FORMATIVE ASSESSMENT
I. Answer the following questions 2 × 1 = 2 M
2. if sin A = cos B, then prove that A + B = 90°
II Answer the following questions 3 × 2 = 6 M
5. the sides of a right angle triangle ABC are AB = 7 cm, BC = 25 Cm and A = 90° respectively. Then find tan B - tan C
III. Answer the following questions 2 × 4 = 8 M
6. A chord of circle of radius 6 cm, is making an angle 60° at the center. Find the length of the chord
7. If cot B = 
, prove that tan2 B - sin2 B = sin4B. sec2B.
IV. Choose correct option for the following questions and write capital letters in the brackets provided against them. 8 × 
= 4 M
8. The value of tan 24°. tan 42°. tan 48°. tan 66°= .............. ( )
A) 
B) 
C) 0 4) 1
( )
A) 
B) 
C) 
D) 49
10. If cos(A + B) = 
and sin(A - B) = 
0 < B < A < 90° then A = B =....... ( )
A) 60°, 45° B) 52.5°, 7.5° C) 30°, 45° D) 60°, 15°
11. If 3 tan A = 4 then sin A = ........ ( )
A) 
B) 
C) 
D) 
12. 
( )
A) 
B) 
C) 
D) 3
13. If tan θ - cot θ = 1, then tan2 θ + cot2 θ = ....... ( )
A) 3 B) 4 C) 1 D) 2
A) sec A + cosec A B) secA.cosecA
C) cos A + sin A D) 1
15. If sin2 X, sin2 Y, sin2 Z are in A.P. then X, Y, Z are ......
A) 45°, 45°, 90° B) 30°, 60°,90°
C) 0°, 45°, 90° D) 30°, 45°,60°
PROJECT WORK
Preliminary Information:
Class: 10th
Subject: Mathematics
Name of the unit: Trigonometry
No.of the project:
Medium: English
Allotment of work: Individual
Detailed information of the project:
Tittle of the project: Solution to a Problem Indifferent Methods.
Objectives of the Project:
1. Finding solutions of a problem in different methods
Material used: Paper, Pen
Tools: 1) Identify a problem
2) Comparing different solutions of a problem
Procedure: Collection of data: Let us collect a problem from Trigonometry
Problem: Prove the identity tan2 θ + cot2 θ + 2 = sec2 θ. cosec2 θ.
Solutions:
Method: 3) L.H.S = tan2 θ + cot2 θ + 2
= 1 + tan2 θ + cot2 θ + 1
Method: 4) R.H.S = sec2 θ. cosec2 θ
= (1 + tan2 θ)(1 + cot2 θ)
= 1 + cot2 θ + tan2 θ + tan2 θ.cot2 θ
= 1 + tan2 θ + cot2 θ + 1 (... tan2 θ. cot2 θ = 1)
= tan2 θ + cot2 θ + 2
Method: 5) L.H.S = tan2 θ + cot2 θ + 2
Method: 6) L.H.S = tan2 θ + cot2 θ + 2
Analysis: In method 1, I used both trigonometric principles and the algebraic identity
a2 + 2ab + b2 = (a + b)2
In method 2, I used only trigonometric principles
In method 3, I used trigonometric principles and factorization
In method 4, I used trigonometric principles and factorization observe that it is also reciprocal to method 3
In method 5, I used trigonometric principles
In method 6, I used trigonometric principles and the algebraic identify
a2 + 2ab + b2 = (a + b)2.
conclusion: From this project, I conclude that I can solve a problem is different methods with logic.
Experiences of the students:
i) I enjoyed while doing this project.
ii) I learnt more methods and applications of different identities both in trigonometry and algebra.
Doubts and Questions:
i) can I solve all problems in different methods?
Acknowledgement:
I convey our sincere thanks to my guide teacher.
Reference Books/ Resources:
i) APSCERT, X - Maths Text book
ii) NCERT, X - Maths Text book
iii) Mathematics Class - X by R.D. Sharma
Signature of the Student:
