exercise 8.2 Archives – Mathemafia https://mathemafia.com/tag/exercise-8-2/ Mathematics is like music - Enjoy it!!! Thu, 30 Oct 2025 04:21:19 +0000 en-US hourly 1 https://wordpress.org/?v=7.1.3 https://mathemafia.com/wp-content/uploads/2020/08/cropped-M__Logo__SM-1-32x32.png exercise 8.2 Archives – Mathemafia https://mathemafia.com/tag/exercise-8-2/ 32 32 Exercise 8.2 (NCERT) – Introduction to Trigonometry, Class 10 – All Solutions https://mathemafia.com/exercise-8-2-ncert-introduction-to-trigonometry-class-10-all-solutions-ncert/ Thu, 13 Feb 2025 05:09:38 +0000 https://mathemafia.com/?p=11077 Complete solution of Exercise 8.2, Chapter 8, Introduction to Trigonometry of NCERT book of Mathematics for Class 10. This exercise is more about finding the standard values the trigonometric ratios and hence really easy to solve. The solutions are written in pure mathematical way with mobile-friendly view, fully responsive. The free YouTube video tutorial of every question is embedded after the written solution. This educational content has been created with utmost care to make it easily readable and hence understandable by the students of class 10, especially preparing for the final board exams.

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Exercise 8.2 (NCERT) - Trigonometry, Class 10

Chapter 8 Exercise 8.2 Class 10

Below are the Quick links for all questions of Exercise 8.2, Trigonometry, Class 10.

Click a link to view the solution of corresponding question.

Below are the solutions of every question of Exercise 8.2 of Chapter 8, Introduction to Trigonometry, NCERT of Class 10. YouTube video tutorial of every question is also embedded along with the written solution.  

This exercise is primarily about the standard values of trigonometric ratios and that’s why the questions are quite easy to solve.

Very Important: Before you start solving the questions of this exercise, it is very important that you remember the table displayed below.

Trigonometric ratios of standard angles

∠A 0° 30° 45° 60° 90°
sin A
0
\(\begin{aligned} \frac {1}{2}\end{aligned} \)
\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \)
\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \)
1
cos A
1
\(\begin{aligned} \frac {\sqrt{3}}{2}\end{aligned} \)
\(\begin{aligned} \frac {1}{\sqrt{2}}\end{aligned} \)
\(\begin{aligned} \frac {1}{2}\end{aligned} \)
0
tan A
0
\(\begin{aligned} \frac {1}{\sqrt {3}}\end{aligned} \)
1
\(\begin{aligned} \sqrt {3} \end{aligned} \)
N.D.
cosec A
N.D.
2
\(\begin{aligned} \sqrt {2} \end{aligned} \)
\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \)
1
sec A
1
\(\begin{aligned} \frac {2}{\sqrt{3}}\end{aligned} \)
\(\begin{aligned} \sqrt{2}\end{aligned} \)
2
N.D.
cot A
N.D.
\(\begin{aligned} \sqrt{3}\end{aligned} \)
1
\(\begin{aligned} \frac {1}{\sqrt{3}}\end{aligned} \)
1

N.D. – Not Defined

How to quickly make this table in exams

Let’s start using these trigonometric ratios of angles 0°, 30°, 45°, 60° and 90° to solve the questions. 

Question 1

Evaluate the following :

(𝑖)  𝑠𝑖𝑛60°𝑐𝑜𝑠30°+𝑠𝑖𝑛30°𝑐𝑜𝑠60°
=3‾√2×3‾√2+12×12=34+14=1
(𝑖𝑖)  2𝑡𝑎𝑛245°+𝑐𝑜𝑠230°–𝑠𝑖𝑛260°
=2×12+(3‾√2)2–(3‾√2)2=2+34–34=2
(𝑖𝑖𝑖)  𝑐𝑜𝑠45°𝑠𝑒𝑐30°+𝑐𝑜𝑠𝑒𝑐30°
=12√23√+2=12√2+23√3√=12‾√×3‾√2+23‾√=3‾√22‾√+22‾√3‾√=3‾√22‾√+26‾√

Note: As per my understanding, this should be treated as the final answer, but in your book, the denominator has been rationalized. To achieve that, we need to further solve it as follows. 

Rationalizing the denominator 

=3‾√22‾√+26‾√×22‾√–26‾√22‾√–26‾√=22‾√3‾√–23‾√6‾√4(2‾√)2–4×6=22‾√3‾√–23‾√2‾√3‾√4×2–4×6=26‾√–62‾√8−24=2(6‾√–32‾√)−16=6‾√–32‾√−8=32‾√–6‾√8
(𝑖𝑣)  𝑠𝑖𝑛30°+𝑡𝑎𝑛45°–𝑐𝑜𝑠𝑒𝑐60°𝑠𝑒𝑐30°+𝑐𝑜𝑠60°+𝑐𝑜𝑡45°
=12+1–23√23√+12+1=3√+23√–423√4+3√+23√23√=33√–423√4+33√23√=33‾√–423‾√×23‾√4+33‾√=33‾√–44+33‾√

Rationalizing the denominator

=33‾√–44+33‾√×4–33‾√4–33‾√=123‾√–16–27+123‾√16−27=243‾√–43−11=43–243‾√11
(𝑣)  5𝑐𝑜𝑠260°+4𝑠𝑒𝑐230°–𝑡𝑎𝑛245°𝑠𝑖𝑛230°+𝑐𝑜𝑠230°
=5×(12)2+4×(23√)2−12(12)2+(3√2)2=5×14+4×43–114+34=54+163–144=15+64–1212=6712
Click the following video to watch it’s solution.

