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NCERT Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions
Go through the questions and answers from Exercise 8.1, 8.2 and 8.3 from NCERT Class 10 Chapter 8 Introduction to Trigonometry.
EXERCISE 8.1
Q. In ABC, right-angled at B, AB = 24 cm, BC = 7 cm.
Determine : (i) sin A, cos A (ii) sin C, cos C
Sol. In triangle ΔABC with right angle at B,
∠B = 90°
Given: AB = 24 cm and BC = 7 cm
Using Pythagoras theorem,
AC²=AB²+BC²
AC²=(24)²+7²
AC²=(576+49)
AC²=625 cm²
AC=√625=25
(i) Sin A Cos A
Sin A=Opposite side/hypotenuse=BC/AC=7/25
Cos A= Adjacent side/hypotenuse=AB/AC=24/25
(ii) Sin C Cos C
Sin C= AB/AC=24/25
Cos C=BC/AC=7/25
Q. Find tan P – cot R.
Sol. In triangle ΔPQR with right angle at Q,
Given,
PR = 13cm, PQ = 12cm
Using Pythagoras theorem,
PR²=QR²+PQ²
13²=QR²+12²
169=QR²+144
QR²=169-144
QR²=25
QR=√25=5
To find tan P – cot R:
tan (P) = Opposite side /Adjacent side = QR/PQ = 5/12
Cot (R) = Adjacent side/Opposite side = QR/PQ = 5/12
Therefore, tan (P) – cot (R) = 5/12 – 5/12 = 0
Therefore, tan(P) – cot(R) = 0
Q. If sin A = 3/4, calculate cos A and tan A.
Sol. Given a triangle ΔABC with right angle at B,
Given: Sin A = 3/4
Sin A = Opposite side /Hypotenuse= ¾
Let BC be 3k and AC will be 4k where k is a positive real number.
Using Pythagoras theorem,
AC²=AB²+BC²
(4k)²=AB²+(3k)²
16k²-9k²=AB²
AB=7k²
Therefore, AB = √7k
cos A=Adjacent side/Hypotenuse
AB/AC = √7k/4k = √7/4
Therefore, cos (A) = √7/4
tan(A) = Opposite side/Adjacent side
BC/AB = 3k/√7k = 3/√7
Therefore, tan A = 3/√7
Q. Given 15 cot A = 8, find sin A and sec A.
Sol. Given a triangle ΔABC with right angle at B,
Given: 15 cot A = 8
Cot A = 8/15
cot A = Adjacent side/Opposite side = AB/BC = 8/15
Let AB be 8k and BC will be 15k Where, k is a positive real number.
Using Pythagoras theorem,
AC²=AB²+BC²
AC²=(8k)²+(15k)2
AC²=64k²+225k²
AC²=289k²
Therefore, AC = 17k
sin A= Opposite side /Hypotenuse
Sin A = BC/AC = 15k/17k = 15/17
Therefore, sin A = 15/17
Sec (A) = Hypotenuse/Adjacent side
AC/AB = 17k/8k = 17/8
Therefore sec (A) = 17/8
Q. Given sec θ = 13/12, calculate all other trigonometric ratios.
Sol. Given a ΔABC with right angle at B,
sec θ =13/12 = Hypotenuse/Adjacent side = AC/AB
Let AC be 13k and AB will be 12k Where, k is a positive real number.
Using Pythagoras theorem,
AC²=AB²+BC²
(13k)²=(12k)²+BC²
169k²=144k²+BC²
BC²=169k²-144k²
BC²=25k²
Therefore, BC = 5k
Sin θ = Opposite Side/Hypotenuse = BC/AC = 5/13
Cos θ = Adjacent Side/Hypotenuse = AB/AC = 12/13
tan θ = Opposite Side/Adjacent Side = BC/AB = 5/12
Cosec θ = Hypotenuse/Opposite Side = AC/BC = 13/5
cot θ = Adjacent Side/Opposite Side = AB/BC = 12/5
Q. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Sol. Given ΔABC in which CD⊥AB
Cos (A) = cos (B)
AD/AC = BD/BC
AD/BD = AC/BC
Taking a constant value,
AD/BD = AC/BC = k
AD = k BD …(1)
AC = k BC …(2)
Using Pythagoras theorem,
CD²=BC²-BD²...(3)
CD²=AC²-AD²... (4)
AC²-AD²=BC²-BD²
K²(BC²-BD²)=(BC²-BD²) k²-1
AC=BC
∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)
Q. If cot θ = 7/8, evaluate:
(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2 θ
Sol. Given ΔABC in which ∠B = 90° and ∠C = θ
Given, cot θ = BC/AB = ⅞
Let BC = 7k and AB = 8k, where k is a positive real number
Using Pythagoras theorem,
AC²=AB²+BC²
AC²=(8k)²+(7k)²
AC²=64k²+49k²
AC²=113k²
AC=√113k
sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113
cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113
(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)=1-sin²θ/1-cos²θ
=1-(8/√113)²/1-(7/√113)²=49/64
(ii) cot²θ=(7/8)²=49/64
Q. If 3 cot A = 4, check whether (1-tan²A)/(1+tan² A) = cos² A – sin² A or not.
Sol. Given △ABC in which ∠B=90°
cot(A) = AB/BC = 4/3
Let AB = 4k an BC =3k, where k is a positive real number.
Using Pythagoras theorem,
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 = ⅗
cos (A) = AB/AC = 4/5
LHS=1-tan²A/1+tan²A=1-(3/4)²/1+(3/4)²=1-9/16/1+9/16=7/25
RHS=cos² A-sin² A=(4/5)²-(3/5)²=16/25-9/25=7/25
Since, R.H.S. =L.H.S.
Hence, (1-tan²A)/(1+tan² A) = cos² A – sin² A is proved.
Q. 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. Given ΔABC in which ∠B=90°
tan A = BC/AB = 1/√3
Let BC = 1k and AB = √3 k,
Where k is the positive real number of the problem
Using Pythagoras theorem,
AC²=AB²+BC²
AC²=(√3 k)²+(k)²
AC²=3k²+k²
AC²=4k²
AC = 2k
Sin A = BC/AC = ½
Cos A = AB/AC = √3/2
Sin C = AB/AC = √3/2
Cos C = BC/AC = ½
(i) sin A cos C + cos A sin C = (1/2) ×(1/2 )+ √3/2 ×√3/2 = 1/4 + 3/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2 )(1/2) – (1/2) (√3/2 ) = 0
Q. In ∆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. Given that, PR + QR = 25 , PQ = 5
Let PR be x. ∴ QR = 25 - x
Using Pythagoras theorem,
PR² = PQ² + QR²
(25- x)²= 5²+ x²
25² + x² – 50x = 25 + x²
625 + x² -50x -25 – x²= 0
-50x = -600
x= -600/-50
x = 12 = QR
PR = 25- QR
PR = 25-12
PR = 13
(i) sin p = Opposite Side/Hypotenuse = QR/PR = 12/13
(ii) Cos p = Adjacent Side/Hypotenuse = PQ/PR = 5/13
(iii) tan p =Opposite Side/Adjacent side = QR/PQ = 12/5
Q. 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 =12/15 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. (i) False
ΔMNC in which ∠N = 90º,
MN = 3, NC = 4 and MC = 5
Value of tan M = 4/3 which is greater than 1.
Using Pythagoras theorem,
MC²=MN²+NC²
5²=3²+4²
25=9+16
25=25
(ii) True
ΔMNC in which ∠N = 90º,
MC=12k and MB=5k, where k is a positive real number.
By Pythagoras theorem,
MC²=MN²+NC²
(12k)²=(5k)²+NC²
NC²+25k²=144k²
NC²=119k²
(iii) False
Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.
(iv) False
cot M is not the product of cot and M. It is the cotangent of ∠M.
(v) False
sin θ = Opposite/Hypotenuse
∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.
EXERCISE 8.2
Q. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan² 45° + cos² 30° – sin² 60°
(iii) cos 45°/sec 30°+ cosec 30°
(iv) sin 30° + tan 45° – cosec 60°/sec 30° + cos 60° + cot 45°
(v) 5 cos² 60° + 4 sec² 30°- tan² 45°/sin² 30°+ cos² 30°
Sol. (i) sin 30° = ½
cos 30° = √3/2
sin 60° = 3/2
cos 60°= 1/2
sin 60° cos 30° + sin 30° cos 60° = √3/2 ×√3/2 + (1/2) ×(1/2 ) = 3/4+1/4 = 4/4 =1
(ii) sin 60° = √3/2
cos 30° = √3/2
tan 45° = 1
2 tan² 45° + cos² 30° – sin² 60 = 2(1)²+ (√3/2)² -(√3/2)²
2 tan² 45° + cos² 30° – sin² 60 = 2 + 0
2 tan² 45° + cos² 30° – sin² 60 = 2
(iii) cos 45° = 1/√2
sec 30° = 2/√3
cosec 30° = 2
cos 45°/sec 30°+ cosec 30°=1/√2/2/√3+2=1/√2/2+2√3/√3
=1/√2×√3/2+2√3
=1/√2×√3/2(1+√3)=√3/2√2(1+√3)=√3/2√2(√3+1)
= √3/2√2(√3+1)×√3-1/√3-1=3-√3/2√2 (3-1)=3-√3/2√2(2)
3-√3/2√2(2)×√2/√2=3√2-√3√2/8=3√2-√6/8
Therefore, cos 45°/(sec 30°+cosec 30°) = (3√2 – √6)/8
(iv) sin 30° = ½
tan 45° = 1
cosec 60° = 2/√3
sec 30° = 2/√3
cos 60° = ½
cot 45° = 1
sin 30°+tan 45°-cosec 60°/sec 30°+cos 60°+ cot 45°=½+1-2/√3/2/√3+½+1=√3+2 √3-4/2√3/4+√3+2√3/2√3
√3+2 √3-4/4+√3+2√3=3√3-4/3√3+4
3√3-4/3√3+4×3√3-4/3√3-4
27-12√3-12√3+16/27-12√3+12√3+16=27-24√3+16/11=43-24√3/11
sin 30°+tan 45°-cosec 60°/sec 30°+cos 60°+ cot 45°=43-24√3/11
(v) cos 60° = 1/2
sec 30° = 2/√3
tan 45° = 1
sin 30° = ½
cos 30° = √3/2
(5cos²60° + 4sec²30° – tan²45°)/(sin² 30° + cos² 30°)
= 5(1/2)²+4(2/√3)² -1² /(1/2)²+(√3/2)²
= (5/4+16/3-1)/(1/4+3/4)
= (15+64-12)/12/(4/4)
= 67/12
Q. Choose the correct option and justify your choice :
(i) 2 tan 30°/1+tan² 30° = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan² 45°/1+tan² 45° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2 tan30°/1-tan² 30° =
cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Sol. (i) (A) is correct.
tan 30° = 1/√3
2 tan 30°/1+tan² 30° = 2(1/√3)/1+(1/√3)²
= (2/√3)/(1+1/3) = (2/√3)/(4/3)
= 6/4√3 = √3/2 = sin 60°
(ii) (D) is correct.
tan 45° = 1
1-tan² 45°/1+tan² 45° = (1-1²)/(1+1²)
= 0/2 = 0
(iii) (A) is correct.
sin 2A = 2 sin A is true when A = 0°
As sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2 × 0 = 0
or, Apply the sin 2A formula, to find the degree value
sin 2A = 2sin A cos A
⇒2sin A cos A = 2 sin A
⇒ 2cos A = 2 ⇒ cos A = 1
When 0 degree is applied to cos value, i.e., cos 0 =1
Therefore, ⇒ A = 0°
(iv) (C) is correct.
tan 30° = 1/√3
2 tan 30°/1-tan² 30° = 2(1/√3)/1-(1/√3)²
= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°
Q. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.
Sol. tan (A + B) = √3
Since √3 = tan 60°
⇒ tan (A + B) = tan 60°
(A + B) = 60° … (i)
tan (A – B) = 1/√3
Since 1/√3 = tan 30°
⇒ tan (A – B) = tan 30°
(A – B) = 30° … equation (ii)
A + B + A – B = 60° + 30°
2A = 90°
A= 45°
45° + B = 60°
B = 60° – 45°
B = 15°
Therefore A = 45° and B = 15°
Q. 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. (i) False.
Let us take A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1,
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
(iii) False.
cos 0° = 1
cos 30° = √3/2
cos 45° = 1/√2
cos 60° = ½
cos 90° = 0
(iv) False
sin θ = cos θ, when a right triangle has 2 angles of (π/4).
(v) True.
cot A = cos A/sin A
cot 0° = cos 0°/sin 0° = 1/0 = undefined.
EXERCISE 8.3
Q. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Sol. cosec² A– cot² A = 1
cosec² A = 1 + cot² A
1/sin² A = 1 + cot² A
sin² A = 1/(1+cot² A)
sin A = ±1/(√(1+cot² A)
sin² A = 1/ (1+cot² A)
1 – cos² A = 1/ (1+cot² A)
cos² A = 1 – 1/(1+cot² A)
⇒cos² A = (1-1+cot² A)/(1+cot² A)
⇒ 1/sec² A = cot² A/(1+cot² A)
⇒ sec A = ±√ (1+cot² A)/cotA
tan A = sin A/cos A and cot A = cos A/sin A
Since cot function is the inverse of tan function, it is rewritten as tan A = 1/cot A.
Q. Write all the other trigonometric ratios of ∠A in terms of sec A.
Sol. Cos A function in terms of sec A
sec A = 1/cos A
⇒ cos A = 1/sec A
cos² A + sin² A = 1
sin² A = 1 – cos² A
sin² A = 1 – (1/sec² A)
sin² A = (sec² A-1)/sec² A
sin A = ± √(sec² A-1)/sec A
sin A = 1/cosec A
⇒cosec A = 1/sin A
cosec A = ± sec A/√(sec2A-1)
sec² A – tan² A = 1
⇒ tan² A = sec² A – 1
tan A = √(sec² A – 1)
tan A = 1/cot A
⇒ cot A = 1/tan A
cot A = ±1/√(sec² A – 1)
Q. Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan² A = (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) (sec A + tan A) (1 – sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A
(iv) 1+tan² A/1+cot² A = (A) sec² A (B) -1 (C) cot² A (D) tan² A
Sol. (i) (B) is correct.
9 sec² A – 9 tan² A
= 9 (sec² A – tan² A)
= 9×1 = 9 (∵ sec2 A – tan2 A = 1)
Therefore, 9 sec² A – 9 tan² A = 9
(ii) (C) is correct
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
sec θ = 1/ cos θ
cot θ = cos θ/sin θ
cosec θ = 1/sin θ
= (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ)
= (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ
= (cos θ+sin θ)² -1² /(cos θ sin θ)
= (cos² θ + sin² θ + 2cos θ sin θ -1)/(cos θ sin θ)
= (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos² θ + sin² θ = 1)
= (2cos θ sin θ)/(cos θ sin θ) = 2
Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2
(iii) (D) is correct.
Sec A= 1/cos A
Tan A = sin A / cos A
(secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1+sin A/cos A) (1 – sinA)
= (1 – sin² A)/cos A
= cos² A/cos A = cos A
Therefore, (secA + tanA) (1 – sinA) = cos A
(iv) (D) is correct.
tan² A =1/cot² A
1+tan² A/1+cot² A
= (1+1/cot² A)/1+cot² A
= (cot² A+1/cot² A)×(1/1+cot² A)
= 1/cot² A = tan² A
So, 1+tan² A/1+cot² A = tan² A
Q. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ)²= (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin² A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) ( cos A–sin A+1)/( cos A +sin A–1) = cosec A + cot A, using the identity cosec² A = 1+cot² A.
