Class 10 Math Chapter 8 Introduction to Trigonometry Solutions

Here, Class 10 Math Chapter 8 Introduction to Trigonometry Solutions are provided. Subject matter experts have created these solutions to help students in their board exam. They have created solutions in such a way that students can revise the entire solutions in very little time and can score good marks in the board exam. Every solution is provided in detail and in very simple language.

Class 10 Math Chapter 8 Introduction to Trigonometry Solutions are very important for board exam. Solving these Introduction to Trigonometry solutions will boost the confidence and students can score better in exam. Solutions are provided according to latest NCERT syllabus. Experts created these step-by-step solutions, which will definitely be helpful for students.

Introduction to Trigonometry Solutions

Exercise 8.1

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Exercise 8.2

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Exercise 8.3

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Chapter 7: Coordinate Geometry Solutions
Chapter 9: Some Applications of Trigonometry Solutions

We have given solutions of all exercise of class 10 math chapter 8. These solutions will be helpful for students understanding core concept of Introduction to Trigonometry.

To understand the chapter thoroughly, it is essential to practice all the questions provided within it, as well as any additional questions. Solving previous years’ question papers also proves to be very helpful in exam preparation. You can also find other study material resources on our website.

Important Questions for NCERT Chapter 8: Introduction to Trigonometry

Question : Evaluate the following

sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ

Answer :

sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ
=(32)(32)+(12)(12)= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)
=34+14=44=1= \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1

Question : Evaluate the following

2tan2⁡45∘+cos2⁡30∘−sin2⁡60∘2\tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ

Answer :

2tan2⁡45∘+cos2⁡30∘−sin2⁡60∘2\tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ
=2(1)2+(32)2−(32)2= 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2
=2+34−34=2= 2 + \frac{3}{4} – \frac{3}{4} = 2

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