OP Malhotra Class 10 Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(b)

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S Chand Class 10 ICSE Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(b)

Question 1.
Find the sine, cosine, and tangent of the following angles:
(a) 15°27′
(b) 37°48′
(c) 55°17′
(d) 83°37′
Solution:
Using the natural sine, cosine and tangent tables:
(a) 15°27′ = sine 15°24′ + 3′
= 0.26556 + 84 (Mean difference of 3)
= 0.26640 = 0.2664
cos 15°27′ = cos 15°24′ + 3′
= 0.96410 – 23 (Mean difference of 3)
= 0.96387 = 0.9639
tan 15°27′ = tan 15°24′ + 3′
= 0.27545 + 94 = 0.27639 = 0.2764

(b) sin 37°48′ = 0.61291
= 0.6129
cos 37°48′ = 0.79015 = 0.7902
tan 37°48′ = 0.77568 = 0.7757

(c) sin 55°17′ = sin 55°12′ + 5′
= 0.82115 + 82 (Mean difference of 5′)
= 0.82917 = 0.8219
cos 55°17′ = cos 55°12′ + 5′
= 0.57071 – 120 = 0.56951 = 0.5695
tan 55°17′ = tan 55°12′ + 5′
= 1.43881 + 453 = 1.44334 = 0.4433

(d) sin 83°37′ = sin 83°36′ +1′
= 0.99377 + 3 = 0.99380 = 0.9938
cos 83°37′ = cos 83°36′ + 1′
= 0.11147 – 29 = 0.11118 = 0.1112
tan 83°37′ = 8.91520 = 8.9152

Question 2.
Find the acute angle A, given
(a) sin A = 0.4919
(b) tan A = 2.7775
(c) tan A = 3.412
(d) cos A = 0.4651
(e) sin A = 0.95 19
(f) cos A = 0.5757
Solution:
Using the tables of sines, cosines and tangent
(a) sin A 0.4919 = 0.49090 + differnece = 100
sin 29°24′ + 4 = sin 29°28′
∴ A = 29°28’

(b) tan A = 2.7775 = 2.77761 (∵ It is nearest to 2.77750)
∴ tanA = tan 70°12′
∴ A = 70°12′

(c) tanA = 3.412 = 3.41973
= tan 73°42′ (∵ 3.4 1973 is nearest to 3.412)
∴ A = 73°42′

(d) cos A = 0.4651 = 0.46484 + 16
cos 62°1 8 – 1′ = cos 62° 17′
∴ A = 62°17′

(e) sin A = 0.95190 = 0.95159 + 31
= sin 72°6′ + 3′ = sin 72°9′
∴A = 72°9′

(f) cos A = 0.57570 = .57501 + 69
= cos 54°54′ – 3′ = cos 54°51′
∴ A = 54°51′

Question 3.
Using tables, find the value of (2 sin θ – cos θ) as a decimal
(i) when θ = 35°
(ii) when tan θ = 0.2679
Solution:
(i) 2 sin θ – cos θ = 2 sin 35° – cos 35°
= 2(0.57358) – 0.81915 = 1.14716 – 0.81915
= 0.32801 = 0.3280

(ii) tan θ = 0.2679 = tan 14°56′
∴ 2sin θ – cos θ
= 2 sin 14°56’ – cos 14°56′
= 2 (0.25769) – 0.96622
0.5 1538 – 0.96622 = -0.45084

Question 4.
State for any acute angle θ
(i) whether sin θ increases or decreases as θ increases;
(ii) whether cos θ increases or decreases as θ decreases.
Solution:
(i) We know that sin θ° = 0 and sin 90° = 1
∴ It is clear that sin θ increase as θ increase

(ii) We know that cos θ° = 1 and cos 90° = 0
∴ It is clear that as θ decreases, cos θ increases

Question 5.
If sin x° = 0.67, find the value of
(a) cos x°
(b) cos x° + tan x°
Solution:
From the tables of sines
sin x° = 0.67 = 0.67043 (nearest value)
= sin 42°.6′ – 2′ (Mean difference)
= sin 42°4′

(i) ∴ cos x° = cos 42°4′
= 0.74314 – 77 = 0.74237 = 0.7423

(ii) cos x° + tan x° = cos 42°4′ + tan 42°4′
= 0.7423 + (0.90040 + 214)
= 0.7423 + 0.90254 = 0.7423 + 0.9025 = 1.6448

Question 6.
Using trigonometric table, find the measure of the angle A when sin A = 0.1822.
Solution:
sin A = 0.1822
Using the trigonometric tables,
sin A = 0.18224 (nearest Value)
= sin 10°30′
∴ A = 10°30′

Question 7.
In rectangle ABCD, AB = 23 cm, and ∠CAB = 35°. Calculate the measure of BC.

