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Detailed Chapter 8 Introduction to Trigonometry NCERT Solutions for Class 10 Mathematics
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Class 10 Mathematics Chapter 8 Introduction to Trigonometry NCERT Solutions PDF
Exercise 8.1
Q.1) In Δ ABC, right-angled at 𝐵, 𝐴𝐵 = 24 𝑐𝑚, 𝐵𝐶 = 7 𝑐𝑚.
Determine : (i) sin 𝐴 , cos 𝐴 (ii) sin 𝐶 , cos 𝐶
Sol.1) In 𝛥 𝐴𝐵𝐶, ∠𝐵 = 90°
By Applying Pythagoras theorem , we get
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
= (24)2 + 72 = (576 + 49)𝑐𝑚2 = 625 𝑐𝑚2
⇒ 𝐴𝐶 = 25
(i) 𝑠𝑖𝑛 𝐴 = 𝐵𝐶/𝐴𝐶 = 7/25
𝑐𝑜𝑠 𝐴 = 𝐴𝐵/𝐴𝐶 = 24/25
(ii) 𝑠𝑖𝑛 𝐶 = 𝐴𝐵/𝐴𝐶 = 24/25
𝑐𝑜𝑠 𝐶 = 𝐵𝐶/𝐴𝐶 = 7/25
Q.2) In Fig. , find 𝑡𝑎𝑛 𝑃 – 𝑐𝑜𝑡 𝑅.
Sol.2) By Applying Pythagoras theorem in 𝛥𝑃𝑄𝑅 , we get
𝑃𝑅2 = 𝑃𝑄2 + 𝑄𝑅2
= (13)2 = (12)2 + 𝑄𝑅2
= 169 = 144 + 𝑄𝑅2
⇒ 𝑄𝑅2 = 25
⇒ 𝑄𝑅 = 5 𝑐𝑚
Now, 𝑡𝑎𝑛 𝑃 = 𝑄𝑅/𝑃𝑄 = 5/12
𝑐𝑜𝑡 𝑅 = 𝑄𝑅/𝑃𝑄 = 5/12
A/q
𝑡𝑎𝑛 𝑃 – 𝑐𝑜𝑡 𝑅 = 5/12 − 5/12 = 0
Q.3) If 𝑠𝑖𝑛 𝐴 = 3/4 , calculate 𝑐𝑜𝑠 𝐴 and 𝑡𝑎𝑛 𝐴.
Sol.3) Let 𝛥𝐴𝐵𝐶 be a right-angled triangle, right-angled at B.
We know that, 𝑠𝑖𝑛 𝐴 = 𝐵𝐶/𝐴𝐶 = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
(4𝑘)2 = 𝐴𝐵2 + (3𝑘)2
16𝑘2 − 9𝑘2 = 𝐴𝐵2
𝐴𝐵2 = 7𝑘2
𝐴𝐵 = √7𝑘
Q.4) Given 15 𝑐𝑜𝑡 𝐴 = 8, find 𝑠𝑖𝑛 𝐴 and 𝑠𝑒𝑐 𝐴.
Sol.4) Let ΔABC be a right-angled triangle, right-angled at B.
We know that 𝑐𝑜𝑡 𝐴 = 𝐴𝐵/𝐵𝐶 = 8/15
(Given)
Let 𝐴𝐵 be 8𝑘 and BC will be 15𝑘 where 𝑘 is a positive real number.
By Pythagoras theorem we get,
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝐴𝐶2 = (8𝑘)2 + (15𝑘)2
𝐴𝐶2 = 64𝑘2 + 225𝑘2
𝐴𝐶2 = 289𝑘2
𝐴𝐶 = 17 𝑘
𝑠𝑖𝑛 𝐴 = 𝐵𝐶/𝐴𝐶 = 15𝑘/17𝑘 = 15/17
𝑠𝑒𝑐 𝐴 = 𝐴𝐶/𝐴𝐵 = 17𝑘/8𝑘 = 17/8
Q.5) Given 𝑠𝑒𝑐 𝜃 = 13/12, calculate all other trigonometric ratios.
Sol.5) Let 𝛥𝐴𝐵𝐶 be a right-angled triangle, right-angled at B.
We know that 𝑠𝑒𝑐 𝜃 = 𝑂𝑃/𝑂𝑀 = 13/12
(Given)
Let 𝑂𝑃 be 13𝑘 and OM will be 12𝑘 where 𝑘 is a positive real number.
