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Important Formula (Trigonometry class ten) and Chapter Summary

In the right angle triangle ABC, the right angle is ∠ B.
Class ten Math-Trigonometry-NCERT Solution1 Class ten Math-Trigonometry-NCERT Solution2

Class ten Math-Trigonometry-NCERT Solution3
Class ten Math-Trigonometry-NCERT Solution4 

Class ten Math-Trigonometry-NCERT Solution5 

The value of sin or cos never exceeds 1, but the value of sec and cosec is always greater than or equal to 1.
Class ten Math-Trigonometry-NCERT Solution6 

Exercise 8.1
Question – 1 - In triangle ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
Class ten Math-Trigonometry-NCERT Solution7
(a) Sin A, Cos A
Solution: AB = 24 cm, BC = 7 cm, AC = ?
The value of AC can be calculated by using Pythagoras Theorem:
Class ten Math-Trigonometry-NCERT Solution7
(b) Sin C, Cos C
Solution:
Class ten Math-Trigonometry-NCERT Solution7
Question – 2 - In the given figure, find tan P – cot R.
Class ten Math-Trigonometry-NCERT Solution8
Solution: Value of QR can be calculated by using Pythagoras theorem:
Class ten Math-Trigonometry-NCERT Solution8
Now;
Class ten Math-Trigonometry-NCERT Solution9
Question – 3 - If sin A = ¾, calculate cos A and tan A.
Solution: Sin A = ¾ = p/h
We can calculate b by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution11
Now;
Class ten Math-Trigonometry-NCERT Solution12
Question – 4 - Given 15 cot A = 8, find sin A and sec A.
Solution: 15 cot A = 8
Class ten Math-Trigonometry-NCERT Solution13
This means, b = 8 and p = 15
We can calculate h by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution14
Now;
Class ten Math-Trigonometry-NCERT Solution15
Question – 5 - Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution:
Class ten Math-Trigonometry-NCERT Solution16
This means, h = 13 and b = 12.
We can calculate p by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution17
Other trigonometric ratios can be calculated as follows:
Class ten Math-Trigonometry-NCERT Solution18
Question – 6 - If ∠ A and ∠ B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B
Solution: For no two different angles the cos ratio is same. (Ref: Table of trigonometric ratio).
Question – 7 - If cot θ = 7/8, evaluate:
Solution:
Class ten Math-Trigonometry-NCERT Solution19
This means, b = 7 and p = 8.
We can calculate h by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution20 

Class ten Math-Trigonometry-NCERT Solution21
Solution:
Class ten Math-Trigonometry-NCERT Solution22
Solution:
Class ten Math-Trigonometry-NCERT Solution23
Question – 8 - If 3 cot A = 4, check whether Class ten Math-Trigonometry-NCERT Solution24 or not.
Solution: 3 cot A = 4 means cot A = 4/3 = b/p
Hence, p = 3 and b = 4.
We can calculate h by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution25
Now; the equation can be checked as follows:
LHS:
Class ten Math-Trigonometry-NCERT Solution26
RHS:
Class ten Math-Trigonometry-NCERT Solution27
It is clear that LHS = RHS.
Question – 9 - In triangle ABC, right-angled at B, if tan A = Class ten Math-Trigonometry-NCERT Solution28 find the value of:
Solution:
Class ten Math-Trigonometry-NCERT Solution29 

Class ten Math-Trigonometry-NCERT Solution30
We can calculate h by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution31
(a) Sin A Cos C + Cos A Sin C
Solution:
Sin A Cos C + Cos A Sin C
Class ten Math-Trigonometry-NCERT Solution32
(b) Cos A Cos C – Sin A Sin C
Solution:
Cos A Cos C – Sin A Sin C
Class ten Math-Trigonometry-NCERT Solution33
Question – 10 - 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.
Class ten Math-Trigonometry-NCERT Solution33
Solution: Given; PR + QR = 25 cm and PQ = 5 cm.
Hence, PR = 25 – QR
We can calculate PR and QR by using Pythagoras theorem;
Class ten Math-Trigonometry-NCERT Solution33
Now;
Class ten Math-Trigonometry-NCERT Solution34
Question – 11 - State whether the following are true or false. Justify your answer.
(a) The value of tan A is always less than 1.
Solution: False; value of tan begins from zero and goes on to become more than 1.
(b) sec A = 12/5 for some value of angle A.
Solution: True, value of cos is always more than 1.
(c) cos A is the abbreviation used for the cosecant of angle A.
Solution: False, cos is the abbreviation of cosine.
(d) cot A is the product of cot and A.
Solution: False, cot A means cotangent of angle A.
(e) sin A = 4/3 for some angle A.
Solution: False, value of sin is less than or equal to 1, while this value is more than 1.