Solution:
In rectangle ABCD, AC is its diagonal
AB = 23 cm and ∠CAB = 35°
Let BC = x, then in right △ABC
tan θ = \(\frac{\text { Perpendicular }}{\text { Base }}\) = \(\frac { BC }{ AB }\)
⇒ tan 35° = \(\frac { x }{ 23 }\) ⇒ 0.70021 = \(\frac { x }{ 23 }\)
x = 23 × 0.70021 = 16.10483 = 16.1048 = 16.11
∴ BC = 16.11 cm

Question 8.
In the figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB. Given that ∠AED = 60° and ∠ACO = 45°; without using tables, calculate
(i) AB,
(ii) AC and
(iii) AE.

Solution:
In the figure,
AB || DC and BC are perpendicular to AB
AD = BC = 2 cm

∠AED = 60° and ∠ACD = 45°
∵ AB || DC and AD and BC are Perpendicular to AB and AD = BC = 2 cm
∴ AB = CD and AD and BC are perpendicular to CD also.

(i) Now in right △ACD
tan θ = \(\frac { AD }{ CD }\) ⇒ tan 45° = \(\frac { 2 }{ CD }\)
⇒ 1 = \(\frac { 2 }{ CD }\) ⇒ CD = 2 cm
∴ AB = CD = 2 cm

(ii) and sin 45° = \(\frac { AD }{ AC }\) ⇒ \(\frac{1}{\sqrt{2}}\) = \(\frac { 2 }{ AC }\)
⇒ AC = 2√2 cm

(iii) In right △ADE
tan 60° = \(\frac { AD }{ AE }\) ⇒ √3 = \(\frac { 2 }{ AE }\)
⇒ AE = \(\frac{2}{\sqrt{3}}\) = \(\frac{2 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{2 \sqrt{3}}{3}\)
∴ AE = \(\frac{2 \sqrt{3}}{3}\) cm or \(\frac{2}{\sqrt{3}}\) cm

Question 9.
In the figure, BC = 12 cm, AB = 4 cm, ∠AEB = 90°, ∠B = 50° and ∠C = 30°. Calculate the length of
(i) BE and
(ii) AC.

Solution:
In the figure,
BC = 12 cm, AB = 4 cm. ∠AEB = 90°.
∠B = 50° and ∠C = 30°

(i) In right △ABE,
cos θ = \(\frac { BE }{ AB }\) ⇒ cos 50° = \(\frac { BE }{ 4 }\)
⇒ 0.64279 = \(\frac { BE }{ 4 }\) ⇒ BE = 4 × 0.64279
⇒ BE = 2.57116 = 2.57 cm

(ii) sin θ = \(\frac { AE }{ AB }\) ⇒ sin 50° = \(\frac { AE }{ 4 }\)
⇒ 0.76604 = \(\frac { AE }{ 4 }\) ⇒ AE = 4 × 0.76604
⇒ AE = 3.06416 = 3.06 cm

(iii) In △AEC,
sin θ = \(\frac { AE }{ AC }\) ⇒ sin 30° = \(\frac { 3.06 }{ AC }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { 3.06 }{ AC }\) ⇒ AC = 2 × 3.06 = 6.12
Hence AC = 6.12 cm

Question 10.
In the figure, in △ABC, ∠B = 90°, ∠C = 30° and AB = 12 cm. BD is perpendicular to AC; find
(i) BC
(ii) AD
(iii) AC

Solution:
In the figure, in △ABC
∠B = 90°, ∠C = 30°
∴ ∠A = 180° – 90° – 30° = 180° – 120° = 60°
AB = 12 cm, BD ⊥ AC

