Find the best tutors and institutes for Class 10 Tuition
Q7(i):
If $\cot \theta = \frac{7}{8}$, evaluate : (i) $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$
Solution :
Given: $\cot \theta = \frac{7}{8}$
To find: The value of $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$
Step 1: Define the trigonometric ratio in terms of sides.
In a right-angled triangle, $\cot \theta = \frac{\text{Adjacent side}}{\text{Opposite side}}$.
Given $\cot \theta = \frac{7}{8}$, let the adjacent side $AB = 7k$ and the opposite side $BC = 8k$, where $k$ is a positive constant.
Step 2: Calculate the hypotenuse using the Pythagorean Theorem.
According to the Pythagorean Theorem: $AC^2 = AB^2 + BC^2$.
$AC^2 = (7k)^2 + (8k)^2$
$AC^2 = 49k^2 + 64k^2$
$AC^2 = 113k^2$
$AC = \sqrt{113}k$
Step 3: Determine $\sin \theta$ and $\cos \theta$.
$\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{8k}{\sqrt{113}k} = \frac{8}{\sqrt{113}}$
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{7k}{\sqrt{113}k} = \frac{7}{\sqrt{113}}$
Step 4: Simplify the expression using algebraic identities.
The expression is $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$.
Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$:
Numerator: $(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta$
Denominator: $(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2 \theta$
Expression = $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$
Step 5: Substitute the values and calculate.
$1 - \sin^2 \theta = 1 - (\frac{8}{\sqrt{113}})^2 = 1 - \frac{64}{113} = \frac{113 - 64}{113} = \frac{49}{113}$
$1 - \cos^2 \theta = 1 - (\frac{7}{\sqrt{113}})^2 = 1 - \frac{49}{113} = \frac{113 - 49}{113} = \frac{64}{113}$
Expression = $\frac{49/113}{64/113} = \frac{49}{113} \times \frac{113}{64} = \frac{49}{64}$
Final Answer: \frac{49}{64}
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1
- Q1(i): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (i) $\sin A, \cos A$
- Q1(ii): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (ii) $\sin C, \cos C$
- Q10: In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.
- Q11(i): State whether the following are true or false. Justify your answer. (i) The value of $\tan A$ is always less than 1.
- Q11(ii): State whether the following are true or false. Justify your answer. (ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
- Q11(iii): State whether the following are true or false. Justify your answer. (iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
- Q11(iv): State whether the following are true or false. Justify your answer. (iv) $\cot A$ is the product of cot and $A$.
- Q11(v): State whether the following are true or false. Justify your answer. (v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
- Q2: In Fig. 8.13, find $\tan P – \cot R$.
- Q3: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
- Q4: Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.
- Q5: Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.
- Q6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.
- Q7(ii): If $\cot \theta = \frac{7}{8}$, evaluate : (ii) $\cot^2 \theta$
- Q8: If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.
- Q9(i): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (i) $\sin A \cos C + \cos A \sin C$
- Q9(ii): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (ii) $\cos A \cos C – \sin A \sin C$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
Chapters in CBSE - Class 10 Mathematics
Top Tutors who teach Introduction to Trigonometry
Nanmangalam, Chennai
3 yrs of
Exp
2 students
88 hrs of classes completed
260
per hour
Online Classes
Tutor's home
Hi, my name is M.Praveshika. I am a graduate in Master of Philosophy (Mathematics) from Stella Maris College. During my degree, I have developed an easy algorithm for solving certain graphs in my specialization paper. I spent my spare time by tutoring my cousins and working with them gave me a new understanding how students are facing problems in understanding the concepts. So i started my career as a Subject Matter Expert (EzeeScore) which enabled me to utilize my skills and knowledge for the growth of the students as well as to the organisation. My key skills are Leadership, Time maganement, Decision making and Problem solving.
Top Tutor
Chitragupta Nagar, Patna Sadar
5 yrs of
Exp
9 students
5038 hrs of classes completed
300
per hour
Online Classes
Hello all, I am Utkarsh kumar. As an experienced, qualified teacher and tutor with over 8 years of experience in teaching Maths, Physics, Chemistry and Biology across different boards including CBSE, and State Board. Passionate about solving mathematical problems and chemical equations. Over the years I have helped thousands of students overcome their fear of Maths, Physics Chemistry and Biology. If u wish to learn in interactive and fun manner, feel free to make a contact.
Super Tutor
T Nagar, Chennai
10 yrs of
Exp
48 students
1130 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.
Dr Nandhini ma'am's methodology is appreciable, and she is approachable too. Her repeated questioning of the kids made them feel lively in class, and it helped a lot. Thank you, ma'am, for your kind, encouraging words to my kid.
Top Tutor
BRS Nagar, Ludhiana
27 hrs of classes completed
Online Classes
Student's Home
Tutor's home
I was a student in one of the top colleges of Ludhiana. I have proficiency in English . I can teach all subjects to primary wing .
Top Tutor
Sector 150, Noida
5 yrs of
Exp
800
per hour
Student's Home
Tutor's home
Having 2 years experience in teaching class 6th to 10th across different boards.
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 8.1 worksheets
Download Now
