NCERT Solutions Class 10 Maths Chapter 8 Introduction To Trigonometry
NCERT Solutions Class 10 Maths Chapter 8 Introduction To Trigonometry: National Council of Educational Research and Training Class 10 Maths Chapter 8 Solutions – Introduction To Trigonometry. NCERT Solutions Class 10 Maths Chapter 8 PDF Download.
NCERT Solutions Class 10 Maths Chapter 8: Overview
Board | NCERT |
Class | 10 |
Subject | Maths |
Chapter | 8 |
Chapter Name | Introduction To Trigonometry |
Topic | Exercise Solutions |
NCERT Solutions
Class 10 Maths
Chapter 8 – Introduction To Trigonometry
EXERCISE 8.1
(1) In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) Sin A, cos A
(ii) Sin C, cos C
Solution:
In ∆ABC, given ∠B = 90°
AB = 24 cm, BC = 7 cm
We know that,
AC2 = AB2 + BC2 [Pythagoras Theorem]
AC2 = (24)2 + (7)2
AC2 = 576 + 49
AC2 = 625 cm2
AC = √625 = 2
Hence AC = 25cm
(i) Sin A, Cos A
Sin A = Opposite side/hypotenuse = BC/AC = 7/25
Cos A = Adjacent side/Hypotenuse = AB/AC = 24/25
(ii) Similarity, we have to find = Sin C, Cos C
Cos C = Adjacent side/Hypotenuse = BC/AC = 7/25
Sin C = Opposite side/Hypotenuse = AB/AC = 24/25
(2) In Fig. 8.13, find tan P – cot R.
Solution:
In ∆PQR, given ∠B = 90°, PQ = 12 PR = 13.
Since ∆PQR is right angle triangle applying Pythagoras theorem,
PR2 = PQ2 + OR2
132 = 122 + QR2 [Putting the value of PA & PR]
QR2 = 169 – 144
QR2 = 25
Q = 5 cm
∴ Tan P = Opposite side/adjacent side = QR/PQ = 5/12
Cot R = adjacent side/opposite side = 5/12
Than tan P – cot R
= 5/12 – 5/12
= 0
(3) If sin A = ¾, calculate cos A and tan A.
Solution:
Sin A = 3/4
Opposite side (Perpendicular)/Hypotenuse = 3/4
Let perpendicular = P
Hypotenuse = H
P/1+ = 3/4
Let P = 3x, H = 4x
Applying Pythagoras theorem,
AC2 = AB2 + BC2
(4K)2 = (3x)2 + BC2 (Putting the value)
BC2 = 16x2 – 9x2
BC2 = 7x2
BC = √7 x
∴ Cos A = Adjacent side (base)/Hypotenuse = AB/AC
= √7 x/4x = √7/4
Tan A = Opposite side (Perpendicular)/Adjacent side (Base)
= BC/AB = 3/√7
(4) Given 15 cot A = 8, find sin A and sec A.
