = \tan(90^{\circ} - 20^{\circ}) = \cot 20^{\circ} Step 3: Substitute these back into the original expression: cos⁡54∘cos⁡54∘−cot⁡20∘cot⁡20∘\frac{\cos 54^{\circ}}{\cos 54^{\circ}} - \frac{\cot 20^{\circ}}{\cot 20^{\circ}} Step 4: Simplify the fractions: 1−1=01 - 1 = 0

Explanation:

This solution uses the complementary angle identities to convert the numerator of each fraction to match its denominator, allowing them to be simplified to 1.

Problem 2:

Find the value of θ\theta if sin⁡3θ=cos⁡(θ−6∘)\sin 3\theta = \cos(\theta - 6^{\circ}), where 3θ3\theta and (θ−6∘)(\theta - 6^{\circ}) are acute angles.

Solution:

Step 1: Use the identity cos⁡A=sin⁡(90∘−A)\cos A = \sin(90^{\circ} - A) to make the trigonometric ratios on both sides the same. sin⁡3θ=sin⁡(90∘−(θ−6∘))\sin 3\theta = \sin(90^{\circ} - (\theta - 6^{\circ})) Step 2: Since the sine ratios are equal and the angles are acute, we can equate the angles: 3θ=90∘−(θ−6∘)3\theta = 90^{\circ} - (\theta - 6^{\circ}) Step 3: Solve the linear equation: 3θ=90∘−θ+6∘3\theta = 90^{\circ} - \theta + 6^{\circ} 3θ+θ=96∘3\theta + \theta = 96^{\circ} 4θ=96∘4\theta = 96^{\circ} Step 4: Divide by 4: θ=96∘4=24∘\theta = \frac{96^{\circ}}{4} = 24^{\circ}

Explanation:

By converting the cosine term into a sine term using complementary angle properties, we can create an algebraic equation to solve for the unknown angle.

Problem 3:

Show that tan⁡15∘×tan⁡30∘×tan⁡75∘=13\tan 15^{\circ} \times \tan 30^{\circ} \times \tan 75^{\circ} = \frac{1}{\sqrt{3}}.

Diagram highlighting the product of complementary tangents as 1.

Solution:

We know that tan⁡(90∘−θ)=cot⁡θ\tan (90^{\circ} - \theta) = \cot \theta. Step 1: Express tan⁡75∘\tan 75^{\circ} in terms of its complement. tan⁡75∘=tan⁡(90∘−15∘)=cot⁡15∘\tan 75^{\circ} = \tan (90^{\circ} - 15^{\circ}) = \cot 15^{\circ} Step 2: Substitute this into the original expression. tan⁡15∘×tan⁡30∘×cot⁡15∘\tan 15^{\circ} \times \tan 30^{\circ} \times \cot 15^{\circ} Step 3: Since tan⁡θ×cot⁡θ=1\tan \theta \times \cot \theta = 1, the expression becomes: 1×tan⁡30∘1 \times \tan 30^{\circ} Step 4: Use the standard value tan⁡30∘=13\tan 30^{\circ} = \frac{1}{\sqrt{3}}. 1×13=131 \times \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}

Explanation:

This problem uses the complementary identity for tangent to pair up angles that sum to 90∘90^{\circ}, simplifying the product to 1.

Problem 4:

Evaluate: 3cos⁡80∘sin⁡10∘+2cos⁡59∘csc⁡31∘3 \frac{\cos 80^{\circ}}{\sin 10^{\circ}} + 2 \cos 59^{\circ} \csc 31^{\circ}

Visualizing pairs of angles that sum to 90 degrees.

Solution:

Step 1: Simplify the first term using cos⁡80∘=sin⁡(90∘−80∘)=sin⁡10∘\cos 80^{\circ} = \sin(90^{\circ} - 80^{\circ}) = \sin 10^{\circ}. 3sin⁡10∘sin⁡10∘=3(1)=33 \frac{\sin 10^{\circ}}{\sin 10^{\circ}} = 3(1) = 3 Step 2: Simplify the second term. Note that csc⁡31∘=sec⁡(90∘−31∘)=sec⁡59∘\csc 31^{\circ} = \sec(90^{\circ} - 31^{\circ}) = \sec 59^{\circ}. 2cos⁡59∘sec⁡59∘2 \cos 59^{\circ} \sec 59^{\circ} Step 3: Since sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}, then cos⁡59∘sec⁡59∘=1\cos 59^{\circ} \sec 59^{\circ} = 1. 2(1)=22(1) = 2 Step 4: Add the results from Step 1 and Step 3. 3+2=53 + 2 = 5

Explanation:

The complementary angle identities for sine-cosine and secant-cosecant are used to reduce fractions and products to unity.