Find the best tutors and institutes for Class 10 Tuition
Q4(viii):
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
Solution :
Given: An identity involving trigonometric functions of an acute angle $A$: $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$.
To Prove: $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.
Step 1: Expanding the Left Hand Side (LHS)
We start with the expression: $LHS = (\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2$.
Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$, we expand both terms:
$(\sin A + \text{cosec } A)^2 = \sin^2 A + 2\sin A \cdot \text{cosec } A + \text{cosec}^2 A$
$(\cos A + \sec A)^2 = \cos^2 A + 2\cos A \cdot \sec A + \sec^2 A$
Step 2: Simplifying the expanded terms
Recall the reciprocal identities: $\text{cosec } A = \frac{1}{\sin A}$ and $\sec A = \frac{1}{\cos A}$.
Substituting these into the middle terms:
$2\sin A \cdot \text{cosec } A = 2\sin A \cdot \left(\frac{1}{\sin A}\right) = 2$
$2\cos A \cdot \sec A = 2\cos A \cdot \left(\frac{1}{\cos A}\right) = 2$
Now, substitute these back into the expanded expression:
$LHS = \sin^2 A + 2 + \text{cosec}^2 A + \cos^2 A + 2 + \sec^2 A$
Step 3: Grouping terms using Trigonometric Identities
Rearrange the terms to group $\sin^2 A$ and $\cos^2 A$ together:
$LHS = (\sin^2 A + \cos^2 A) + 2 + 2 + \text{cosec}^2 A + \sec^2 A$
Using the Pythagorean identity $\sin^2 A + \cos^2 A = 1$:
$LHS = 1 + 4 + \text{cosec}^2 A + \sec^2 A$
$LHS = 5 + \text{cosec}^2 A + \sec^2 A$
Step 4: Converting to $\tan^2 A$ and $\cot^2 A$
We use the trigonometric identities: $\text{cosec}^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$.
Substitute these into the expression:
$LHS = 5 + (1 + \cot^2 A) + (1 + \tan^2 A)$
$LHS = 5 + 1 + 1 + \tan^2 A + \cot^2 A$
$LHS = 7 + \tan^2 A + \cot^2 A$
Step 5: Conclusion
Since the simplified LHS is equal to the Right Hand Side (RHS):
$LHS = RHS = 7 + \tan^2 A + \cot^2 A$
Final Answer: The identity $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ is proven.
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3
- Q1: Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.
- Q2: Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
- Q3(i): Choose the correct option. Justify your choice. (i) $9 \sec^2 A – 9 \tan^2 A =$
- Q3(ii): Choose the correct option. Justify your choice. (ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta – \text{cosec } \theta) =$
- Q3(iii): Choose the correct option. Justify your choice. (iii) $(\sec A + \tan A) (1 – \sin A) =$
- Q3(iv): Choose the correct option. Justify your choice. (iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
- Q4(i): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) $(\text{cosec } \theta – \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
- Q4(ii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
- Q4(iii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$ [Hint : Write the expression in terms of $\sin \theta$ and $\cos \theta$]
- Q4(iv): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ [Hint : Simplify LHS and RHS separately]
- Q4(ix): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ix) $(\text{cosec } A – \sin A) (\sec A – \cos A) = \frac{1}{\tan A + \cot A}$ [Hint : Simplify LHS and RHS separately]
- Q4(v): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.
- Q4(vi): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$
- Q4(vii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
- Q4(x): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (x) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
Chapters in CBSE - Class 10 Mathematics
Top Tutors who teach Introduction to Trigonometry
Top Tutor
Burul Rd Burul, Burul
5 yrs of
Exp
100
per hour
Online Classes
Student's Home
Tutor's home
I am a teacher of math. I am giving home/online/tutor home tuition since 2016. My key skills are computer language. I am b. Tech graduate. I can teach short tricks that can help students in their future life.
Top Tutor
Talwandi Sector-B, Kota
10 yrs of
Exp
1 student
699 hrs of classes completed
1850
per hour
Online Classes
Tutor's home
My teaching exp.11 year in most reputed coaching in india. Mentor of jee top rank air-12 yatis agrawal, 19, 28 many more in top 100 in jee adavnced. Many of students got 100%tile in jee mains along with 100% marks in board exams. I also provide complete relevent study material like worksheet, practice sheet , dpps , assignment etc. Daily quizz based on assignment. Periodically test papers topic wise, minor and major as per shedule. I have helped thousands of students overcome their fear of maths.
Top Tutor
Sikandra, Agra
4 yrs of
Exp
1 student
350
per hour
Online Classes
Student's Home
I can teach every topic very easily and make it stronger for the students.
Super Tutor
T Nagar, Chennai
10 yrs of
Exp
48 students
1174 hrs of classes completed
500
per hour
Online Classes
Teaching Experience in Detail (Class 10 Tuition) I have over 10 years of experience teaching CBSE, ICSE, and State Board Class 10 students, helping them achieve excellent results in their board examinations through a structured and concept-oriented teaching approach. I specialize in Science, Mathematics, Social Science, and English, ensuring students develop a thorough understanding of each subject rather than relying on rote learning. My teaching methodology begins with strengthening fundamental concepts, followed by chapter-wise practice, NCERT-based learning, important board questions, competency-based questions, case study questions, and previous years' board papers. Regular assignments, periodic tests, and detailed performance analysis help students identify weak areas and improve consistently. I provide personalized attention to every student by adapting my teaching style to their learning pace. My classes include interactive discussions, doubt-clearing sessions, revision plans, and exam-oriented strategies that build confidence and improve time management. My objective is to help students not only score high marks in the Class 10 Board Examination but also develop strong analytical and problem-solving skills that prepare them for higher secondary education. Through consistent guidance, disciplined practice, and individual mentoring, I strive to make learning enjoyable, effective, and result-oriented.
We are very happy with the way Dr. Nandhini teaches. The concepts are explained clearly and patiently, making even difficult topics easy to understand. The teacher is friendly, approachable, and encourages students to ask questions. There is a good focus on both understanding the concepts and preparing well for exams. Overall, we are very satisfied with the teaching and guidance provided.
Top Tutor
Melakalkandarkottai, Tiruchirappalli
10 yrs of
Exp
1 student
15 hrs of classes completed
300
per hour
Online Classes
Student's Home
I specialize in Class 10 Mathematics tuition with an exam-oriented and concept-based teaching approach. I simplify complex mathematical concepts using step-by-step explanations, making them easy for students to understand and apply confidently. I assess each student's learning level, identify weak areas, and provide personalized guidance to improve both accuracy and speed. My classes include regular practice, chapter-wise tests, previous years' board exam questions, doubt-clearing sessions, and effective exam strategies. I maintain a friendly and motivating learning environment where students feel comfortable asking questions. My objective is to help every student strengthen their mathematical foundation, gain confidence, and achieve excellent results in the Class 10 board examinations.
Experienced Mathematics tutor with a strong focus on concept-based learning. I help students improve their understanding, problem-solving skills, and academic performance through personalized guidance.
Find more Tutor for Introduction to Trigonometry in your City
- Bangalore Mathematics Tutors
- Delhi Mathematics Tutors
- Chennai Mathematics Tutors
- Gurgaon Mathematics Tutors
- Noida Mathematics Tutors
- Hyderabad Mathematics Tutors
- Mumbai Mathematics Tutors
- Ghaziabad Mathematics Tutors
- Chandigarh Mathematics Tutors
- Pune Mathematics Tutors
- Jaipur Mathematics Tutors
- Surat Mathematics Tutors
Download free CBSE - Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3 worksheets
Download Now