Solution:
Let ∆ABC is a right angled triangle, ∠B = 90°
Given, 15 cot A = 8
Or, Cot A = 8/15
Adjacent side (Base)/Opposite side (perpendicular) = AB/BC = 8/15
Let, AB = 8x
BC = 15
Applying Pythagoras theorem
AC2 = AB2 + BC2
AC2 = (8x)2 + (15x)2
AC2 = 64x2 + 225x2
AC = √289x2 = 17x
∴ Sin A = Opposite side (Perpendicular)/Hypotenuse
= BC/AC = 15x/17x = 15/17
Sec A = Hypotenuse/Adjacent side (Base)
= AC/BC = 17K/8K = 17/8
(5) Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution:
Let, ∆ABC is a right angled triangle, ∠B = 90°
Given, Sec θ = 13/12
Hypotenuse/Adjacent side (Base) = 13/12
AC/BC = 13/12
Let, AC = 13x, BC = 12x
Applying Pythagoras theorem, we get
AC2 = AB2 + BC2, BC = 12x
AC2 = AB2 + BC2
(13x)2 = AB2 + (12x)2 [Putting the value]
AB2 = 169x2 – 144x
AB2 = 25x2
AB = 5x
So, Sin θ = Perpendicular/Hypotenuse = AB/AC = 5x/13x = 5/13
Cos θ = Base/Hypotenuse = BC/AC = 12x/13x = 12/13
Tan θ = Perpendicular/Base = AB/BC = 5x/12x = 5/12
Cosec θ = Hypotenuse/Perpendicular = AC/AB = 13x/5x = 13/5
Cot θ = Base/Perpendicular = BC/AB = 12x/5x = 12/5
(6) If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let, ∆ABC has ∠A and ∠B are acute angle
Given that,
Cos A = Cos B
We can write from ∆ABC
Cos A = AC/AB & Cos B = BC/AB
AB Cos A = Cos B
AC/AB = BC/AB
Or, AC = BC
Or, ∠B = ∠A
(An angles opposite to equal side of the triangle are equal)
(7) If cot θ = 7/8, evaluate:
(i) (1+sin θ) (1-sinθ)/(1+cosθ) (1-cosθ)
(ii) Cot2θ
Solution:
Let, ∆ABC is a right angled triangle ∠B = 90°
Given, cot θ = 7/8
BC/AB = 7/8
Let, BC = 7x
AB = 8k
Applying Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = (8k)2 + (7k)2 (Putting the value)
AC2 = 64x2 + 49x2
AC2 = 113x2
AC = √113x
So, Sin θ AB/AC = Perpendicular/Hypotenuse = 8K/√113x = 8k/√113
Cos θ = BC/AC = Base/Hypotenuse = 7k/√113x = 7/√113
(i) (1+sinθ) (1-sinθ)/(1+cosθ) (1-cosθ)
= 1 – sin2 θ/1 – cos2 θ
= 1 – (8/√113)2/1 – 7/√113)2
= 1 – (64/113)/1 – (49/113)
= (113-64/113)/(113-49/113) = 49/64
(ii) cot2 θ = (7/8)2 = 49/64
(8) If 3 cot A = 4, check whether 1-tan2A/1+tan2A = cos2A – sin2A or not.
Solution:
Let ∆ABC is a right angled triangle ∠B = 90°
Given,
Cot A = 4/3
AB/BC = 4/3
Let AB = 4x, BC = 3x
Applying Pythagoras theorem
AC2 = AB2 + BC2
AC2 = (4x)2 + (3x)2 (Putting the value)
= AC2 = 16x2 + 9x2
AC2 = 25x2
AC = 5x
Sin A = Perpendicular/Hypotenuse = BC/AC = 3k/5x = 3/5
Cos A = Base/hypotenuse = AB/AC = 4x/5x = 4/5
Tan A = Perpendicular/Base = BC/AB = 3x/4x = 3/4
R.H.S
Cos2A – sin2A = (4/5)2 – (3/5)2 = 16/25 – 9/25
= 16-9/25 = 7/25
L.H.S
1-tan2A/1+tan2A = 1-(3/4)2/1+(3/4)2 = (16-9/16)/(16+9/16)
= 7/25
Since R.H.S = L.H.S = 7/25
Hence (1-tan2A)/(1+tan2A) = cos2A – sin2A (Proved)
(9) In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of:
(i) SinA cosC + cosA sinC
(ii) cosA cosC – sinA sinC
Solution:
Let us assume ∆ABC is a right angled triangle. ∠B = 90°
tan A = 1/√3
Or, Perpendicular/Base = 1/√3
Or, BC/AB = 1/√3
Let, BC = K, AB = √3x
From the Pythagoras theorem,
AC2 = AB2 + BC2
AC2 = (√3x)2 + (x)2 (Putting the value)
AC2 = 3x2 + x2
AC2 = 4x2
AC = 2x
Now Sin A = BC/AC = 1/2
Cos A = AB/AC = √3/2
Cos C = BC/AC = 1/2
Sin C = AB/AC = √3/2
(i) Sin A Cos C + cos A sin C
= ½ × 1/2 + √3/2 . √3/2
= 1/4 + 3/4 = 1+3/4
= 4/4
= 1
(ii) Cos A Cos C – sin A sin C
= √3/2 . 1/2 – 1/2 . √3/2
= √3/4 – √3/4 = 0
(10) 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.