By Pythagoras theorem we get,
𝑂𝑃2 = 𝑂𝑀2 + 𝑀𝑃2
(13𝑘)2 = (12𝑘)2 + 𝑀𝑃2
169𝑘2 − 144𝑘2 = 𝑀𝑃2
𝑀𝑃2 = 25𝑘2
𝑀𝑃 = 5
Q.6) If ∠𝐴 and ∠𝐵 are acute angles such that 𝑐𝑜𝑠 𝐴 = 𝑐𝑜𝑠 𝐵, then show that ∠𝐴 = ∠𝐵.
Sol.6) Let 𝛥𝐴𝐵𝐶 in which 𝐶𝐷 ⊥ 𝐴𝐵.
A/q,
𝑐𝑜𝑠 𝐴 = 𝑐𝑜𝑠 𝐵
⇒ 𝐴𝐷/𝐴𝐶 = 𝐵𝐷/𝐵𝐶
⇒ 𝐴𝐷/𝐵𝐷 = 𝐴𝐶/𝐵𝐶
Let 𝐴𝐷/𝐵𝐷 = 𝐴𝐶/𝐵𝐶 = 𝑘
⇒ 𝐴𝐷 = 𝑘𝐵𝐷 .... (i)
⇒ 𝐴𝐶 = 𝑘𝐵𝐶 .... (ii)
By applying Pythagoras theorem in 𝛥𝐶𝐴𝐷 and 𝛥𝐶𝐵𝐷 we get,
𝐶𝐷2 = 𝐴𝐶2 − 𝐴𝐷2 …. (iii)
and also
𝐶𝐷2 = 𝐵𝐶2 − 𝐵𝐷2 …. (iv)
From equations (iii) and (iv) we get,
𝐴𝐶2 − 𝐴𝐷2 = 𝐵𝐶2 − 𝐵𝐷2
⇒ (𝑘𝐵𝐶)2 − (𝑘 𝐵𝐷)2 = 𝐵𝐶2 − 𝐵𝐷2
⇒ 𝑘2 (𝐵𝐶2 − 𝐵𝐷2) = 𝐵𝐶2 − 𝐵𝐷2
⇒ 𝑘2 = 1 ⇒ 𝑘 = 1
Putting this value in equation (ii), we obtain
𝐴𝐶 = 𝐵𝐶
⇒ ∠𝐴 = ∠𝐵 (Angles opposite to equal sides of a triangle are equal-isosceles triangle)
Q.7) If 𝑐𝑜𝑡 𝜃 = 7/8,
evaluate : (i) (1+𝑠𝑖𝑛 𝜃 )(1−𝑠𝑖𝑛 𝜃)/(1+𝑐𝑜𝑠 𝜃)(1−𝑐𝑜𝑠 𝜃)
(ii) cot2 𝜃
Sol.7) Let 𝛥𝐴𝐵𝐶 in which ∠𝐵 = 90𝑜 and ∠𝐶 = 𝜃
A/q,
𝑐𝑜𝑡 𝜃 = 𝐵𝐶/𝐴𝐵 = 7/8
Let 𝐵𝐶 = 7𝑘 and 𝐴𝐵 = 8𝑘, where k is a positive real number.
By Pythagoras theorem in 𝛥𝐴𝐵𝐶 we get.
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝐴𝐶2 = (8𝑘)2 + (7𝑘)2
𝐴𝐶2 = 64𝑘2 + 49𝑘2
𝐴𝐶2 = 113𝑘2
𝐴𝐶 = √113 𝑘
Q.8) If 3 𝑐𝑜𝑡 𝐴 = 4/3, check whether 1−𝑡𝑎𝑛 2𝐴/1+𝑡𝑎𝑛 2𝐴 = cos2 𝐴 – sin2 𝐴 or not
Sol.8) Let 𝛥𝐴𝐵𝐶 in which ∠𝐵 = 90°,
A/q,
𝑐𝑜𝑡 𝐴 = 𝐴𝐵/𝐵𝐶 = 4/3
Let 𝐴𝐵 = 4𝑘 and 𝐵𝐶 = 3𝑘, where 𝑘 is a positive real number.
By Pythagoras theorem in 𝛥𝐴𝐵𝐶 we get.