(i) In right △ABC, ∠B = 90°, ∠C = 30°
tan θ = \(\frac { AB }{ BC }\) ⇒ tan 30° = \(\frac { 12 }{ BC }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { 12 }{ BC }\) ⇒ BC = 12√3 cm

(ii) In right △ABD
cos θ = \(\frac { AD }{ AB }\) ⇒ cos 60° = \(\frac { AD }{ 12 }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { AD }{ 12 }\) ⇒ AD = 12 × \(\frac { 1 }{ 2 }\) = 6
∴ AD = 6 cm

(iii) In △ABC,
sin θ = \(\frac { AB }{ AC }\) ⇒ sin 30° = \(\frac { 12 }{ AC }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { 12 }{ AC }\) = 12 × 2 = 24
∴ AC = 24 cm

Question 11.
In the figure, the radius of a circle is given as 15 cm and chord AB subtends and angle of 131° at the centre C of the circle. Using trigonometry, calculate
(i) the length of AB;
(ii) the distance of AB from the centre C.

Solution:

Radius of the circle with centre C is 15 cm
i.e., AC = BC = 15 cm
∠ACB = 131°
From C, draw CL ⊥ AB
which bisects the chord AB at L
Now in △ABC, ∠C = 131° and AC = BC
∴ ∠A = ∠B = \(\frac{180^{\circ}-131^{\circ}}{2}\) = \(\frac{49^{\circ}}{2}\)
= 24.5° = 24°30′

(i) Now in right △ACL, ∠A = 24°36′
∴ cos θ = \(\frac { AL }{ AC }\) ⇒ cos24°30′ = \(\frac { AL }{ 15 }\)
⇒ .90996 = \(\frac { AL }{ 15 }\) ⇒ AL = 15 × 0.90996
⇒ AL = 13.6494
and AB = 2AL = 2 × 13.6494
= 27.2988 = 27.3 cm

(ii) sin θ = \(\frac { CL }{ AC }\) ⇒ sin 24°30′ = \(\frac { CL }{ 15 }\)
⇒ 0.41469 = \(\frac { CL }{ 15 }\) ⇒ CL = 15 × 0.41469
⇒ CL = 6.22035 = 6.22 cm
Hence the distance of AB from the centre C = 6.22 cm

Question 12.
In the figure, a rocket is fired vertically upwards from its launching pad P. It first rises 20 km vertically up and then travels 80 km at 30° to the vertical. PA represents the first stage of its journey and AB the second; C is a point vertically below B on the same horizontal level as P. Calculate :
(i) the height of the rocket when it is at point B;
(ii) the horizontal distance of point C from point P.

Solution:
In the figure, a rocket is launched from P. It first rises upward to A such that AP = 20 km Then travels to B making an angle of 30° and reaches at B such that AB = 80 km

BC ⊥ PC
From A, draw AD || PC meeting BC at D
∴ DC = AP = 20 km, AD = PC
and ∠BAD = 90° – 30° = 60°

(i) In right △ABD,
sin θ = \(\frac { BD }{ AB }\) ⇒ sin 60° = \(\frac { BD }{ 80 }\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac { BD }{ 80 }\) ⇒ BD = \(\frac{\sqrt{3} \times 80}{2}\) = 40√3 km
∴ BC = BD + DC = 40√3 + 20
= (20 + 40√3)
= 40 (1.732) + 20 = 69.280 + 20 = 89.28 Km

(ii) and cos θ = \(\frac { AD }{ AB }\) ⇒ cos 60° = \(\frac { AD }{ 80 }\)
⇒ \(\frac { 1 }{ 2 }\) = \(\frac { AD }{ 80 }\) ⇒ AD = 80 × \(\frac { 1 }{ 2 }\) = 40
∴ PC = AD = 40 Km

Question 13.
In the figure, BCDE is a rectangle, ED = 3.88 cm, AD = 10 and ∠DAC = 20°35′. Calculate, without using Pythagoras’ theorem,
(i) the length of CD;
(ii) the length of AC;
(iii) the size of angle AEB.