Solution of NCERT Exercise 8.2

Exercise 8.2
Question – 1 - Evaluate the following:
Class ten Math-Trigonometry-NCERT Solution1 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution2 Exercise 8.2 

Class ten Math-Trigonometry-NCERT Solution3 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution4 Exercise 8.2 

Class ten Math-Trigonometry-NCERT Solution5 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution6 Exercise 8.2 

Class ten Math-Trigonometry-NCERT Solution7 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution8 Exercise 8.2 

Class ten Math-Trigonometry-NCERT Solution9 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution10 Exercise 8.2
Question – 2 - Choose the correct option and justify your choice:
Class ten Math-Trigonometry-NCERT Solution11 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution12 Exercise 8.2 

Class ten Math-Trigonometry-NCERT Solution13 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution14 Exercise 8.2
(iii) Sin 2A = 2 Sin A is true when A = ?
  1. 30°
  2. 45°
  3. 60°
Solution: Option (a) is correct answer.
Sin 0° = 0 and Sin 2 x 0° = Sin 0° = 0
Sin 30° = ½ But Sin 2 x 30° = Sin 60° is not equal to Sin 30°. The same holds true for sin 45°.
Class ten Math-Trigonometry-NCERT Solution15 Exercise 8.2
Solution:
Class ten Math-Trigonometry-NCERT Solution16 Exercise 8.2
Question – 3 - If Class ten Math-Trigonometry-NCERT Solution17 Exercise 8.2, find A and B.
Solution: We know;
Class ten Math-Trigonometry-NCERT Solution18 Exercise 8.2
Hence, A + B = 60°
Similarly,
Class ten Math-Trigonometry-NCERT Solution19 Exercise 8.2
Hence, A – B = 30°
Adding equation (1) and (2), we get;
Class ten Math-Trigonometry-NCERT Solution20 Exercise 8.2
Question – 4 - State whether the following are true or false. Justify your answer.
(i) Sin (A+B) = Sin A + Sin B
Solution: False; this can be proved by assuming different values for A and B.
(ii) The value of sin θ increases as θ increases.
Solution: True, the value of sin θ increases from zero for zero degree and goes up to 1 for 90 degree.
(iii) The value of cos θ increases are θ increases.
Solution: False, the value of cos decreases from one to zero.
(iv) Sin θ = Cos θ for all values of θ.
Solution: False, values of sin and cos are equal only for 45 degree.
(v) Cot A is not defined for A = 0°
Solution: True, refer to values for trigonometric ratios.

NCERT Solution of

Exercise 8.3

Question – 1 - Evaluate
class ten math trigonometry ncert solution1 of Exercise 8.3
Solution:
class ten math trigonometry ncert solution2 of Exercise 8.3
Solution:
class ten math trigonometry ncert solution3 of Exercise 8.3
Solution:
class ten math trigonometry ncert solution4 of Exercise 8.3
Solution:
class ten math trigonometry ncert solution5 of Exercise 8.3
Question – 2 - Show that:
class ten math trigonometry ncert solution6 of Exercise 8.3
Solution:
class ten math trigonometry ncert solution7 of Exercise 8.3
Solution:
class ten math trigonometry ncert solution8 of Exercise 8.3
Question – 3 - If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution: tan 2A = cot (90° – 2A)
This means;
90° – 2A = A – 18°
Or, 108° – 2A = A
Or, 3A = 108°
Or, A = 108/3 = 36°
Question – 4 - If tan A = cot B, prove that A + B = 90°
Solution: tan A = cot (90°- A) = cot B
class ten math trigonometry ncert solution9 of Exercise 8.3
Or, A + B = 90° proved.
Question – 5 - If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution: sec 4A = cosec (90° - 4A) = cosec (A - 20°)
This means;
90° – 4A = A - 20°
Or, 110°– 4A = A
Or, 5A = 110°
Or, A = 22°
Question – 6 - If A, B and C are interior angles of a triangle ABC, then show thatclass ten math trigonometry ncert solution10 of Exercise 8.3
Solution: Since, A + B + C = 90°
So, B + C = 90°- A
Hence,
class ten math trigonometry ncert solution11 of Exercise 8.3
Question – 7 - Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution: sin 67° + cos 75° can be written as follows:
Sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°