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝐴𝐶2 = (4𝑘)2 + (3𝑘)2
𝐴𝐶2 = 16𝑘2 + 9𝑘2
𝐴𝐶2 = 25𝑘2
𝐴𝐶 = 5𝑘
𝑡𝑎𝑛 𝐴 = 𝐵𝐶/𝐴𝐵 = 3/4
𝑠𝑖𝑛 𝐴 = 𝐵𝐶/𝐴𝐶 = 3/5
Q.9) In triangle ABC, right-angled at B, if 𝑡𝑎𝑛 𝐴 = 1/√3 find the value of: (i) 𝑠𝑖𝑛 𝐴 𝑐𝑜𝑠 𝐶 + 𝑐𝑜𝑠 𝐴 𝑠𝑖𝑛 𝐶 (ii) 𝑐𝑜𝑠 𝐴 𝑐𝑜𝑠 𝐶 – 𝑠𝑖𝑛 𝐴 𝑠𝑖𝑛 𝐶
Sol.9) Let 𝛥𝐴𝐵𝐶 in which ∠𝐵 = 90°,
A/q,
𝑡𝑎𝑛 𝐴 = 𝐵𝐶/𝐴𝐵 = 1/√3
Let 𝐴𝐵 = √3 𝑘 and 𝐵𝐶 = 𝑘, where 𝑘 is a positive real number.
By Pythagoras theorem in 𝛥𝐴𝐵𝐶 we get.
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
𝐴𝐶2 = (√3𝑘)2 + (𝑘)2
𝐴𝐶2 = 3𝑘2 + 𝑘2
𝐴𝐶2 = 4𝑘2
𝐴𝐶 = 2𝑘
Q.10) In 𝛥 𝑃𝑄𝑅, right-angled at 𝑄, 𝑃𝑅 + 𝑄𝑅 = 25 𝑐𝑚 and 𝑃𝑄 = 5 𝑐𝑚. Determine the values of 𝑠𝑖𝑛 𝑃, 𝑐𝑜𝑠 𝑃 and 𝑡𝑎𝑛 𝑃.
Sol.10) Given that, 𝑃𝑅 + 𝑄𝑅 = 25 , 𝑃𝑄 = 5
Let 𝑃𝑅 be 𝑥.
∴ 𝑄𝑅 = 25 − 𝑥
By Pythagoras theorem ,
𝑃𝑅2 = 𝑃𝑄2 + 𝑄𝑅2
𝑥2 = (5)2 + (25 − 𝑥)2
𝑥2 = 25 + 625 + 𝑥2 − 50𝑥
50𝑥 = 650
𝑥 = 13
∴ 𝑃𝑅 = 13 𝑐𝑚
𝑄𝑅 = (25 − 13) 𝑐𝑚 = 12 𝑐𝑚
𝑠𝑖𝑛 𝑃 = 𝑄𝑅/𝑃𝑅 = 12/13
𝑐𝑜𝑠 𝑃 = 𝑃𝑄/𝑃𝑅 = 5/13
𝑡𝑎𝑛 𝑃 = 𝑄𝑅/𝑃𝑄 = 12/5
Q.11) State whether the following are true or false. Justify your answer.
(i) The value of 𝑡𝑎𝑛 𝐴 is always less than 1.
(ii) 𝑠𝑒𝑐 𝐴 = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) 𝑠𝑖𝑛 𝜃 = 4/3 for some angle 𝜃.
Sol.11) (i) False.
In 𝛥𝐴𝐵𝐶 in which ∠𝐵 = 90°, 𝐴𝐵 = 3, 𝐵𝐶 = 4 and 𝐴𝐶 = 5
Value of 𝑡𝑎𝑛 𝐴 = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
52 = 32 + 42
25 = 9 + 16
25 = 25
(ii) True.
Let a 𝛥𝐴𝐵𝐶 in which ∠𝐵 = 90°, 𝐴𝐶 be 12𝑘 and 𝐴𝐵 be 5𝑘, where 𝑘 is a positive real number. By Pythagoras theorem we get,
𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2
(12𝑘)2 = (5𝑘)2 + 𝐵𝐶2
𝐵𝐶2 + 25𝑘2 = 144𝑘2
𝐵𝐶2 = 119𝑘2
Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) False
Abbreviation used for cosecant of angle A is 𝑐𝑜𝑠𝑒𝑐 𝐴. 𝑐𝑜𝑠 𝐴 is the abbreviation used for cosine of angle A.
