## Trigonometric Ratios for Complementary angles - NCERT Solutions for class 10 maths Ex 8.3

Trigonometric ratios for complementary angles, example problems on trigonometric ratios for complementary angles and Solution to NCERT class 10 Ex 8.3.

We'll cover now,

**What are trigonometric ratios for complementary angles.****Example problems on trigonometric ratios for complementary angles.****Solution to NCERT Class 10 maths Ex 8.3 on trigonometric ratios for complementary angles.**

*You may move directly to any of the above sections by clicking its link and return by clicking on browser back button.*

### Trigonometric ratios for complementary angles

First let us recall the concept of complementary angles,

If sum of two angles is $90^0$ the angles are complementary to each other.

In the right-angled $\triangle ABC$ right-angled at B, as $\angle A+\angle C=90^0$, these two angles must always be complementary to each other for any pair of their values.

**Question is:** What is the relationship between values of trigonometric ratios, on $\angle A$ and on $\angle C$ that are complementary to each other?

When trigonometric ratios are taken on $\angle A$, BC opposite to $\angle A$ is the **height** and AB adjacent to $\angle A$ is the **base**.

But when trigonometric ratios are taken on $\angle C$, AB opposite to $\angle C$ is the **height** and BC adjacent to $\angle C$ is the **base**.

The third side CA is the **hypotenuse.**

As sum of three angles in a triangle is $180^0$, sum of $\angle A$ and $\angle C$ is $90^0$, $\angle B$ being $90^0$. That is,

$\angle A+\angle C=90^0$.

Or, $\angle A=90^0-\angle C$, and

$\angle C=90^0-\angle A$.

Taking the the first trigonometric ratio on $\angle A$,

$\sin A=\sin (90^0-C)=\displaystyle\frac{BC}{CA}$

But if you take trigonometric ratio on $\angle C$, what trigonometric ratio is equal to $\displaystyle\frac{BC}{CA}$? It is simply $\cos C$, as **base** in this case is BC.

So,

$\displaystyle\frac{BC}{CA}=\cos C=\sin (90^0-C)$.

Though we started by taking the trigonometric ratio $\sin A$ on $\angle A$, we have arrived at the first complementary angle relationship of trigonometric ratios on $\angle C$ as,

$\sin (90^0-C)=\cos C=\displaystyle\frac{BC}{CA}$.

As $\angle C$ and $\angle A$ have same role in the triangle, it is also true that,

$\sin (90^0-A)=\cos A=\displaystyle\frac{AB}{CA}$.

In the same way,

$\cos(90^0-A)=\sin A=\displaystyle\frac{BC}{CA}$, and

$\cos(90^0-C)=\sin C=\displaystyle\frac{AB}{CA}$.

#### Complementary angle relationship between trigonometric ratios $\tan$ and $\text{cot}$

Just as $\sin$ and $\cos$ form a complementary angle trigonometric ratio pair, For $\tan$, $\text{cot}$ is its complementary angle trigonometric ratio pair member.

$\tan(90^0-A)=\tan C=\displaystyle\frac{AB}{BC}=\text{cot }A$,

$\tan(90^0-C)=\tan A=\displaystyle\frac{BC}{AB}=\text{cot }C$,

$\text{cot }(90^0-A)=\text{cot }C=\displaystyle\frac{BC}{AB}=\tan A$, and

$\text{cot }(90^0-C)=\text{cot }A=\displaystyle\frac{AB}{BC}=\tan C$.

#### Complementary angle relationship between trigonometric ratios $\text{sec}$ and $\text{cosec}$

The last two, $\text{sec}$ and $\text{cosec}$ form the third complementary angle pair of trigonometric ratios,

$\text{sec }(90^0-A)=\text{sec }C=\displaystyle\frac{CA}{BC}=\text{cosec }A$,

$\text{sec }(90^0-C)=\text{sec }A=\displaystyle\frac{CA}{AB}=\text{cosec }C$,

$\text{cosec }(90^0-A)=\text{cosec }C=\displaystyle\frac{CA}{AB}=\text{sec }A$, and,

$\text{cosec }(90^0-C)=\text{cosec }A=\displaystyle\frac{CA}{BC}=\text{sec }C$.

#### Summing up the concept of trigonometric ratios for a pair of complementary angles

- In a right-angled triangle, the two angles other than the right angle form a pair of complementary angles. Say $\angle A$ and $\angle C$ are such a pair of angles for which $\angle A+\angle C=90^0$.
- The trigonometric ratios for $\sin$ and $\cos$ form the first complementary pair of trigonometric ratios, $\tan$ and $\text{cot}$ form the second pair and $\text{sec}$ and $\text{cosec}$ form the third pair.
- $\sin(90^0-A)=\cos A$ and $\cos(90^0-A)=\sin A$ that holds good for $\angle C$ also. Same transformation happens between the other two pairs of trigonometric ratios, when ratios are taken on $\angle (90^0-A)$, or $\angle (90^0-C)$.