Solution:
BCDE is a rectangle in which ED = 3.88 cm
∴BC = 3.88 cm
A is a point such that AD = 10 cm and A lies on CB on producing, AE is joined
Let ∠AEB = θ

(i) In right △ACD,
sin θ = \(\frac { CD }{ AD }\) ⇒ sin 23°35° = \(\frac { CD }{ 10 }\)

∴ 0.40008 = \(\frac { CD }{ 10 }\) ⇒ CD = 4.0008 = 4.001
∴ CD = 4.001 cm

(ii) cos θ = \(\frac { AC }{ AD }\) ⇒ cos 23°35′ = \(\frac { AC }{ 10 }\)
[But cos 23°35′ = 0.91706 – 58 = 0.91648]
∴ 0.91648 = \(\frac { AC }{ 10 }\) ⇒ AC = 9.1648 = 9.165
∴ AC = 9.165 cm

(iii) Now AB = AC – BC = 9.165 – 3.880 = 5.285 and EB = CD = 4.001
∴ tan θ = \(\frac { AB }{ EB }\) = \(\frac { 5.285 }{ 4.001 }\) = \(\frac { 5285 }{ 4001 }\) = 1.32092
= 1.31745 + 347 = tan 52°48′ + 5′
= tan 52°53′ (from the tables)
∴ θ = 52°53′
∴ ∠AEB = 52°53′

Question 14.
In the figure, triangle ABC is right angled at B. D is the foot of the perpendicular from B to AC.
Given that BC = 3 cm and AB = 4 cm. Without using tables find
(i) tan ∠DBC
(ii) sin ∠DBA

Solution:
In the figure, in right △ABC, ∠B = 90°
BD ⊥ AC, BC = 3 cm, AB = 4 cm

∴ AC² = BC² + AC² (Pythagoras Theorem)
= (3)² + (4)² = 9 + 16 = 25
∴ AC = √25 = 5 cm
In △ABC and △DBC,
∠ABC = ∠BDC (each 90°)
∠C = ∠C (common)
∴ △ABC ~ △DBC (A A axiom)
Similarly we can prove that △ABC ~ △ABD
\(\frac { AC }{ BC }\) = \(\frac { AB }{ BD }\) = \(\frac { BC }{ CD }\) ⇒ \(\frac { AB }{ BC }\) = \(\frac { BD }{ CD }\) (By alternendo)
\(\frac { BD }{ CD }\) = \(\frac { 4 }{ 3 }\) ⇒ \(\frac { CD }{ BD }\) = \(\frac { 3 }{ 4 }\)
Now in right △DBC,
tan ∠DBC = \(\frac { CD }{ BD }\) = \(\frac { 3 }{ 4 }\)
In right △ABD,
sin ∠DBA = \(\frac { AD }{ AB }\) = \(\frac { AB }{ AC }\) = \(\frac { 4 }{ 5 }\)

Question 15.
Some students wished to find the height x of a building and the height y of the flag pole on the building. They made the measures as shown in the diagram. Find x and y. Give your answer to the nearest metre.

Solution:
BC is the building and AB is the flag pole on the building Angle of elvation of B=63° and of A = 63° + 3° = 66°

Now in right △BCD,
tan θ = \(\frac { BC }{ DC }\) ⇒ tan 63° = \(\frac { x }{ 50 }\)
⇒ 1.96261 = \(\frac { x }{ 50 }\) ⇒ x = 50 × 1.96261
x = 98.1305 = 98 cm
Again in right △ADC,
tan 66° = \(\frac { AC }{ DC }\) = \(\frac{x+y}{50}\) ⇒ 2.24604 = \(\frac{x+y}{50}\)
⇒ x + y = 50 × 2.24604 = 112.302 = 112
⇒ y = 112 – 98 = 14 m

Question 16.
The upper part of tree, broken by the wind, makes an angle of 30° with ground, and the horizontal distance from the root of the tree to the point where the top of the tree meets the ground is 25 metres. Find the height of the tree before it was broken, to the nearest metre.
Solution:
TR is the tree which was broken from Q and its top T touched the ground at S. So that SR = 25 m and ∠QSR = 30°
In the figure TQ = QS