(iv) False.
cot A is not the product of cot and A. It is the cotangent of ∠𝐴.
(v) False.
𝑠𝑖𝑛 𝜃 = 𝐻𝑒𝑖𝑔ℎ𝑡/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
We know that in a right angled triangle, Hypotenuse is the longest side.
∴ 𝑠𝑖𝑛 𝜃 will always less than 1 and it can never be 4/3 for any value of 𝜃.
Exercise 8.2
Q.1) Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan 245° + cos 230° – sin260°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos 260° + 4sec 230° - tan 245°)/(sin230° + cos 230°)
Sol.1) (i) 𝑠𝑖𝑛 60° 𝑐𝑜𝑠 30° + 𝑠𝑖𝑛 30° 𝑐𝑜𝑠 60°
= 67/12
Q.2) Choose the correct option and justify your choice :
(i) 2𝑡𝑎𝑛 30°/1+𝑡𝑎𝑛 230° =
(A) 𝑠𝑖𝑛 60° (B) 𝑐𝑜𝑠 60° (C) 𝑡𝑎𝑛 60° (D) 𝑠𝑖𝑛 30°
(ii) 1−𝑡𝑎𝑛 245°/1+𝑡𝑎𝑛2 45° =
(A) 𝑡𝑎𝑛 90° (B) 1 (C) 𝑠𝑖𝑛 45° (D) 0
(iii) sin2 𝐴 = 2 𝑠𝑖𝑛 𝐴 is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2𝑡𝑎𝑛30°/1−𝑡𝑎𝑛 230° =
(A) 𝑐𝑜𝑠 60° (B) 𝑠𝑖𝑛 60° (C) 𝑡𝑎𝑛 60° (D) 𝑠𝑖𝑛 30°
Sol.2) (i) (A) is correct.
(iii) (A) is correct.
sin2 𝐴 = 2 𝑠𝑖𝑛 𝐴 is true when A =
= As sin2 𝐴 = 𝑠𝑖𝑛 0° = 0
2 𝑠𝑖𝑛 𝐴 = 2𝑠𝑖𝑛 0° = 2 × 0 = 0
or, sin2 𝐴 = 2𝑠𝑖𝑛 𝐴 𝑐𝑜𝑠 𝐴
⇒ 2𝑠𝑖𝑛 𝐴 𝑐𝑜𝑠 𝐴 = 2 𝑠𝑖𝑛 𝐴
⇒ 2𝑐𝑜𝑠 𝐴 = 2
⇒ 𝑐𝑜𝑠 𝐴 = 1
⇒ 𝐴 = 0°
(iv) (C) is correct.
Q.3) If 𝑡𝑎𝑛 (𝐴 + 𝐵) = √3 and 𝑡𝑎𝑛 (𝐴 – 𝐵) = 1/√3 ; 0° < 𝐴 + 𝐵 ≤ 90°; 𝐴 > 𝐵, find A and B.
Sol.3) 𝑡𝑎𝑛 (𝐴 + 𝐵) = √3
⇒ 𝑡𝑎𝑛 (𝐴 + 𝐵) = 𝑡𝑎𝑛 60°
⇒ (𝐴 + 𝐵) = 60° ... (i)
𝑡𝑎𝑛 (𝐴 – 𝐵) = 1/√3
⇒ 𝑡𝑎𝑛 (𝐴 − 𝐵) = 𝑡𝑎𝑛 30°
⇒ (𝐴 − 𝐵) = 30° ... (ii)
Adding (i) and (ii), we get
𝐴 + 𝐵 + 𝐴 − 𝐵 = 60° + 30°
2𝐴 = 90°
𝐴 = 45°
Putting the value of A in equation (i)
45° + 𝐵 = 60°
⇒ 𝐵 = 60° − 45°
⇒ 𝐵 = 15°
Thus, 𝐴 = 45° and 𝐵 = 15°
Q.5) State whether the following are true or false. Justify your answer.
(i) 𝑠𝑖𝑛 (𝐴 + 𝐵) = 𝑠𝑖𝑛 𝐴 + 𝑠𝑖𝑛 𝐵.
(ii) The value of 𝑠𝑖𝑛 𝜃 increases as 𝜃 increases.
(iii) The value of 𝑐𝑜𝑠 𝜃 increases as 𝜃 increases.