We'll solve a few example problems using the concept of trigonometric ratio relations for complementary angles before solving the exercise problems.

### Example problems on trigonometric ratios for complementary angles

#### Problem example 1.

Evaluate $\tan 5^0.\tan 25^0.\tan 45^0.\tan 65^0.\tan 85^0$.

#### Solution example 1:

By applying complementary angle trigonometric ratio relations,

$\tan5^0=\text{cot }(90^0-5^0)=\text{cot }85^0$, and,

$\tan 25^0=\text{cot }(90^0-25^0)=\text{cot }65^0$.

The given expression is thus transformed to,

$\text{cot }85^0.\text{cot }65^0.\tan 45^0.\tan 65^0.\tan 85^0$

$=(\text{cot }85^0.\tan 85^0)(\text{cot }65^0.\tan 65^0)\tan 45^0$

$=\tan 45^0=1$.

#### Problem example 2.

If $\text{cosec }4A=\text{sec }5A$, find $A$, where $5A$ is an acute angle.

#### Solution example 2.

As $5A$ is acute, $4A$ is also an acute angle so that complementary angle trigonometric ratio relations can be applied on these two angles.

Applying complementary angle trigonometric ratio relations on the LHS, the given equation is transformed to,

$\text{cosec }4A=\text{sec }(90^0-4A)=\text{sec }5A$

So,

$(90^0-4A)=5A$,

Or, $9A=90^0$,

Or $\angle A=10^0$.

#### Problem example 3.

Evaluate $\cos 32^0\cos 58^0-\sin 32^0\sin 58^0$.

#### Solution example 3.

By complementary angle trigonometric ratio relations,

$\cos 32^0=\sin(90^0-32^0)=\sin 58^0$, and,

$\cos 58^0=\sin(90^0-58^0)=\sin 32^0$.

Substituting in the given expression,

$\cos 32^0\cos 58^0-\sin 32^0\sin 58^0$

$=\sin 58^0\sin 32^0-\sin 32^0\sin 58^0$

It is time to solve the exercise problems.

### Solution to NCERT Class 10 maths Ex 8.3 on Trigonometric ratios for Complementary angles

#### Problem 1.i.

Evaluate $\displaystyle\frac{\sin 18^0}{\cos 72^0}$.

#### Solution to Problem 1.i.

$\sin 18^0=\cos (90^0-18^0)=\cos 72^0$.

So given expression,

$\displaystyle\frac{\sin 18^0}{\cos 72^0}=\frac{\cos 72^0}{\cos 72^0}=1$.

**Answer:** 1.

#### Problem 1.ii.

Evaluate $\displaystyle\frac{\tan 26^0}{\text{cot } 64^0}$.

#### Solution to Problem 1.ii.

$\tan 26^0=\text{cot }(90^0-26^0)=\text{cot }64^0$.

The given expression is transformed to,

$\displaystyle\frac{\tan 26^0}{\text{cot } 64^0}$

$=\displaystyle\frac{\text{cot } 64^0}{\text{cot } 64^0}$

$=1$.

**Answer:** 1.

#### Problem 1.iii.

Evaluate $\cos 48^0-\sin 42^0$.

#### Solution to Problem 1.iii.

$\cos 48^0=\sin (90^-48^0)=\sin 42^0$.

So the value of the given expression is,

$\cos 48^0-\sin 42^0=\sin 42^0-\sin 42^0=0$

**Answer:** 0.

#### Problem 1.iv.

Evaluate $\text{cosec }31^0-\text{sec }59^0$.

#### Solution to Problem 1.iv.

$\text{cosec }31^0=\text{sec }(90^0-31^0)=\text{sec }59^0$.

The value of the given expression is,

$\text{cosec }31^0-\text{sec }59^0=\text{sec }59^0-\text{sec }59^0=0$.

**Answer:** 0.

#### Problem 2.i.

Show that, $\tan 48^0 \tan 23^0 \tan 42^0 \tan 67^0=1$.

#### Solution to Problem 2.i.

$\tan 48^0=\text{cot }(90^0-48^0)=\text{cot }42^0$, and,

$\tan 23^0=\text{cot }(90^0-23^0)=\text{cot }67^0$.

So the value of LHS of the given expression is,

$\tan 48^0 \tan 23^0 \tan 42^0 \tan 67^0$

$=(\text{cot }42^0 .\tan 42^0 )(\text{cot }67^0. \tan 67^0)$

$=1$ **Proved.**

#### Problem 2.ii.

Show that, $\cos 38^0 \cos 52^0 - \sin 38^0 \sin 52^0=0$.

#### Solution to Problem 2.ii.

$\cos 38^0=\sin(90^0-38^0)=\sin 52^0$, and,

$\cos 52^0=\sin(90^0-52^0)=\sin 38^0$.

So the value of the LHS of the given expression is,

$\cos 38^0 \cos 52^0 - \sin 38^0 \sin 52^0$

$=\sin 52^0 \sin 38^0 - \sin 38^0 \sin 52^0$

$=0$. **Proved.**

#### Problem 3.

If $\tan 2A=\text{cot }(A-18^0)$, where $\angle 2A$ is an acute angle, find the value of $\angle A$.

#### Solution to Problem 3.

As $2A$ and $A$ are both acute angles, by complementary angle trigonometric ratio relationships,

$\text{cot }(A-18^0)=\tan[90^0-(A-18^0)]=\tan(108^0-A)$.

So from the given expression,,

$\tan 2A=\text{cot }(A-18^0)=\tan(108^0-A)$

Thus,

$2A=108^0-A$,

Or, $3A=108^0$,

Or, $A=36^0$.

**Answer:** $\angle A=36^0$.

#### Problem 4.

If $\tan A=\text{cot }B$, prove that $A+B=90^0$.

#### Solution to Problem 4

By complementary angle trigonometric ratio relationships,

$\text{cot }B=\tan(90^0-B)$.

So by the given expression,

$\tan A=\text{cot }B=\tan(90^0-B)$.

So,

$A=90^0-B$,

Or, $A+B=90^0$. **Proved.**

#### Problem 5.

If $\text{sec }4A=\text{cosec }(A-20^0)$, where $4A$ is an acute angle, find the value of A.

#### Solution to Problem 5.

As $4A$ and $A$ are both acute angles, by trigonometric ratio relationships of complementary angles,

$\text{cosec }(A-20^0)=\text{sec }[90^0-(A-20^0)]=\text{sec }(110^0-A)$.

So the given expression is transformed to,

$\text{sec }4A=\text{cosec }(A-20^0)=\text{sec }(110^0-A)$.

So,

$4A=110^0-A$,

Or, $5A=110^0$,

Or, $A=22^0$.

**Answer:** $\angle A=22^0$.

#### Problem 6.

If A, B and C are interior angles of a $\triangle ABC$, then show that,

$\sin \left(\displaystyle\frac{B+C}{2}\right)=\cos \displaystyle\frac{A}{2}$.

#### Solution to Problem 6.

Being interior angles of a triangle, the sum of three angles A, B, and C is $180^0$.

So,

$A+B+C=180^0$,

Or, $\displaystyle\frac{A+B+C}{2}=90^0$,

Or, $\displaystyle\frac{B+C}{2}=90^0-\displaystyle\frac{A}{2}$.

Taking Sine of both equal angles,

$\sin \left(\displaystyle\frac{B+C}{2}\right)=\sin \left(90^0-\displaystyle\frac{A}{2}\right)=\cos \displaystyle\frac{A}{2}$.

Proved.

#### Problem 7.

Express $\sin 67^0+\cos 75^0$ in terms of ratio of angles between $0^0$ and $45^0$.

#### Solution to Problem 7.

By trigonometric ratio relationships for complementary angles,

$\sin 67^0=\cos (90^0-67^0)=\cos 23^0$, and

$\cos 75^0=\sin (90^0-75^0)=\sin 15^0$.

So the given expression is transformed to,

$\sin 67^0+\cos 75^0=\cos 23^0+\sin 15^0$, both the new ratio angles within the desired range of $0^0$ and $45^0$.

**Answer:** $\cos 23^0+\sin 15^0$.

### NCERT Solutions for Class 10 Maths

#### Chapter 1: Real Numbers

**NCERT Solutions for Class 10 Maths on Real numbers part 1, Euclid’s division lemma puzzle solutions**

#### Chapter 2: Polynomials

#### Chapter 3: Linear Equations

**NCERT solutions for class 10 maths Chapter 3 Linear equations 7 Problem Collection**

**NCERT solutions for class 10 maths Chapter 3 Linear equations 6 Reducing non-linear to linear form**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 4 Algebraic solution by Elimination**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 3 Algebraic solution by Substitution**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 2 Graphical solutions**

**NCERT solutions for class 10 maths Chapter 3 Linear Equations 1 Graphical representation.**

#### Chapter 4: Quadratic equations

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 1 What are quadratic equations**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 2 Solving by factorization**

**NCERT solutions for class 10 maths Chapter 4 Quadratic Equations 3 Solution by Completing the square**

#### Chapter 6: Triangles

**NCERT solutions for class 10 maths chapter 6 Triangles 1 Similarity of Triangles and Polygons**

**Solutions to Exercise 2 Chapter 6 NCERT X Maths, Characteristics of Similar triangles**

#### Chapter 8: Introduction to Trigonometry, Concepts and solutions to exercise problems

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 1 Trigonometric Ratios*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 3 Complementary angle relations*

*NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 4 Trigonometric identities*

#### Chapter 8: Introduction to Trigonometry, only solutions to selected problems

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 6**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 5**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 4**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 3**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 2**

**NCERT Solutions for Class 10 Maths on Trigonometry, solution set 1**