CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 08

Here is the CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set 08 for your studies. Get chapter-wise Value Based Questions (VBQs) for the 2026-27 session, tailored for Class 10 Mathematics learners. These questions make it easy to learn ethics and match standard test patterns from CBSE, NCERT, and KVS.

VBQ for Class 10 Mathematics Chapter 8 Introduction to Trigonometry

Check out these Value Based Questions for Chapter 8 Introduction to Trigonometry designed for Class 10 learners. They teach you how textbook ideas apply to real-world situations. Reviewing these competency-based questions with answers boosts your Class 10 exam scores and moral growth.

Chapter 8 Introduction to Trigonometry VBQ Solutions for Class 10 Mathematics

Question. Prove that: \( \frac{1 + \sin \theta}{1 - \sin \theta} = (\sec \theta + \tan \theta)^2 \)
Answer: \( \frac{1 + \sin \theta}{1 - \sin \theta} = (\sec \theta + \tan \theta)^2 \)

 

Question. Prove that: \( \frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \cdot \tan B \)
Answer: \( \frac{\cot A + \tan B}{\cot B + \tan A} = \cot A \cdot \tan B \)

 

Question. Prove that: \( \frac{\tan^2 \theta}{\tan^2 \theta - 1} + \frac{\csc^2 \theta}{\sec^2 \theta - \csc^2 \theta} = \frac{1}{\sin^2 \theta - \cos^2 \theta} \)
Answer: \( \frac{\tan^2 \theta}{\tan^2 \theta - 1} + \frac{\csc^2 \theta}{\sec^2 \theta - \csc^2 \theta} = \frac{1}{\sin^2 \theta - \cos^2 \theta} \)

 

Question. Prove that: \( \frac{\cos \theta}{\csc \theta + 1} + \frac{\cos \theta}{\csc \theta - 1} = 2 \tan \theta \)
Answer: \( \frac{\cos \theta}{\csc \theta + 1} + \frac{\cos \theta}{\csc \theta - 1} = 2 \tan \theta \)

 

Question. Prove that: \( 2 \sec^2 \theta - \sec^4 \theta - 2 \csc^2 \theta + \csc^4 \theta = \cot^4 \theta - \tan^4 \theta \)
Answer: \( 2 \sec^2 \theta - \sec^4 \theta - 2 \csc^2 \theta + \csc^4 \theta = \cot^4 \theta - \tan^4 \theta \)

 

Question. Prove that: \( \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 2 \sec \theta \)
Answer: \( \frac{\cos \theta}{1 - \sin \theta} + \frac{\cos \theta}{1 + \sin \theta} = 2 \sec \theta \)

 

Question. Prove that: \( \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta \)
Answer: \( \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta \)

 

Question. Prove that: \( \frac{1 + \cos \theta}{1 - \cos \theta} = (\csc \theta + \cot \theta)^2 \)
Answer: \( \frac{1 + \cos \theta}{1 - \cos \theta} = (\csc \theta + \cot \theta)^2 \)

 

Question. Prove that: \( \left( 1 + \frac{1}{\tan^2 \theta} \right) \left( 1 + \frac{1}{\cot^2 \theta} \right) = \frac{1}{\sin^2 \theta - \sin^4 \theta} \)
Answer: \( \left( 1 + \frac{1}{\tan^2 \theta} \right) \left( 1 + \frac{1}{\cot^2 \theta} \right) = \frac{1}{\sin^2 \theta - \sin^4 \theta} \)

 

Question. Prove that: \( \frac{1}{\sec \theta - \tan \theta} = \sec \theta + \tan \theta \)
Answer: \( \frac{1}{\sec \theta - \tan \theta} = \sec \theta + \tan \theta \)

 

Question. Prove that: \( \frac{1}{\sec x - \tan x} - \frac{1}{\cos x} = \frac{1}{\cos x} - \frac{1}{\sec x + \tan x} \)
Answer: \( \frac{1}{\sec x - \tan x} - \frac{1}{\cos x} = \frac{1}{\cos x} - \frac{1}{\sec x + \tan x} \)

 

Question. Prove that: \( \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} = 2 \csc \theta \)
Answer: \( \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} = 2 \csc \theta \)

 

Question. Prove that: \( \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} + \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = 2 \csc \theta \)
Answer: \( \sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}} + \sqrt{\frac{1 - \cos \theta}{1 + \cos \theta}} = 2 \csc \theta \)

 