(iv) 𝑠𝑖𝑛 𝜃 = 𝑐𝑜𝑠 𝜃 for all values of 𝜃.
(v) 𝑐𝑜𝑡 𝐴 is not defined for 𝐴 = 0°.
Sol.5) (i) False.
Let 𝐴 = 30° 𝑎𝑛𝑑 𝐵 = 60°, then
𝑠𝑖𝑛 (𝐴 + 𝐵) = 𝑠𝑖𝑛 (30° + 60°) = 𝑠𝑖𝑛 90° = 1 and,
𝑠𝑖𝑛 𝐴 + 𝑠𝑖𝑛 𝐵 = 𝑠𝑖𝑛 30° + 𝑠𝑖𝑛 60° = 1/2 + √3/2 = 1 + √3/2
(ii) True.
𝑠𝑖𝑛 0° = 0
𝑠𝑖𝑛 30° = 1/2
𝑠𝑖𝑛 45° = 1/√2
𝑠𝑖𝑛 60° = √3/2
𝑠𝑖𝑛 90° = 1
Thus the value of sin θ increases as θ increases.
(iii) False.
𝑐𝑜𝑠 0° = 1
𝑐𝑜𝑠 30° = √3/2
𝑐𝑜𝑠 45° = 1/√2
𝑐𝑜𝑠 60° = 1/2
𝑐𝑜𝑠 90° = 0
Thus the value of cos θ decreases as θ increases.
(iv) True.
𝑐𝑜𝑡 𝐴 = 𝑐𝑜𝑠 𝐴/sin 𝐴
𝑐𝑜𝑡 0° = 𝑐𝑜𝑠 0°/𝑠𝑖𝑛 0° = 1/0 = undefined.
Exercise 8.3
Q.1) Evaluate : (i) 𝑠𝑖𝑛 18°/𝑐𝑜𝑠 72°
(ii) 𝑡𝑎𝑛 26°/𝑐𝑜𝑡 64°
(iii) 𝑐𝑜𝑠 48° – 𝑠𝑖𝑛 42°
(iv) 𝑐𝑜𝑠𝑒𝑐 31° – 𝑠𝑒𝑐 59°
Sol.1)
(iii) 𝑐𝑜𝑠 48° − 𝑠𝑖𝑛 42°
= 𝑐𝑜𝑠 (90° − 42°) − 𝑠𝑖𝑛 42°
= 𝑠𝑖𝑛 42° − 𝑠𝑖𝑛 42° = 0
(iv) 𝑐𝑜𝑠𝑒𝑐 31° − 𝑠𝑒𝑐 59°
= 𝑐𝑜𝑠𝑒𝑐 (90° − 59°) − 𝑠𝑒𝑐 59°
= 𝑠𝑒𝑐 59° − 𝑠𝑒𝑐 59° = 0
Q.2) Show that :
(i) 𝑡𝑎𝑛 48° 𝑡𝑎𝑛 23° 𝑡𝑎𝑛 42° 𝑡𝑎𝑛 67° = 1
(ii) 𝑐𝑜𝑠 38° 𝑐𝑜𝑠 52° – 𝑠𝑖𝑛 38° 𝑠𝑖𝑛 52° = 0
Sol.2) (i) 𝑡𝑎𝑛 48° 𝑡𝑎𝑛 23° 𝑡𝑎𝑛 42° 𝑡𝑎𝑛 67°
= 𝑡𝑎𝑛 (90° − 42°) 𝑡𝑎𝑛 (90° − 67°) 𝑡𝑎𝑛 42° 𝑡𝑎𝑛 67°
= 𝑐𝑜𝑡 42° 𝑐𝑜𝑡 67° 𝑡𝑎𝑛 42° 𝑡𝑎𝑛 67°
= (𝑐𝑜𝑡 42° 𝑡𝑎𝑛 42°) (𝑐𝑜𝑡 67° 𝑡𝑎𝑛 67°) = 1 × 1 = 1
(ii) 𝑐𝑜𝑠 38° 𝑐𝑜𝑠 52° − 𝑠𝑖𝑛 38° 𝑠𝑖𝑛 52°
= 𝑐𝑜𝑠 (90° − 52°) 𝑐𝑜𝑠 (90° − 38°) − 𝑠𝑖𝑛 38° 𝑠𝑖𝑛 52°
= 𝑠𝑖𝑛 52° 𝑠𝑖𝑛 38° − 𝑠𝑖𝑛 38° 𝑠𝑖𝑛 52° = 0
Q.3) If 𝑡𝑎𝑛 2𝐴 = 𝑐𝑜𝑡 (𝐴 – 18°), where 2𝐴 is an acute angle, find the value of A.
Sol.3) A/q,
𝑡𝑎𝑛 2𝐴 = 𝑐𝑜𝑡 (𝐴 − 18°)
⇒ 𝑐𝑜𝑡 (90° − 2𝐴) = 𝑐𝑜𝑡 (𝐴 − 18°)
Equating angles,
⇒ 90° − 2𝐴 = 𝐴 − 18°
⇒ 108° = 3𝐴
⇒ 𝐴 = 36°
Q.4) If 𝑡𝑎𝑛 𝐴 = 𝑐𝑜𝑡 𝐵, prove that 𝐴 + 𝐵 = 90°.
Sol.4) A/q,
𝑡𝑎𝑛 2𝐴 = 𝑐𝑜𝑡 (𝐴 − 18°)
⇒ 𝑐𝑜𝑡 (90° − 2𝐴) = 𝑐𝑜𝑡 (𝐴 − 18°)
Equating angles,
⇒ 90° − 2𝐴 = 𝐴 − 18°
⇒ 108° = 3𝐴
⇒ 𝐴 = 36°
Q.5) If sec 4A = cosec (A – 20°), where 4𝐴 is an acute angle, find the value of 𝐴.
Sol.5) A/q,
𝑠𝑒𝑐 4𝐴 = 𝑐𝑜𝑠𝑒𝑐 (𝐴 − 20°)
⇒ 𝑐𝑜𝑠𝑒𝑐 (90° − 4𝐴) = 𝑐𝑜𝑠𝑒𝑐 (𝐴 − 20°)
Equating angles, 90° − 4𝐴 = 𝐴 − 20°
⇒ 110° = 5𝐴
⇒ 𝐴 = 22°
Q.6) If A, B and C are interior angles of a triangle ABC, then show that 𝑠𝑖𝑛 (𝐵 + 𝐶/2) = 𝑐𝑜𝑠 𝐴/2
Sol.6) In a triangle, sum of all the interior angles
𝐴 + 𝐵 + 𝐶 = 180°
⇒ 𝐵 + 𝐶 = 180° − 𝐴
Q.7) Express 𝑠𝑖𝑛 67° + 𝑐𝑜𝑠 75° 𝑖n terms of trigonometric ratios of angles between 0° and 45°.
Sol.7) 𝑠𝑖𝑛 67° + 𝑐𝑜𝑠 75°
= 𝑠𝑖𝑛 (90° − 23°) + 𝑐𝑜𝑠 (90° − 15°)
= 𝑐𝑜𝑠 23° + 𝑠𝑖𝑛 15°
Exercise 8.4
Q.1) Express the trigonometric ratios 𝑠𝑖𝑛 𝐴, 𝑠𝑒𝑐 𝐴 and 𝑡𝑎𝑛 𝐴 in terms of 𝑐𝑜𝑡 𝐴.
Sol.1) 𝑐𝑜𝑠𝑒𝑐2𝐴 − cot2 𝐴 = 1
⇒ 𝑐𝑜𝑠𝑒𝑐2 𝐴 = 1 + cot2 𝐴
Q.2) Write all the other trigonometric ratios of ∠𝐴 in terms of 𝑠𝑒𝑐 𝐴
Sol.2) We know that,
𝑠𝑒𝑐 𝐴 = 1/cos 𝐴
⇒ 𝑐𝑜𝑠 𝐴 = 1/sec 𝐴
also, cos2 𝐴 + sin2 𝐴 = 1
⇒ sin2 𝐴 = 1 − 𝑐𝑜𝑠2𝐴
(ii) 𝑠𝑖𝑛 25° 𝑐𝑜𝑠 65° + 𝑐𝑜𝑠 25° 𝑠𝑖𝑛 65°
= 𝑠𝑖𝑛(90° − 25°) 𝑐𝑜𝑠 65° + 𝑐𝑜𝑠(90° − 65°) 𝑠𝑖𝑛 65°
= 𝑐𝑜𝑠 65° 𝑐𝑜𝑠 65° + 𝑠𝑖𝑛 65° 𝑠𝑖𝑛 65°
= 𝑐𝑜𝑠 265° + 𝑠𝑖n265° = 1
Q.4) Choose the correct option. Justify your choice.
(i) 9 sec2 𝐴 − 9 𝑡𝑎𝑛 2𝐴 =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + 𝑡𝑎𝑛 𝜃 + 𝑠𝑒𝑐 𝜃) (1 + 𝑐𝑜𝑡 𝜃 − 𝑐𝑜𝑠𝑒𝑐 𝜃)
(A) 0 (B) 1 (C) 2 (D) - 1
(iii) (𝑠𝑒𝑐𝐴 + 𝑡𝑎𝑛𝐴) (1 − 𝑠𝑖𝑛𝐴) =
(A) sec 𝐴 (B) sin 𝐴 (C) 𝑐𝑜𝑠𝑒𝑐 𝐴 (D) 𝑐𝑜𝑠 𝐴
(iv) 1+tan2 𝐴/1+cot2 𝐴 =
(A) 𝑠𝑒c2𝐴 (B) -1 (C) 𝑐𝑜t2𝐴 (D) 𝑡𝑎n2𝐴
Sol.4) (i) (B) is correct.
9 sec2 𝐴 − 9 tan2 𝐴
= 9 (sec2 𝐴 − tan2 𝐴)
= 9 × 1 = 9 (∵ sec2 𝐴 − tan2 𝐴 = 1)
(ii) (C) is correct
(1 + 𝑡𝑎𝑛 𝜃 + 𝑠𝑒𝑐 𝜃) (1 + 𝑐𝑜𝑡 𝜃 − 𝑐𝑜𝑠𝑒𝑐 𝜃)
Q.5) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
L.H.S. = (𝑐𝑜𝑠 𝐴– 𝑠𝑖𝑛 𝐴 + 1)/(𝑐𝑜𝑠 𝐴 + 𝑠𝑖𝑛 𝐴– 1)
Dividing Numerator and Denominator by sin A,
= (𝑐𝑜𝑠 𝐴– 𝑠𝑖𝑛 𝐴 + 1)/𝑠𝑖𝑛 𝐴/(𝑐𝑜𝑠 𝐴 + 𝑠𝑖𝑛 𝐴– 1)/𝑠𝑖𝑛 𝐴
= (𝑐𝑜𝑡 𝐴 − 1 + 𝑐𝑜𝑠𝑒𝑐 𝐴)/(𝑐𝑜𝑡 𝐴 + 1 – 𝑐𝑜𝑠𝑒𝑐 𝐴)
= (𝑐𝑜𝑡 𝐴 − 𝑐𝑜𝑠𝑒𝑐2𝐴 + cot2 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴)/(𝑐𝑜𝑡 𝐴 + 1 – 𝑐𝑜𝑠𝑒𝑐 𝐴)
(using 𝑐𝑜𝑠𝑒𝑐2𝐴 − cot2 𝐴 = 1)
= [(𝑐𝑜𝑡 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − (𝑐𝑜𝑠𝑒𝑐2𝐴 − cot2 𝐴)]/(𝑐𝑜𝑡 𝐴 + 1 – 𝑐𝑜𝑠𝑒𝑐 𝐴)
= [(𝑐𝑜𝑡 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴) − (𝑐𝑜𝑠𝑒𝑐 𝐴 + 𝑐𝑜𝑡 𝐴)(𝑐𝑜𝑠𝑒𝑐 𝐴 − 𝑐𝑜𝑡 𝐴)]/(1 – 𝑐𝑜𝑠𝑒𝑐 𝐴 + 𝑐𝑜𝑡 𝐴)
= (𝑐𝑜𝑡 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴)(1 – 𝑐𝑜𝑠𝑒𝑐 𝐴 + 𝑐𝑜𝑡 𝐴)/(1 – 𝑐𝑜𝑠𝑒𝑐 𝐴 + 𝑐𝑜𝑡 𝐴)
= 𝑐𝑜𝑡 𝐴 + 𝑐𝑜𝑠𝑒𝑐 𝐴 = R.H.S.
| NCERT Solutions Class 10 Mathematics Chapter 8 Introduction to Trigonometry |
Free study material for Chapter 8 Introduction to Trigonometry
NCERT Solutions Class 10 Mathematics Chapter 8 Introduction to Trigonometry
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