Solution:
Given ∆PQR, right angled at a,
PQ = 5cm
PR + QR = 25cm
Let, QR = x
PR = 25 – QR
PR = 25 – x
Applying Pythagoras theorem
PR2 = PQ2 + QR2
(25 – K)2 = 52 + x2
625 – 50K + x2 = 52 + x2
50x = 625 – 25
x = 600/50 = 12
∴ x = 12 = QR
So, PR = 25 – QR = 25 – 12 = 13
(i) Sin P = Perpendicular/Hypotenuse = QR/PR = 12/13
(ii) Cos P = Base/Hypotenuse = Pa/PR = 5/13
(iii) Tan P = Perpendicular/Base = QR/P = 12/5
(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
Solution:
(i) The value of tan A always less than 1. = It is wrong statement.
Let, In ∆ABC, ∠B = 90°
AB = 6, BC = 8, AC = 10
∴ Tan A = 8/6 which is greater than 1.
According Pythagoras theorem,
AC2 = AB2 + BC2
(10)2 = (8)2 + (6)2
100 = 64 + 36
100 = 100
(ii) Sec A = 12/5 for some value of angle A.
Solution:-
-> It is a true statement.
Sec A = 12/5
Hypotenuse/Base = 12/5
AC/AB = 12/5
Let, AC = 12k, AB = 5k
By Pythagoras theorem,
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
BC2 = 144x2 – 25x2
BC2 = 119k2
∴ BC = √119 k
Such a triangle is possible.
(iii) Cos A is the abbreviation used for the cosecant of angle A –> It is a false statement.
Abbreviation used for cosecant of angle A is cosec A.
(iv) Cot A is the product of cot and A
–> False
=> Cot A is not the product of cot and A. ∠A is the angle of cotangent.
(v) Sin θ = 4/3 for some angle θ
–> False
–> Sinθ = Perpendicular/Hypotenuse
–> Hypotenuse is the longest side.
∴ Sinθ is always less than 1.
It can never be 4/3.
Exercise – 8.2
(1) Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
Solution:
sin 60° cos 30° + sin 30° cos 60°
= √3/2 × √3/2 + 1/2 × 1/2
= 3/4 + 1/4
= 3+1/4 = 4/4 = 1 (Ans)
[Putting the value of sin 60° = √3/2 = co 30° Sin 30° = ½ = cos 60°]
(ii) 2 tan2 45° + cos2 30° – sin2 60°
Solution:
2 tan2 45° + cos2 30° – sin2 60°
Putting the value
= 2 (1)2 + (√3/2)2 – (√3/2)2
= 2 + 3/4 – 3/4
= 2 (Ans)
[We know that tan45° = 1 Sin 60° = √3/2 = cos 30°]
(iii) Cos 45°/Sec 30° + cosec 30°
Solution:
Cos 45°/Sec 30° + cosec 30°
We know that
Cos 45° = 1/√2, Sec 30° = 2/√3, Cosec 30° = 2
Putting the value, we get
= (1/√2)/(2√3 + 2)
= (1√2)/(2+2√3/√3)
= √3/√2(2+2√3)
= √3/(2√2 + 2√6) × (2√2 – 2√6)/(2√2 – 2√6)
= 2 (√6 – √18)/(2√2)2 – (2√6)
= 2(√6 – √13)/8-24
= 2 (√6 – √18)/-16
= √6 – √18/8
= √18 – √6/8
= √2×3×3 – √6/8
= 3√2 – √6/8
(iv) Sin 30° + tan 45° – cosec 60°/sec 30° + cos 60° + cot 45°
Solution:
Sin 30° + tan 45° – cosec 60°/sec 30° + cos 60° + cot 45°
We know that
Sin 30° = 1/2, tan 45° = 1, cosec 60° = 2/√3
Sec 30° = 2/√3, cos 60° = ½, cot 45° = 1
Putting this value,
(v) 5 cos2 60° + 4 sec2 30° – tan2 45°/sin2 30° + cos2 30°
Solution:
5 cos2 60° + 4 sec2 30° – tan2 45°/sin2 30° + cos2 30°
We know that,
Cos 60° = 1/2, sec 30° = 2/√3, tan 45° = 1
Sin 30° = 1/2, cos 30° = √3/2
Putting this value,
(2) Choose the correct option and justify your choice:
(i) 2tan30°/1+tan230° =
(A) sin 60°
(B) cos 60°
(C) tan 60°
(D) sin 30°
Solution:
2tan30°/1+tan230°
We know that
tan 30° = 1/√3 and tan2 30° = 1/3
Putting this value
= 2×(1/√3)/(1+1/3)
= (2/√3)/(3+1/3)
= 2/√3 × ¾
= 3/√3×2 × √3/√3
= 3√3/3×2
= √3/2 = sin 30°
Hence D option is correct answer.