Question 2

Choose the correct option and justify your choice :

(𝑖)  2𝑡𝑎𝑛30°1+𝑡𝑎𝑛230°(𝐴)  𝑠𝑖𝑛60°  (𝐵)  𝑐𝑜𝑠60°(𝐶)  𝑡𝑎𝑛60°  (𝐷)  𝑠𝑖𝑛30°
2𝑡𝑎𝑛30°1+𝑡𝑎𝑛230°=2×13√1+(13√)2=23√1+(13)=23√43=23‾√×34=323‾√=3‾√×3‾√23‾√=3‾√2=𝑠𝑖𝑛60°

Hence (A) is the answer. 

(𝑖𝑖)  1−𝑡𝑎𝑛245°1+𝑡𝑎𝑛245°(𝐴) tan90°  (𝐵)  1(𝐶)  𝑠𝑖𝑛45°  (𝐷)  0
=1−𝑡𝑎𝑛245°1+𝑡𝑎𝑛245°=1−121+12=02=0

Hence (D) is the answer. 

(iii) sin 2A = 2 sin A is true when A = 

(𝐴)  0°  (𝐵)  30°(𝐶)  45°  (𝐷)  60°

Let us take A = 0°

⇒𝑠𝑖𝑛𝐴⇒2𝑠𝑖𝑛𝐴𝐴𝑙𝑠𝑜  𝑠𝑖𝑛2𝐴⇒2𝑠𝑖𝑛𝐴=𝑠𝑖𝑛0°=0=0=𝑠𝑖𝑛(2×0)=𝑠𝑖𝑛0°=0=𝑠𝑖𝑛2𝐴

Hence (A) is the answer.

(𝑖𝑣)  2𝑡𝑎𝑛30°1–𝑡𝑎𝑛230°(𝐴)  𝑐𝑜𝑠60°  (𝐵)  𝑠𝑖𝑛60°(𝐶)  𝑡𝑎𝑛60°  (𝐷)  𝑠𝑖𝑛30°
2𝑡𝑎𝑛30°1–𝑡𝑎𝑛230°=2×13√1–(13√)2=23√1–13=23√23=23‾√×32=33‾√=3‾√=𝑡𝑎𝑛60°

Hence (C) is the answer.

Click the following video to watch it’s solution.

Question 3

𝐼𝑓  𝑡𝑎𝑛(𝐴+𝐵)=3‾√  𝑎𝑛𝑑  𝑡𝑎𝑛(𝐴–𝐵)=13‾√

; 0° < A + B ≤ 90°; A > B, find A and B.

𝑡𝑎𝑛(𝐴+𝐵)⇒𝐴+𝐵𝐴𝑙𝑠𝑜  𝑡𝑎𝑛(𝐴–𝐵)⇒𝐴−𝐵Adding (1) and (2)2𝐴⇒𝐴=3‾√=60  …(1)=13‾√=30  …(2)=90=45

Put the value of A in (1)

𝐴+𝐵⇒45+𝐵⇒𝐵=60=60=15

Answer: A = 45° and B = 15°

Click the following video to watch it’s solution.

Question 4

State whether the following are true or false. Justify your answer.

(i) sin (A + B) = sin A + sin B. 

Answer: False

Consider A = 30° and B = 60°

⇒  𝑎𝑛𝑑  𝑠𝑖𝑛𝐴+𝑠𝑖𝑛𝐵=𝑠𝑖𝑛30°+𝑠𝑖𝑛60°=12+3‾√2=1+3‾√2  …  …(1)𝑠𝑖𝑛(𝐴+𝐵)=𝑠𝑖𝑛(30°+60°)=𝑠𝑖𝑛90°=1  …  …  …(2)

From (1) and (2)

𝑠𝑖𝑛𝐴+𝑠𝑖𝑛𝐵≠𝑠𝑖𝑛(𝐴+𝐵)

Hence the statement of not true in general.

(ii) The value of sin θ increases as θ increases.

Answer: Actually it’s answer is False but it is given as True in your book.

The reason is, the value of angle θ can be more than 90 degrees as well. The value of sin θ increases as θ increases from 0 to 90 degrees but sin θ starts decreasing as the value of θ increases from 90 to 180 degrees. 

Now talk to your class teacher to know what to write in exam 😉

Ideally the question should be as follows: 

The value of sin θ increases as θ increases from 0 to 90 degrees.

(iii) The value of cos θ increases as θ increases.

Answer: False

Reason: cos 0° = 1 and cos 90° = 0

(iv) sin θ = cos θ for all values of θ.

Answer: False

Reason: sin 0° = 0 and cos 0° = 1

(v) cot A is not defined for A = 0°.

Answer: True

Reason: Just remember the value of cot 0° or try to evaluate cos 0°/sin 0° 😉

Click the following video to watch it’s solution.

Some general tips for solving trigonometry questions easily and quickly are:

  1. Prepare a table of all standard values of sin, cos, tan, cosec, sec and cot. This will help you memorize them to be applied quickly to your questions.
  2. Another important tip is that if you get stuck somewhere, try to change everything into sin and cos. In some questions it may help you progress further towards the solution. 
  3. Some important relations are sin(x) = cos(90-x) , tan(x) = cot(90-x) , cosec(x) = sec(90-x) and vice versa.
  4. It is important to know which side of the triangle is base and which is perpendicular. The one opposite to the angle under consideration is known as perpendicular and the one adjacent is the base. (the longest one is the hypotenuse)
  5. In case you are not able to prove LHS equal to RHS, then progress both of them separately and bring them to a common point. This will be considered a valid solution to the question. 

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