NCERT Solution of

Exercise 8.4 

Question – 1 - Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Solution: sin A can be expressed in terms of cot A, as follows:
class ten math trigonometry ncert solution11 of Exercise 8.4
sec A can be expressed in terms of cot A, as follows:
class ten math trigonometry ncert solution12 of Exercise 8.4
tan A can be expressed in terms of cot A, as follows:
class ten math trigonometry ncert solution13 of Exercise 8.4
Question – 2 - Write all the other trigonometric ratios of A in terms of sec A.
Solution: Sin A can be written in terms of sec A as follows:
class ten math trigonometry ncert solution14 of Exercise 8.4
cos A can be written in terms of sec A as follows:
class ten math trigonometry ncert solution15 of Exercise 8.4
tan A can be written in terms of sec A as follows:
class ten math trigonometry ncert solution16 of Exercise 8.4
cosec A can be written in terms of sec A as follows:
class ten math trigonometry ncert solution17 of Exercise 8.4
cot A can be written in terms of sec A as follows:
class ten math trigonometry ncert solution18 of Exercise 8.4
Question – 3 - Evaluate:
class ten math trigonometry ncert solution9 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution10 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution11 of Exercise 8.4
Question – 4 - Choose the correct option. Justify your choice.
class ten math trigonometry ncert solution12 of Exercise 8.4
Solution: (b) is the correct option. Following is the explanation:
class ten math trigonometry ncert solution13 of Exercise 8.4
Solution: (c) is the correct option. Following is the explanation:
class ten math trigonometry ncert solution14 of Exercise 8.4  

class ten math trigonometry ncert solution15 of Exercise 8.4
Solution: (d) is the correct option. Following is the explanation:
class ten math trigonometry ncert solution16 of Exercise 8.4  

class ten math trigonometry ncert solution17 of Exercise 8.4
Solution: (d) is the correct option. Following is the explanation:
class ten math trigonometry ncert solution18 of Exercise 8.4  

Question – 5 - Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
class ten math trigonometry ncert solution19 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution20 of Exercise 8.4
LHS = RHS proved
class ten math trigonometry ncert solution21 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution22 of Exercise 8.4  

class ten math trigonometry ncert solution23 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution24 of Exercise 8.4  
class ten math trigonometry ncert solution25 of Exercise 8.4  

class ten math trigonometry ncert solution26 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution27 of Exercise 8.4
using the identity class ten math trigonometry ncert solution28 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution29 of Exercise 8.4
After dividing numerator and denominator by sin A, we get
class ten math trigonometry ncert solution30 of Exercise 8.4
After multiplying with [cotA – (1-cosecA)] in numerator and denominator both we get
class ten math trigonometry ncert solution31 of Exercise 8.4  
class ten math trigonometry ncert solution32 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution33 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution34 of Exercise 8.4  

class ten math trigonometry ncert solution35 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution36 of Exercise 8.4  
class ten math trigonometry ncert solution37 of Exercise 8.4
Hence; LHS = RHS proved
class ten math trigonometry ncert solution38 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution39 of Exercise 8.4
Hence; LHS = RHS proved
class ten math trigonometry ncert solution40 of Exercise 8.4
Solution:
class ten math trigonometry ncert solution41 of Exercise 8.4  
class ten math trigonometry ncert solution42 of Exercise 8.4
Hence; LHS = Middle Term = RHS proved

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