[We know that sin 30° = √3/2]
(ii) 1-tan2 45°/1+tan2 45° =
(A) tan 90°
(B) 1
(C) sin 45°
(D) 0
Solution:
1-tan2 45°/1+tan2 45°
We know that
tan 45° = 1 Hence, tan2 45 = 1
Putting this value
= 1-1/1+1
= 0/2 = 0 (Ans)
Hence, (D) option is correct answer.
(iii) Sin 2A = 2 sin A is true when A =
(A) 0°
(B) 30°
(C) 45°
(D) 60°
Solution:
Sin 2A = 2 sin A
R.H.S
Sin 2A
Putting A = 0°
Sin θ = 0
Putting A = 30°
Sin 2×30° = sin 60°
= √3/2
Putting A = 45°
Sin 2×45° = Sin 90°
= 1
L.H.S
2 sin A
Putting A = 0°
2 Sin θ = 0°
Putting A = 30°
2 Sin 30° = 2×1/2 = 1
Putting A = 45°
2 sin 45° = 2 × 1/√2
= 2/√2 × √2/√2
= √2
Hence when A = 0°, the sin 2A = 2 sinA is Satisfy
Hence option (A) is correct answer.
(iv) 2 tan30°/1-tan230°
(A) cos 60°
(B) sin 60°
(C) tan 60°
(D) sin 30°
Solution:
2 tan30°/1-tan230°
We know that
Tan 30° = 1/√3 and tan2 30 = (1/√3)2 = 1/3
Putting this value
= (2×1/√3)/(1-1/3)
= (2/√3)/(3-1/3)
= (2√3)/(2/3)
= 2/√3 × 3/2 = 3/√3 × √3/√3
= 3√3/3
= √3 Tan 60° [We know that tan60° = √3]
Hence (c) option is correct answer.
(3) If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° <A + B ≤ 90°; A > B, find A and B.
Solution:
Given tan (A + B) = √3
Or, tan (A+B) = tan60° [We know that tan60° = √3 & 0° < A + B ≤ 90°]
Or, A+B = 60° —— eq (1)
Also given,
tan (A-B) = 1/√3
Or, tan (A – B) = tan30° [We know that tan 30° = 1/√3]
(A – B) = 30° —- eq(2)
From eq (1) & (2) we can write
A = 45°
Putting this value in eq (1)
B + 45° = 60°
B = 15°
Solving eq (1) & (2), we get
A = 45° & B = 15°
(4) 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°.
Solution:
Sin (A+B) = sinA + sinB
It is a false statement.
In a triangle, A = B ≠ 0
If we put A = 45°, and B = 45°
R.H.S
Sin (A+B)
= Sin (45° + 45°)
= sin 90°
= 1
L.H.S
SinA + sin B
= Sin 45° + sin 45°
= 1/√2 + 1/√2
= 2/√2 × √2/√2
= R√2/2 = √2
Hence R.H.S ≠ L.H.S
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(ii) The value of sin θ increases as θ increases. The statement is correct (True)
We know that
If θ = 30°
If θ = 60°
If θ = 90°
Sin 30° = 1/2
Sin 60° = √3/2
Sin 90° = 1
Increasing the θ value, the sin θ also increases.