Question. Prove that: \( \left( \tan \theta + \frac{1}{\cos \theta} \right)^2 + \left( \tan \theta - \frac{1}{\cos \theta} \right)^2 = 2 \left( \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \right) \)
Answer: \( \left( \tan \theta + \frac{1}{\cos \theta} \right)^2 + \left( \tan \theta - \frac{1}{\cos \theta} \right)^2 = 2 \left( \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \right) \)

 

Question. Prove that: \( \frac{\cos^2 \theta}{1 - \tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta} = 1 + \sin \theta \cos \theta \)
Answer: \( \frac{\cos^2 \theta}{1 - \tan \theta} + \frac{\sin^3 \theta}{\sin \theta - \cos \theta} = 1 + \sin \theta \cos \theta \)

 

Question. Prove that: \( \tan^2 A \sec^2 B - \sec^2 A \tan^2 B = \tan^2 A - \tan^2 B \)
Answer: \( \tan^2 A \sec^2 B - \sec^2 A \tan^2 B = \tan^2 A - \tan^2 B \)

 

Question. Prove that: \( \frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \cdot \tan B \)
Answer: \( \frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \cdot \tan B \)

 

Question. Prove that: \( \frac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2 \cos^2 \theta} = \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \)
Answer: \( \frac{(1 + \sin \theta)^2 + (1 - \sin \theta)^2}{2 \cos^2 \theta} = \frac{1 + \sin^2 \theta}{1 - \sin^2 \theta} \)

 

Question. Prove that: \( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{\sin^2 A - \cos^2 A} = \frac{2}{2 \sin^2 A - 1} \)
Answer: \( \frac{\sin A + \cos A}{\sin A - \cos A} + \frac{\sin A - \cos A}{\sin A + \cos A} = \frac{2}{\sin^2 A - \cos^2 A} = \frac{2}{2 \sin^2 A - 1} \)

 

Question. Prove that: \( \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} + \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{2}{1 - 2 \cos^2 \theta} \)
Answer: \( \frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta} + \frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{2}{1 - 2 \cos^2 \theta} \)

 

Question. Prove that: \( \frac{\cot A + \csc A - 1}{\cot A - \csc A + 1} = \frac{1 + \cos A}{\sin A} \)
Answer: \( \frac{\cot A + \csc A - 1}{\cot A - \csc A + 1} = \frac{1 + \cos A}{\sin A} \)

 

Question. Prove that: \( \sin^4 A + \cos^4 A = 1 - 2 \sin^2 A \cos^2 A \)
Answer: \( \sin^4 A + \cos^4 A = 1 - 2 \sin^2 A \cos^2 A \)

 

Question. Prove that: \( (\sec A - \csc A)(1 + \tan A + \cot A) = \tan A \sec A - \cot A \csc A \)
Answer: \( (\sec A - \csc A)(1 + \tan A + \cot A) = \tan A \sec A - \cot A \csc A \)

 

Question. Prove that: \( \csc^6 \theta = \cot^6 \theta + 1 + 3 \cot^2 \theta \csc^2 \theta \)
Answer: \( \csc^6 \theta = \cot^6 \theta + 1 + 3 \cot^2 \theta \csc^2 \theta \)

 

Question. Prove that: \( (\tan A + \csc B)^2 - (\cot B - \sec A)^2 = 2 \tan A \cot B (\csc A + \sec B) \)
Answer: \( (\tan A + \csc B)^2 - (\cot B - \sec A)^2 = 2 \tan A \cot B (\csc A + \sec B) \)

 

Question. Prove that: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \)
Answer: \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta \)

 

Question. Prove that: \( \frac{\cos A}{1 - \sin A} + \frac{\sin A}{1 - \cos A} + 1 = \frac{\sin A \cos A}{(1 - \sin A)(1 - \cos A)} \)
Answer: \( \frac{\cos A}{1 - \sin A} + \frac{\sin A}{1 - \cos A} + 1 = \frac{\sin A \cos A}{(1 - \sin A)(1 - \cos A)} \)

 

Question. Prove that: \( \frac{1 - \cos \theta + \sin \theta}{1 + \cos \theta - \sin \theta} = \frac{1 + \sin \theta}{\cos \theta} \)
Answer: \( \frac{1 - \cos \theta + \sin \theta}{1 + \cos \theta - \sin \theta} = \frac{1 + \sin \theta}{\cos \theta} \)

 