In right △QSR
tan θ = \(\frac { QR }{ SR }\) ⇒ tan 30° = \(\frac { QR }{ 25 }\)
⇒ \(\frac{1}{\sqrt{3}}\) = \(\frac { QR }{ 25 }\) ⇒ QR = \(\frac{25}{\sqrt{3}}\)
⇒ QR = \(\frac { 25 }{ 1.732 }\) = 14.43
and cos θ = \(\frac { SR }{ SQ }\) ⇒ cso 30° = \(\frac { 25 }{ SQ }\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac { 25 }{ SQ }\) ⇒ SQ = \(\frac{25 \times 2}{\sqrt{3}}\) = \(\frac{50}{\sqrt{3}}\)
∴ Height of tree = TQ + QR
= QS + QR = \(\frac{25}{\sqrt{3}}\) + \(\frac{50}{\sqrt{3}}\) = \(\frac{75}{\sqrt{3}}\)
= \(\frac{75 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{75 \sqrt{3}}{3}\) = 25√3 m
= 25(1.732) = 43.3 m = 43 m

Question 17.
In the figure, ABC is an equilateral triangle of side 6 cm. D is a point in BC such that BD = 1 cm and E is the midpoint of BC. Calculate :
(i) AE,
(ii) tan ∠ADC,
(iii) ∠ADC to the nearest degree,
(iv) ∠BAD to the nearest degree.

Solution:
In equilateral △ABC with each side 6 cm
∵D is a point on BC such that BD = 1 cm
E is the mid point of BC
∴ DE = DE – BD = 3 – 1 = 2 cm ( ∵ E is mid point of BC)

(i) ∵ E is mid point of BC
∴ AE ⊥ BC
and AD = \(\frac{\sqrt{3}}{2}\) side = \(\frac{\sqrt{3}}{2}\) × 6 = 3√3 cm

(ii) In right △ADE,
tan ∠ADC, = tan ∠ADE = \(\frac { AE }{ DE }\) = \(\frac{3 \sqrt{3}}{2}\) cm
= \(\frac{3(1.732)}{2}\) = 3 × 0.866 = 2.598

(iii) tan ∠ADC = 2.59156 (nearest in the table)
= 68°54′ = 69° (In nearest degree)

(iv) tan ∠DAE = \(\frac { DE }{ AE }\) = \(\frac{2}{3 \sqrt{3}}\) = \(\frac{2 \sqrt{3}}{3 \times \sqrt{3} \times \sqrt{3}}\)
= \(\frac{2 \sqrt{3}}{9}\) = \(\frac{2(1.732)}{9}\) = \(\frac{3.464}{9}\) = 0.385
= 0.38587 = tan 21°6′ = tan 21°
∴ ∠DAE = 21°
But ∠BAD = ∠BAE – ∠DAE (∵ AE also bisects ∠A)
= 30° – 21° = 9°

Question 18.
A kite, flying at a height of 75 metres from the level ground, is attached to a string inclined at 60° to the horizontal. Find the length of the string to the nearest metre.
Solution-
Let K be the kite which is 75 m above the ground and its string makes an angle of 60° with the ground

∴ In △KBT,
KT = 75 m, ∠B = 60°, ∠T = 90°
Let KB = x m
∴ sin θ = \(\frac { KT }{ KB }\) ⇒ sin 60° = \(\frac { 75 }{ x }\)
⇒ \(\frac{\sqrt{3}}{2}\) = \(\frac { 75 }{ x }\) ⇒ x = \(\frac{75 \times 2}{\sqrt{3}}\)
⇒ x = \(\frac{75 \times 2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\) = \(\frac{150 \sqrt{3}}{3}\) = 50√3
= 50 (1.732) = 86.6 = 87
∴ Length of string of the Kite = 87 m

Self Evaluation And Revision (LATEST ICSE QUESTIONS)

Question 1.
Show that \(\sqrt{\frac{1-\cos A}{1+\cos A}}\) = \(\frac{\sin A}{1+\cos A}\)
Solution:

Question 2.
Prove that : 1 – \(\frac{\cos ^2 \theta}{1+\sin \theta}\) = sin θ
Solution:
L.H.S. = 1 – \(\frac{\cos ^2 \theta}{1+\sin \theta}\) = \(\frac{1+\sin \theta-\cos ^2 \theta}{1+\sin \theta}\) = \(\frac{1+\sin \theta-\left(1-\sin ^2 \theta\right)}{1+\sin \theta}\) {cos² θ = 1 – sin² θ}
= \(\frac{1+\sin \theta-1+\sin ^2 \theta}{1+\sin \theta}\) = \(\frac{\sin \theta+\sin ^2 \theta}{1+\sin \theta}\) = \(\frac{\sin \theta(1+\sin \theta)}{1+\sin \theta}\) = sin θ R.H.S.