(iii) The value of cos θ increases as θ increase. The statement is false
We know that
θ = 0°
θ = 90°
cos 0° = 1
cos 90° = 0
When θ increases, the value of cosθ decreases.
(iv) Sinθ = cos θ for all values of θ it is false statement.
Because sin 90° ≠ cos 90°
(v) CotA is not defined for A = 0°
It is a true statement.
Cot A = CosA/SinA
Putting A = 0°
= cos0°/sin0°
= 1/0
= Not defined
EXERCISE 8.3
(1) Evaluate:
(i) sin18°/cos72°
(ii) Tan26°/cot64°
(iii) cos48°-sin42°
(iv) cosec31° – sec59°
Solution:
(i) We know that
cosA = sin (90° – A)
so, cos72° = sin (90° – 72°)
= sin 18°
Hence, sin18°/cos72° = sin18°/sin18° = 1
(ii) We know that
CotA = tan (90°- A)
So, cot 64° = tan (90° – 64°) = tan 26°
Hence tan26°/cot64° = tan26°/tan26° = 1
(iii) We know that
SinA = cos (90° – A)
SinA 42° = cos (90° – 42°) = cos 48°
Therefore, cos48° – sin42°
= cos48° – cos48° = 0
(iv) We know that
SecA = cosec (90° – A)
So, Sec59° = cosec (90° – 59°) = cosec 31°
Hence,
cosec 31° – sec 59°
= cosec 31° – cosec 31°
= 0
(2) Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) We know that
TanA = 1/cotA
cotA = tanA (90 – A)
(i) tan 48° tan 23° tan 42° tan 67°
= 1/cot48° × 1/cot23° × tan42° tan67° [Using this cot48 = tan (90° – 48°) = tan42°
= 1/tan42° × tan42° × 1/tan67 × tan67× [cot23° = tan (90° – 23)] = tan67
= 1 (Proved)
(ii) We know that
sinA = cos (90° – A)
So, sin38° = cos (90° – 38°) = cos 52°
Similarity sin52° = cos (90° – 52°) = cos 38°
Cos 38° cos52° – sin38° sin52°
= cos 38° cos 52 – cos 38° cos 52° [Putting the value]
= 0
(3) If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
Given that
tan2A = cot (A – 18°) —– (1)
We know that
tanθ = cot (90° – θ)
tan2A = cot (90° – 2A]
We can write (1) as
cot (90° – 2A) = cot (A – 18°)
Therefore, we can write
90° – 2A = A – 18° (2A is acute angle)
3A = 90° + 18°
3A = 108°
A = 36°
(4) If tan A = cot B, prove that A + B = 90°
Solution:
Given that tan A = cot B —– (1)
We know that
tan B = cot (90° – A)
We can write (1) as
cot (90° – A) = cot B
(90° – A) & B are both acute angles hence
90° – A = B
∴ A + B = 90° (Proved)
(5) If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
Given that sec4A = cosec (A – 20°) —– (1)
We know that,
sec A = cosec (90° – A)
sec 4A = cosec (90° – 4A)
We can write (1) as
cosec (A – 20°) = cosec (90° – 4A)
Since (A – 20°) & (90° – 4A) are acute angle.
Hence, A – 20° = 90° – 4A
5A = 110°
A = 22°
(6) 6. If A, B and C are interior angles of a triangle ABC, then show that
Sin (B+C/2) = cos A/2
Solution:
Sum of angle in triangle ABC is A + B + C = 180°
Or, B+C = 180 – A
Or, B+C/2 = 90° – A/2 [Divide both side by 2]
Or, Sin B+C/2 = sin (90° – A/2] [Multiplying both side by sin]
Or, Sin B+C/2 = cos A/2 (Proved)
(7) Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
Sin67° + cos75°
= cos (90° – 67°) + sin (90° – 75°)
= cos23° + sin15° [Using sinA = cos (90 – A) cosA = sin (90 – A)
EXERCISE 8.4
(1) Express the trigonometric ratios sinA, secA and tanA in terms of cotA.
