Find the best tutors and institutes for Class 10 Tuition
Q4(x):
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (x) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
Solution :
Given: An identity involving trigonometric ratios of an acute angle $A$: $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$.
To Prove: 1. $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$ 2. $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$
Step 1: Proving the first part $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$
We use the fundamental trigonometric identities:
$1 + \tan^2 A = \sec^2 A$
$1 + \cot^2 A = \csc^2 A$
[Since $\sec^2 \theta - \tan^2 \theta = 1$ and $\csc^2 \theta - \cot^2 \theta = 1$]
Substitute these into the expression:
$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A}$
Express $\sec A$ and $\csc A$ in terms of $\sin A$ and $\cos A$:
$\sec A = \frac{1}{\cos A}$ and $\csc A = \frac{1}{\sin A}$
Therefore:
$\frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{1}{\cos^2 A} \times \frac{\sin^2 A}{1} = \frac{\sin^2 A}{\cos^2 A}$
Using the identity $\tan A = \frac{\sin A}{\cos A}$:
$\frac{\sin^2 A}{\cos^2 A} = \tan^2 A$
Thus, the first part is proved: $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \tan^2 A$.
Step 2: Proving the second part $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$
Express $\cot A$ in terms of $\tan A$:
$\cot A = \frac{1}{\tan A}$
Substitute this into the expression:
$\left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2$
Simplify the fraction:
$\left((1 - \tan A) \times \frac{\tan A}{\tan A - 1}\right)^2$
Note that $(1 - \tan A) = -( \tan A - 1)$:
$\left(\frac{-( \tan A - 1) \times \tan A}{\tan A - 1}\right)^2 = (-\tan A)^2$
Calculate the square:
$(-\tan A)^2 = \tan^2 A$
Conclusion:
Since both the first expression and the second expression are equal to $\tan^2 A$, the identity is proved:
$\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$
Final Answer: The identity is proven as both sides simplify to $\tan^2 A$.
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(viii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
Chapters in CBSE - Class 10 Mathematics
Top Tutors who teach Introduction to Trigonometry
Super Tutor
Kodambakkam, Chennai
15 yrs of
Exp
57 students
4055 hrs of classes completed
800
per hour
Online Classes
With a decade of experience in teaching mathematics, physics, and chemistry for students from grades 8 through 12 across CBSE, IB, and ICSE boards, I offer a comprehensive educational approach that caters to a diverse range of curricula and learning needs. My academic background includes a BE and an MBA, equipping me with both technical and managerial skills that enhance my teaching methodology. Throughout my career, I have specialized in home tutoring, where I have developed a personalized approach to education that focuses on each student's unique needs and strengths. My sessions are designed to be interactive and engaging, fostering a learning environment where students feel supported and motivated. I conduct both online and offline classes, each lasting one hour, and utilize a variety of teaching tools to facilitate learning. I employ a whiteboard to visually explain concepts and provide question papers to help students practice and prepare for their exams. This hands-on approach ensures that students not only understand theoretical concepts but also develop problem-solving skills and confidence in their subjects. My commitment to delivering high-quality education and my extensive experience make me well-suited to guide students through their academic journey, helping them achieve their full potential in mathematics, physics, and chemistry.
Top Tutor
Sardar Patel Nagar, Dhanbad
6 yrs of
Exp
60 hrs of classes completed
Online Classes
I am an experienced, qualified teacher and tutor with over 5 years of experience in teaching Maths and phy , across different boards and diffrent entrance exam including IIT-JEE , Neet ,B.tech. Passionate about solving mathematical problems and physics problem, over the years I have helped thousands of students overcome their fear of Maths and physics.
Top Tutor
Purasaiwakkam Otteri, Chennai
10 yrs of
Exp
36 students
3569 hrs of classes completed
200
per hour
Online Classes
I have a total of 10 years of teaching experience. I am Teaching Online for Overseas and Local Students. My Students are from INDIA, UK, USA, INDONESIA & THAILAND. My passion is Teaching. I am Focusing on CBSE, IGCSE, IB, ICSE,&NIOS,.I have conducted National level conference and workshop, Hands-on-training, Seminar, etc Teaching with Human touch produces a great result. My students of class 10th had proven with good results.
Top Tutor
Ashok Nagar D' Souza Layout, Bangalore
6 yrs of
Exp
13 students
818 hrs of classes completed
400
per hour
Online Classes
I am a teacher , a statistician and a data analyst. I am willing to give online tution to uh in subject maths and statistics. I am a first ranker in my university in MSc (Statistics) and a gold medalist. I am certified in O level. I have cleared written exam for the post statistical officer in 2022 conducted by UPSC. Also cleared exam of MES but could not make in the interview. My key skills are teaching, solving queries, understanding students, visualising and analysing data. I have 6 years experience in teaching and i was a junior consultant having 4 years of experience in Ministry of labour and employment,Shimla . Currently i am working as a Monitoring Assistant in Department of Fisheries, Ministry of Fisheries,Animal Husbandry and Dairying, New Delhi. I know languages like R, python etc.
Super Tutor
Badarpur, Delhi
12 yrs of
Exp
152 students
3859 hrs of classes completed
200
per hour
Online Classes
Tutor's home
Akshay is an excellent teacher. He has taught my son in 10 th standard science , with which he understood the concepts clearly and he also gave so many past question papers for my son to work out, all these made my son score 95 in board exams . I would definitely recommend Akshay for students who want to improve and get good scores in science .
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