Question. Prove that: \( \frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1} = 2 \sec^2 A \)
Answer: \( \frac{\csc A}{\csc A - 1} + \frac{\csc A}{\csc A + 1} = 2 \sec^2 A \)

 

Question. Prove that: \( \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A \)
Answer: \( \frac{\cos A}{1 - \tan A} + \frac{\sin A}{1 - \cot A} = \sin A + \cos A \)

 

Question. Prove that: \( \frac{\tan^3 \theta}{1 + \tan^2 \theta} + \frac{\cot^3 \theta}{1 + \cot^2 \theta} = \sec \theta \csc \theta - 2 \sin \theta \cos \theta \)
Answer: \( \frac{\tan^3 \theta}{1 + \tan^2 \theta} + \frac{\cot^3 \theta}{1 + \cot^2 \theta} = \sec \theta \csc \theta - 2 \sin \theta \cos \theta \)

 

Question. Prove that: \( (1 + \cot A + \tan A)(\sin A - \cos A) = \frac{\sec A}{\csc^2 A} - \frac{\csc A}{\sec^2 A} \)
Answer: \( (1 + \cot A + \tan A)(\sin A - \cos A) = \frac{\sec A}{\csc^2 A} - \frac{\csc A}{\sec^2 A} \)

 

Question. Prove that: \( \frac{\cot^2 A (\sec A - 1)}{1 + \sin A} = \sec^2 A \left( \frac{1 - \sin A}{1 + \sec A} \right) \)
Answer: \( \frac{\cot^2 A (\sec A - 1)}{1 + \sin A} = \sec^2 A \left( \frac{1 - \sin A}{1 + \sec A} \right) \)

 

Question. Prove that: \( \frac{\tan A}{(1 + \tan^2 A)^2} + \frac{\cot A}{(1 + \cot^2 A)^2} = \sin A \cos A \)
Answer: \( \frac{\tan A}{(1 + \tan^2 A)^2} + \frac{\cot A}{(1 + \cot^2 A)^2} = \sin A \cos A \)

Free study material for Mathematics

CBSE Value-Based Resources: Class 10 Mathematics Chapter 8 Introduction to Trigonometry

Value-Based Questions for Class 10 Mathematics Chapter 8 Introduction to Trigonometry

Find reliable Value-Based Questions (VBQs) for Chapter 8 Introduction to Trigonometry designed for the CBSE syllabus. These resources guide Class 10 pupils through moral concepts, sharpening problem-solving skills and performance in Mathematics school tests.

NCERT-Aligned VBQs for Class 10 Mathematics

Compiled using the official NCERT book for Class 10 Mathematics, these questions provide complete clarity. Pair your practice with our comprehensive NCERT solutions for Class 10 Mathematics to study educator-verified responses.

Master Ethical Topics in Class 10 Mathematics

Daily practice of these Class 10 Mathematics value-based problems will make your concepts better and to help you further we have provided more study materials for Chapter 8 Introduction to Trigonometry on our website. By learning these ethical and value-driven topics you will easily get better marks and also understand the real-life application of Mathematics.

FAQs

Where can I find 2026-27 CBSE Value Based Questions (VBQs) for Class 10 Mathematics Chapter 8 Introduction to Trigonometry?

The latest collection of Value Based Questions for Class 10 Mathematics Chapter 8 Introduction to Trigonometry is available for free on StudiesToday.com. These questions are as per 2026 academic session to help students develop analytical and ethical reasoning skills.

Are answers provided for Class 10 Mathematics Chapter 8 Introduction to Trigonometry VBQs?

Yes, all our Mathematics VBQs for Chapter 8 Introduction to Trigonometry come with detailed model answers which help students to integrate factual knowledge with value-based insights to get high marks.

What is the importance of solving VBQs for Class 10 Chapter 8 Introduction to Trigonometry Mathematics?

VBQs are important as they test student's ability to relate Mathematics concepts to real-life situations. For Chapter 8 Introduction to Trigonometry these questions are as per the latest competency-based education goals.

How many marks are usually allocated to VBQs in the CBSE Mathematics paper?

In the current CBSE pattern for Class 10 Mathematics, Chapter 8 Introduction to Trigonometry Value Based or Case-Based questions typically carry 3 to 5 marks.

Can I download Mathematics Chapter 8 Introduction to Trigonometry VBQs in PDF for free?

Yes, you can download Class 10 Mathematics Chapter 8 Introduction to Trigonometry VBQs in a mobile-friendly PDF format for free.