Solution:
sinA = 1/cosecA = 1/√1+cotA [As cosec2A – cot2A = 1 or cosecA = √1+cotA]
Sec2A – tan2A = 1
Or, sec2A – 1+tan2A
Or, Sec2A = 1 + 1/cot2A
Or, Sec2A = cot2A + 1/cotA
Or, Sec A = √cot2A+1/cotA
TanA = 1/cotA
(2) Write all the other trigonometric ratios of ∠A in terms of sec A.
Solution:
Sin2A + cos2A = 1
Or, sin2A = 1 – cos2A
Or, sin2A = 1 – 1/sec2A
Or, sin2A = sec2A-1/sec2A
Or, sinA = √sec2A – 1/secA
∴ cosec A = 1/sinA = secA/√sec2A – 1
cosA = 1/secA
Sec2A – tan2A = 1
Or, tan2A = 1 – sec2A
Or, Tan2A = sec2A – 1
Or, tanA = √Sec2A – 1
TanA = 1/cotA = 1/√Sec2A – 1
(3) Evaluate:
(i) Sin2 63° + sin2 27°/cos2 17° + cos2 73°
(ii) sin25° cos65° + cos25° sin65°
Solution:
Sin2 63° + sin2 27°/cos2 17° + cos2 73° —– (i)
We know that
SinA = cos (90° – A)
So sin27° = cos (90° – 27°) = cos 63°
Similarly cos17° = sin (90° – 17°) = sin 73
Putting this value in equation (i)
sin263° + cos263°/sin273° + cos273° [We know that sin2 θ + cos2θ = 1]
= 1/1 = 1
(ii) sin25° cos 65° + cos 25° sin 65° —– (1)
We know that,
sinA = cos (90° – A)
So, sin25° = cos (90° – 25°) = cos 65°
Similarly cos 25° = sin (90° – 25) = sin 65°
Putting this value in equation (2)
Cos65°.cos65 + sin65°.sin65°
= cos265° + sin265 [We know that sin2 θ + cos2 θ = 1]
= 1
(4) Choose the correct option. Justify your choice.
(i) 9sec2A – 9 tan2A =
(A) 1
(B) 9
(C) 8
(D) 0
Solution:
9sec2A – 9 tan2A
= 9 (sec2A – tan2A) [We know that sec2A – tan2A = 1]
= 9×1
= 9
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) 0
(B) 1
(C) 2
(D) -1
Solution:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= (1 + sinθ/cotθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)
= (cosθ + sinθ + 1/cosθ) (sinθ + cosθ – 1/sinθ)
= 1/sinθ.cosθ [{(sinθ + cosθ) +1} {(sinθ + cosθ)} – 1}]
= 1/sinθ.cosθ [(sinθ + cosθ)2 – 12)] [(We know that (a+b) (a-b) = a2-b2]
= 1/sinθ.cosθ [Sin2θ + cos2θ + 2sinθ cosθ – 1]
= 1/sinθ.cosθ [1 + 2sinθ.cosθ – 1]
= 2sinθ.cosθ/sinθ.cosθ [sin2θ + cos2θ = 1]
= 2
(iii) (sec A + tan A) (1 – sin A) =
(A) Sec A
(B) Sin A
(C) Cosec A
(D) Cos A
Solution:
(SecA + TanA) (1-sinA)
= (1/cosA + sinA/cosA (1-sinA)
= (1+sinA)/cosA (1-sinA)
= [(1)2 – (sinA)2]/cosA [We know that (A-b) (a+b) = a2 – b2]
= 1 – sin2A/cosA
= cos2A/cosA [(Cos2A = 1-sin2A]
= cosA
(iv) 1+tan2 A/1 + cot2 A =
(A) Sec2A
(B) -1
(C) cot2 A
(D) tan2 A
Solution:
1 + tan2A/1+cot2A
(5) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosecθ – cotθ)2 = 1 – cosθ/1+cosθ
Solution:
L.H.S = (cosecθ – cotθ)2
= (1/sinθ – cosθ/sinθ)2
= (1-cosθ/sinθ)2
= (1-cosθ)2/sin2θ
= (1-cosθ)2/(1-cos2θ [We know that sin2θ + cos2θ = 1 sin2θ = 1 – cos2θ]
= (1 – cosθ)2/(1 – cosθ) (1 + cosθ)
= – 1 – cosθ/1+cosθ =- R.H.S (proved)
(ii) cosA/1+sinA + 1+sinA/cosA = 2sec A
Solution:
L.H.S = cosA/1+sinA + 1+sinA/cosA
= cos2A + (1 + sinA)2/cosA (1 + sinA)
= cos2A + 1 + 2sinA + sin2A/cosA (1 + sinA)
= 1+1+2 sinA/cosA (1 + sinA) [sin2A + cos2A = 1]
= 2/cosA
= 2sec A = R.H.S = (proved)
(iii) tanθ/1-cotθ + cotθ/1-tanθ = 1 + secθ cosecθ
Solution:
L.H.S = tanθ/1-cotθ + cotθ/1-tanθ
= cosecθ.secθ + 1
= 1 + secθ.cosecθ
= R.H.S (Proved)
(iv) 1 + secA/secA = sin2A/1-cosA
Solution:
= (1 + cosA)
= (1 + cosA) × (1-cosA)/(1-cosA)
= 1 – cos2A/1 – cosA
= sin2A/1 – cos A = R.H.S = (Proved)
(As sin2A = 1 – cos2A)
(v) cosA – sinA + 1/cosA + sinA – 1 = cosecA + cot A using the identity cosec2 A = 1 + cot2A.
Solution:
L.H.S = cosA – sinA + 1/cosA + sinA – 1
(vi) √1+sinA/1-sinA = secA + tanA
Solution:
L.H.S = √1+sinA/1-sinA
= 1+sinA/cosA
= 1/cosA + sinA/cosA
= secA + tanA = (R.H.S) (Proved)
(vii) sinθ-2sin3θ/2cos3θ – cosθ = tanθ
Solution:
L.H.S = sinθ-2sin3θ/2cos3θ – cosθ
(viii) (SinA + cosecA)2 + (cosA + secA)2 = 7 + tan2A + cot2A
Solution:
L.H.S = (SinA + cosecA) + (cosA + secA)2
= (sin2A + 2sinA.cosecA + cosec2A) + (cos2A + 2cosA secA + sec2A)
= (sin2A + 2sinA.1/sinA + cosec2A + cos2A + 2.cosA. 1/cosA + sec2A)
= (sin2A + cos2A) + 2 + 2 + cosec2A + sec2A
= 1 + 4 + cosec2A + sec2A
= 5 + 1 + cot2A + 1 + tan2A [As cosec2 – cot2A = 1 sec2A – tan2A = 1]
= 7 + cot2A + tan2A
= R.H.S (Proved)
(ix) (cosecA – sinA) (secA – cosA) = 1/tanA + cotA
Solution:
L.H.S = (cosecA – sinA) (secA – cosA)
= (1/sinA – sinA) (1/cosA – cosA)
= (1 – sin2A/sinA) × (1 – cos2A/cosA) [As sin2A + cos2A = 1]
= cos2A/sinA × sin2A/cosA
= cosA.sinA
R.H.S = 1/tanA + cotA
(x) (1+tan2A/1+cot2A) = (1-tanA/1-cotA)2 = tan2A
Solution:
(1+tan2A/1+cot2A) [As cosec2A – cot2A = 1 sec2A – tan2A = 1]
= tan2A
1+tan2A/1+cot2A = (1-tanA/1-cotA)2 = tan2 (Proved)
In case you are missed :- NCERT Solution for Triangles