(vi) √1+ sin A/1-sin A=sec A+ tan A
(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ
(viii) (sin A + cosec A)²+ (cos A + sec A)² = 7+tan² A+cot² A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan² A/1+cot² A) = (1-tan A/1-cot A)² =tan² A
Sol. (i) L.H.S. = (cosec θ – cot θ)²
Since (a-b)² = a² + b² – 2ab
Here a = cosec θ and b = cot θ
= (cosec²θ + cot²θ – 2cosec θ cot θ)
= (1/sin²θ + cos²θ/sin²θ – 2cos θ/sin²θ)
= (1 + cos²θ – 2cos θ)/(1 – cos²θ)
= (1-cos θ)² /(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
Therefore, (cosec θ – cot θ)²= (1-cos θ)/(1+cos θ)
(ii) (cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
L.H.S. = (cos A/(1+sin A)) + ((1+sin A)/cos A)
= [cos²A + (1+sin A)² ]/(1+sin A)cos A
= (cos²A + sin²A + 1 + 2sin A)/(1+sin A) cos A
= (1 + 1 + 2sin A)/(1+sin A) cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.
L.H.S. = R.H.S.
(cos A/(1+sin A)) + ((1+sin A)/cos A) = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
tan θ =sin θ/cos θ
cot θ = cos θ/sin θ
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
= sin²θ/[cos θ(sin θ-cos θ)] + cos²θ/[sin θ(cos θ-sin θ)]
= sin²θ/[cos θ(sin θ-cos θ)] – cos²θ/[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [(sin²θ/cos θ) – (cos²θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin³θ – cos³θ)/sin θ cos θ]
= [(sin θ-cos θ)(sin²θ+cos²θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.
Therefore, L.H.S. = R.H.S.
(iv) (1 + sec A)/sec A = sin²A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
Therefore, (1 + sec A)/sec A = cos A + 1
R.H.S. = sin²A/(1-cos A)
sin²A = (1 – cos²A),
= (1 – cos²A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
Therefore, sin²A/(1-cos A)= cos A + 1
L.H.S. = R.H.S.
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A, using the identity cosec²A = 1+cot²A
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
= (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 + cot²A + cosec A)/(cot A+ 1 – cosec A) (using cosec²A – cot²A = 1
= [(cot A + cosec A) – (cosec²A – cot²A)]/(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
Therefore, (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A
(vi) √1+ sin A/1-sin A=sec A+ tan A
LHS=√1+ sin A/1-sin A
= √1/cos A+ sin A/cos A/1/cos A-sin A/cos A
1/cos A = sec A and sin A/ cos A = tan A
= √(sec A+ tan A)/(sec A-tan A)
=√sec A+tan A/Sec A-tan A×√sec A+tan A/Sec +tan A
(sec A+ tan A)²/sec² A-tan² A
= (sec A + tan A)/1
= sec A + tan A = R.H.S
(vii) (sin θ – 2sin³θ)/(2cos³θ-cos θ) = tan θ
L.H.S. =(sin θ – 2sin³θ)/(2cos³θ-cos θ)
= [sin θ(1 – 2sin²θ)]/[cos θ(2cos²θ- 1)]
sin²θ = 1-cos²θ
= sin θ[1 – 2(1-cos²θ)]/[cos θ(2cos²θ -1)]
= [sin θ(2cos²θ -1)]/[cos θ(2cos²θ -1)]
= tan θ = R.H.S.
(viii) (sin A + cosec A)²+ (cos A + sec A)² = 7+tan²A+cot²A
L.H.S. = (sin A + cosec A)²+ (cos A + sec A)²
(a+b)² =a²+ b² +2ab
= (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)
= (sin²A + cos²A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan²A + 1 + cot²A
= 1 + 2 + 2 + 2 + tan²A + cot²A
= 7+tan²A+cot²A = R.H.S.
Therefore, (sin A + cosec A)²+ (cos A + sec A)² = 7+tan²A+cot²A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin²A)/sin A][(1-cos²A)/cos A]
= (cos²A/sin A)×(sin²A/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin²A+cos²A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.
(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
(x) (1+tan²A/1+cot²A) = (1-tan A/1-cot A)² =tan²A
L.H.S. = (1+tan²A/1+cot²A)
= (1+tan²A/1+1/tan²A)
= 1+tan²A/[(1+tan²A)/tan²A]
= tan²A
(1+tan²A/1+cot²A) = tan²A
Similarly,
(1-tan A/1-cot A)² =tan²A
To download these solutions in pdf, refer to the below link:
| Download NCERT Class 10 Maths Chapter 8 Introduction to Trigonometry Solutions PDF |
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