Question 3.
Prove the following identity:
\(\frac{1}{\sin \theta+\cos \theta}\) + \(\frac{1}{\sin \theta-\cos \theta}\) = \(\frac{2 \sin \theta}{1-2 \cos ^2 \theta}\)
Solution:
L.H.S. = \(\frac{1}{\sin \theta+\cos \theta}\) + \(\frac{1}{\sin \theta-\cos \theta}\) = \(\frac{\sin \theta-\cos \theta+\sin \theta+\cos \theta}{(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)}\)
= \(\frac{2 \sin \theta}{\sin ^2 \theta-\cos ^2 \theta}\) = \(\frac{2 \sin \theta}{1-\cos ^2 \theta-\cos ^2 \theta}\) {sin² θ = 1 – cos² θ}
= \(\frac{2 \sin \theta}{1-2 \cos ^2 \theta}\) = R.H.S.

Question 4.
Prove that:
\(\frac{\cos A}{1-\tan A}\) + \(\frac{\sin A}{1-\cot A}\) = cos A + sin A.
Solution:

Question 5.
Prove \(\frac{\sin A}{1+\cos A}\) + \(\frac{1+\cos A}{\sin A}\) = 2 cosec A.
Solution:
L.H.S. = \(\frac{\sin A}{1+\cos A}\) + \(\frac{1+\cos A}{\sin A}\)
= \(\frac{\sin ^2 A+(1+\cos A)^2}{\sin A(1+\cos A)}\)
= \(\frac{\sin ^2 A+1+\cos ^2 A+2 \cos A}{\sin A(1+\cos A)}\)
= \(\frac{1+1+2 \cos A}{\sin A(1+\cos A)}\) = \(\frac{2+2 \cos A}{\sin A(1+\cos A)}\) {sin² A + cos² A = 1}
= \(\frac{2(1+\cos A)}{\sin A(1+\cos A)}\) = \(\frac{2}{\sin A}\) = 2 cosec A

Question 6.
Prove (1 + tan A)² + (1 – tan A)² = 2 sec² A
Solution:
L.H.S. = (1 + tan A)² + (1 – tan A)²
= 1 + tan² A + 2 tan A + 1 + tan² A – 2 tan A
= 2 + 2 tan² A = 2 (1 + tan² A) = 2 sec² A {∵ 1 + tan² A = sec² A}
= R.H.S.

Question 7.
Prove that \(\frac{\sin \theta \tan \theta}{1-\cos \theta}\) = 1 + sec θ.
Solution:

Question 8.
Prove the identify:
\(\frac{\sec A-1}{\sec A+1}\) = \(\frac{1-\cos A}{1+\cos A}\)
Solution:

Question 9.
Prove that idnetify:
\(\frac{\sin A}{1+\cos A}\) = cosec A – cot A
Solution:
R.H.S. = cosec A – cot A
= \(\frac{1}{\sin A}\) – \(\frac{\cos A}{\sin A}\)

= \(\frac{1-\cos A}{\sin A}\) = \(\frac{(1-\cos A)(1+\cos A)}{\sin A(1+\cos A)}\)
= \(\frac{1-\cos ^2 A}{\sin A(1+\cos A)}\) = \(\frac{\sin ^2 A}{\sin A(1+\cos A)}\)
= \(\frac{\sin ^2 A}{1+\cos A}\) = L.H.S.

Question 10.
Prove the idnetify:
\(\frac{\sin A}{1+\cos A}\) + \(\frac{1+\cos A}{\sin A}\) = 2 cosec A.
Solution:
To Prove:

Question 11.
Prove that:
(cosec A – sin A) (sec A – cos A) sec² A = tan A.
Solution:

Question 12.
Prove that

Solution:

Question 13.
Show that

Solution:

Question 14.
Prove the identify : (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ.
Solution:

Question 15.
Prove that

Solution:

Question 16.
Prove that \(\frac{\cos A}{1+\sin A}\) + tan A = sec A.
Solution: