Maharashtra Board Class 10 Maths Part II Chapter 6 Trigonometry PDF Download

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MSBSHSE Class 10 Maths Part II Chapter 6 Trigonometry Digital Edition

For Class 10 Maths, this chapter in Maharashtra Board Class 10 Maths Part II Chapter 6 Trigonometry PDF Download provides a detailed overview of important concepts. We highly recommend using this text alongside the MSBSHSE Solutions for Class 10 Maths to learn the exercise questions provided at the end of the chapter.

Part II Chapter 6 Trigonometry MSBSHSE Book Class 10 PDF (2026-27)

Trigonometry

Let's Study

Trigonometric ratios

Trigonometric identities

Angle of elevation and angle of depression

Problems based on heights and distances

Let's Recall

1. Fill in the blanks with reference to figure 6.1.

In the figure, let \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\), \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\), \(\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\)

2. Complete the relations in ratios given below.

(i) \(\frac{\sin \theta}{\cos \theta} = \tan \theta\)

(ii) \(\sin \theta = \cos(90° - \theta)\)

(iii) \(\cos \theta = \sin(90° - \theta)\)

(iv) \(\tan \theta \times \tan(90° - \theta) = 1\)

3. Complete the equation.

\(\sin^2 \theta + \cos^2 \theta = 1\)

4. Write the values of the following trigonometric ratios.

(i) \(\sin 30° = \frac{1}{2}\)

(ii) \(\cos 30° = \frac{\sqrt{3}}{2}\)

(iii) \(\tan 30° = \frac{1}{\sqrt{3}}\)

(iv) \(\sin 60° = \frac{\sqrt{3}}{2}\)

(v) \(\cos 45° = \frac{1}{\sqrt{2}}\)

(vi) \(\tan 45° = 1\)

In std IX, we have studied some trigonometric ratios of some acute angles. Now we are going to study some more trigonometric ratios of acute angles.

Teacher's Note

Trigonometric ratios help us find heights and distances without measuring them. For example, we can find the height of a tall building just by knowing the distance and the angle of elevation from the ground.

Exam Trick

Remember: \(\sin^2 \theta + \cos^2 \theta = 1\) always. This is the most important identity. Use it to find any ratio if you know one ratio.

Points to Remember

Sin, cos and tan are main ratios.
Opposite side is across from the angle.
Hypotenuse is the longest side in a right triangle.
Adjacent side is next to the angle.
These ratios work only in right triangles.

Cosec, Sec And Cot Ratios

Multiplicative inverse or the reciprocal of sine ratio is called cosecant ratio. It is written in brief as cosec. \(\therefore \csc \theta = \frac{1}{\sin \theta}\)

Similarly, multiplicative inverses or reciprocals of cosine and tangent ratios are called secant and cotangent ratios respectively. They are written in brief as sec and cot.

\(\therefore \sec \theta = \frac{1}{\cos \theta}\) and \(\cot \theta = \frac{1}{\tan \theta}\)

In figure 6.2, \(\sin \theta = \frac{AB}{AC}\)

\(\therefore \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{AB}{AC}} = \frac{AC}{AB}\)

It means, \(\csc \theta = \frac{\text{hypotenuse}}{\text{opposite side}}\)

\(\cos \theta = \frac{BC}{AC}\)

\(\sec \theta = \frac{1}{\cos \theta} = \frac{1}{\frac{BC}{AC}} = \frac{AC}{BC}\)

It means, \(\sec \theta = \frac{\text{hypotenuse}}{\text{adjacent side}}\)

\(\tan \theta = \frac{AB}{BC}\)

\(\therefore \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{AB}{BC}} = \frac{BC}{AB} = \frac{\text{adjacent side}}{\text{opposite side}}\)

You know that, \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)

\(\therefore \cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{\sin \theta}{\cos \theta}} = \frac{\cos \theta}{\sin \theta}\)

\(\therefore \cot \theta = \frac{\cos \theta}{\sin \theta}\)

Teacher's Note

Cosec, sec and cot are just reciprocals of sin, cos and tan. If sin is small, then cosec will be big. Remember this relationship in your exam.

Exam Trick

Remember: Cosec is for sine, sec is for cosine, and cot is for tangent. Each one is the flip of the other. Just write upside down (reciprocal) and you get it.

Points to Remember

\(\csc \theta = \frac{1}{\sin \theta}\)
\(\sec \theta = \frac{1}{\cos \theta}\)
\(\cot \theta = \frac{1}{\tan \theta}\)
Cosec, sec and cot are reciprocals of sin, cos and tan.
These three ratios are also called trigonometric ratios.

For More Information

The great Indian mathematician Aryabhata was born in 476 A.D. in Kusumpur which was near Patna in Bihar. He has done important work in Arithmetic, Algebra and Geometry. In the book 'Aryabhatiya' he has written many mathematical formulae. For example:

(1) In an Arithmetic Progression, formulae for nth term and the sum of first n terms.

(2) The formula to approximate \(\sqrt{2}\)

(3) The correct value of \(\pi\) upto four decimals, \(\pi = 3.1416\).

In the study of Astronomy he used trigonometry and the sine ratio of an angle for the first time.

Comparing with the mathematics in the rest of the world at that time, his work was great and was studied all over India and was carried to Europe through Middle East.

Most observers at that time believed that the earth is immovable and the Sun, the Moon and stars move around the earth. But Aryabhata noted that when we travel in a boat on the river, objects like trees, houses on the bank appear to move in the opposite direction. 'Similarly', he said 'the Sun, Moon and the stars are observed by people on the earth to be moving in the opposite direction while in reality the Earth moves!'

On 19 April 1975, India sent the first satellite in the space and it was named 'Aryabhata' to commemorate the great Mathematician of India.

Remember This

The relation between the trigonometric ratios, according to the definitions of cosec, sec and cot ratios

\(\frac{1}{\sin \theta} = \csc \theta \quad \therefore \sin \theta \times \csc \theta = 1\)

\(\frac{1}{\cos \theta} = \sec \theta \quad \therefore \cos \theta \times \sec \theta = 1\)

\(\frac{1}{\tan \theta} = \cot \theta \quad \therefore \tan \theta \times \cot \theta = 1\)

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MSBSHSE Book Class 10 Maths Part II Chapter 6 Trigonometry

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