A SQUARE SERIES [Pick the date] - Webnodefiles.urdu4medn.webnode.com/200000016-d25d8d3587/11...SinA...

[Pick the date]

A Squre Study Material | Arif Baig Cell:9703806542, Email:[email protected], web:urdu4medn.webnode.com

SSC TRIGONOMETRY

SSC TR

IGO

NO

METR

Y

A SQUARE SERIES

A Square Study Material SSC Mathematics Trigonometry

2 Prepared By:ARIF BAIG Cell:9703806542, Email:[email protected]., Web: urdu4medn.webnode.com

11. Trigonometry

Triogonometry: Trigonometry is the study of relationship of three sides and three angles measures of a triangle.

Naming the sides in a right angle triangle: C Hypotenus Opposite side of angle A 𝜃 A B Adjacent side of angle A In triangle ABC ∠CAB = 𝜃(acute angle) AB= Adjacent side of angle A BC= Opposite side of angle A AC= Hypotenus C Trigonometric ratios: SinA CosecA Hypotenus Opposite side of angle A SecA TanA 𝜃 A CosA CotA B Adjacent side of angle A

Sine of ∠𝐴= SinA=𝐵𝐶

𝐴𝐶=

Opposite side to angle A

Hypotenus

Cosecant of ∠𝐴=CosecA=𝐴𝐶

𝐵𝐶=

Hypotenus




Cosine of ∠𝐴=CosA=𝐴𝐵

𝐴𝐶=

Adjacent side to angle A

Hypotenus

Secant of ∠𝐴=SecA=𝐴𝐶

𝐴𝐵=

Hypotenus


Tangent of ∠𝐴=TanA=𝐵𝐶

𝐴𝐵=


Adjacent side to an gle A

Cotangent of ∠𝐴=CotA=𝐴𝐵

𝐵𝐶=



SinA and cosecA , CosA and SecA, TanA and CotA are reciprocal or multiplicative

inverse to each other.

SinA=1

𝐶𝑜𝑠𝑒𝑐𝐴 or CosecA=

1

𝑆𝑖𝑛𝐴

CosA=1

𝑆𝑒𝑐𝐴 or SecA=

1

𝐶𝑜𝑠𝐴

TanA=1

𝐶𝑜𝑡𝐴 or CotA=

1

𝑇𝑎𝑛𝐴

TanA=𝑆𝑖𝑛𝐴

𝐶𝑜𝑠𝐴 or CotA=

𝐶𝑜𝑠𝐴

𝑆𝑖𝑛𝐴

Trigonometric Ratios Of Some Specific Angles:

Degree Angle 𝐴

00 300 450 600 900

SinA 0 1

2

1

2 3

2 1

CosA 1 3

2

1

2

1

2 0

TanA 0 1

3 1 3 ∞ (Not defind)

CotA ∞ (Not defind) 3 1 1

3 0

CosecA 0 2

3 2 2 ∞ (Not defind)

SecA ∞ (Not defind) 2 2 2

3 0

Note:

The values of Sin𝜃 and Cos𝜃 always lies between 0 and 1.



The values of Tan𝜃 always lies between 0 and ∞(not defind).

The values of Cot𝜃 always lies between ∞(not defind) and 0.

The values of Cosec𝜃 always lies between ∞(not defind) and 1.

The values of Sec𝜃 always lies between 1 and ∞(not defind).

Trigonometric Ratios Of Complementary Angles: If the sum of two angles is equal to 900 then the angles are called complementary angles.

In a ∆ABC if ∠𝐵=900 , then sum of other two angles must be 90o. C (∴ Sum of angles in a triangle 180o)

i,e; ∠A + ∠C = 900.

Hence∠A and ∠C are called complementary angles. If 𝜃 is acute anglethen Sin(900-A)=CosA and Cos(900-A)=SinA Tan(900-A)=CotA and Cot(900-A)=TanA Sec(900-A)=CosecA and Cosec(900-A)=SecA 900

A B

Trigonometric Identity:

Sin2A+Cos2A=1 Sec2A-Tan2A=1 Cosec2A-Cot2A=1

(SinA)2=Sin2A≠SinA2

DO THIS Identify “Hypotenuse”, “Opposite side” and “Adjacent side” for the given angles in the given triangles.(PAGE NO.271) 1. For angle R P Sol: in the∆PQR Opposite side=PQ

Adjacent side=QR Hypotenuse=PR Q R

2. (i) For angle X Z (ii) For angle Y Sol: in the ∆XYZ i) for angle X Opposite side=YZ X Y

Adjacent side=XZ Hypotenuse=XY iI) for angle Y Opposite side=XZ

Adjacent side=YZ Hypotenuse=XY

TRY THIS Find lengths of “Hypotenuse”, “Opposite side” and c “Adjacent side” for the given angles in the given triangles. 1. For angle C 4cm 5cm 2. For angle A Sol: By Pythagoras theorem B A



AC2=AB2+BC2 (5)2=AB2+(4)2 25= AB2+16 AB2=25-16 AB2=9 AB2=32 AB=3cm.

DO THIS

1. Find (i) sin C (ii) cos C and C (iii) tan C in the adjacent triangle. Sol: By Pythagoras theorem 5cm 13cm AC2=AB2+BC2 (13)2=AB2+(5)2 169= AB2+25 B A AB2=169-25 AB2=144 AB2=122 AB=12cm.

i) SinC=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒=

𝐴𝐵

𝐴𝐶=

12

13 , ii) CosC=

𝐴𝑑𝑗𝑎𝑐𝑒𝑛 𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐


𝐵𝐶

𝐴𝐶=

5

13 ,

iii) TanC=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑐=

𝐴𝐵

𝐵𝐶=

12

5

2. In a triangle XYZ, ∠Y is right angle, XZ = 17 m and YZ = 15 cm, then find (i) sin X (ii) cos Y (iii) tan X Sol:Given In ∆XYZ , ∠Y is right angle Z XY=17cm, YZ=15cm By Pythagoras theorem XZ2=XY2+YZ2 15cm 17cm (17)2=XY2+(15)2 289= XY2+225 XY2=289-225 Y X XY2=64 XY2=82 XY=8cm.

i) SinX=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑋

ℎ𝑦𝑝𝑜𝑡 𝑒𝑛𝑢𝑠𝑒=

𝑌𝑍

𝑋𝑍=

15

17 , ii) CosZ=

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑍


𝑌𝑍

𝑋𝑍=

15

17 ,

iii) TanX=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑋

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑋=

𝑌𝑍

𝑋𝑌=

15

8

3. In a triangle PQR with right angle at Q, the value of angle P is x, PQ = 7 cm and QR = 24 cm, then find sin x and cos x.

Sol:Given In a triangle PQR with right angle at Q R angle P is x, PQ = 7 cm and QR = 24 cm By Pythagoras theorem PR2=PQ2+QR2 24cm PR2=(7)2+(24)2 PR2=49+576 X PR2=625 Q 7cm P PR2=252 PR=25cm.

Sinx=0𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑥


𝑄𝑅

𝑃𝑅=

24

25 , Cosx=

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝑥


𝑃𝑄

𝑃𝑅=

7

25.

For angle C:

Opposite side=AB=3cm

Adjacent side=BC=4cm Hypotenuse=AC=5cm

For angle C:

Opposite side=BC=4cm

Adjacent side=AB=3cm Hypotenuse=AC=5cm



TRY THESE

In a right angle triangle ABC, right angle is at C. BC + CA = 23 cm and BC - CA = 7cm, then find sin A and tan B.

Sol:Given In a triangle ABC with right angle at C And BC+CA=23cm……(1) , BC-CA=7cm……..(2) From (1) and (2) we get BC+CA=23 B BC - CA =7 2BC=30 BC=15cm Substituting BC=15cm in (1) 15cm BC+CA=23 15+CA=23 CA=23-15=8cm. By Pythagoras theorem C A AB2=AC2+BC2 8cm AB2=(8)2+(15)2 AB2=64+225 AB2=289 AB2=172 AB=17cm.

SinA=0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴


𝐵𝐶

𝐴𝐵=

15

17 , TanB=

0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐵

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐵=

𝐴𝐶

𝐵𝐶=

8

15

THINK AND DISCUSS

Discuss between your friends that

(i) sin x =4

3 for some value of angle x.

Sol: the value of sin𝜃 is always lies between o and 1. But here sin x =4

3> 1so, its does not exist.

(ii) The value of sin A and cos A is always less than 1. Why? Sol: (iii) tan A is product of tan and A. Sol: tanA is an abbreviated form of “tan of angle A”.so, tanA is not a product of tan and A. if tan separated from A has no meaning because of here A is angle.

TRY THIS

What will be the side ratios for sec A and cot A?

Sol: SecA=𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴 , CotA=

𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴

0𝑝𝑝0𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑡𝑜 ∠𝐴

THINK AND DISCUSS

1.is 𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴 is equql to tanA?

Sol: yes, 𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴=tanA

∴sinA=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒, ∴cosA=

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴=

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒




=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒

𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 =tanA.

2.is 𝑐𝑜𝑠𝐴

𝑠𝑖𝑛𝐴 is equql to cotA?

Sol: yes, 𝑐𝑜𝑠𝐴

𝑠𝑖𝑛𝐴=cotA

∴sinA=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒

ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒, ∴cosA=



𝑐𝑜𝑠𝐴

𝑠𝑖𝑛𝐴 =



𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒


=𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 =cotA

EXERCISE - 11.1 1. In right angle triangle ABC, 8 cm, 15 cm and 17 cm are the lengths of AB, BC and CA



respectively. Then, find out sin A, cos A and tan A. sol: Given the length of the sides of triangle A AB=8cm , BC=15cm and AC=17cm We have 17cm 8cm B C

sinA=𝐵𝐶

𝐴𝐶=

15

17, coasA=

𝐴𝐵

𝐴𝐶=

8

17, tanA=

𝐵𝐶

𝐴𝐵=

15

8.

15cm 2. The sides of a right angle triangle PQR are PQ = 7 cm, QR = 25cm and RP = 24 cm respectively. Then find, tan Q -tan R. Sol: Given in ∆𝑃𝑄𝑅 PQ=7cm, QR=25cm and RP=24cm R We have

tanQ=𝑅𝑃

𝑃𝑄=

24

7 25cm

24cm

tanR= 𝑄𝑃

𝑃𝑅=

7

24 𝜃

Q 7cm P

tanQ-tanR=24

7−

7

24=

576−49

168−

527

168

3. In a right angle triangle ABC with right angle at B, in which a = 24 units, b = 25 units and ∠BAC = 𝜃. Then, find cos 𝜃 and tan 𝜃. Sol: Given in ∆𝐴𝐵𝐶 right angle at B. a=24units , b=25units , ∠BAC = 𝜃 A we have 𝜃 by Pythagoreans theorem 25 ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 (25)2 = 𝐴𝐵2 + (24)2 C 24 B 625=576+𝐴𝐵2 625-576=𝐴𝐵2 49=𝐴𝐵2 72=𝐴𝐵2 7cm= AB. Hence

cos 𝜃 =𝐴𝐵

𝐴𝐶=

7

25 tan 𝜃 =

𝐵𝐶

𝐴𝐵=

25

7

4. If cos A =12

13, then find sin A and tan A.

Sol: Given that C

cos A =12

13

we have 13 by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 𝜃 (13)2 = (12)2 + 𝐵𝐶2 B 12 A 169=144+𝐵𝐶2 169-144=𝐵𝐶2 25=𝐵𝐶2 52 = 𝐵𝐶2 5cm=BC We have

sinA=𝐵𝐶

𝐴𝐶=

5

13 , cosA=

𝐵𝐶

𝐴𝐵=

5

12

5. If 3 tan A = 4, then find sin A and cos A. Sol: Given that 3 tan A = 4 C

tanA= 4

3

hence, 4 by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 𝜃 𝐴𝐶2 = 32 + 42 B 3 A 𝐴𝐶2=9+16 𝐴𝐶2=25 𝐴𝐶2=52 AC=5cm We have

sinA=𝐵𝐶

𝐴𝐶=

4

5 , cosA=

𝐴𝐵

𝐴𝐶=

3

5.

6. If ∠A and ∠X are acute angles such that cos A = cos X then show that ∠A = ∠X. Sol: Let ∆𝐴𝐵𝐶 , ∆𝑃𝑄𝑅 X Given cosA=cosX

We have cosA=𝐴𝐵

𝐴𝐶 , cosX=

𝑋𝑌

𝑋𝑍

Then 𝐴𝐵

𝐴𝐶=

𝑋𝑌

𝑋𝑍 C A

by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 and ∴ 𝑋𝑍2 = 𝑋𝑌2 + 𝑌𝑍2

𝐴𝐶 = 𝐴𝐵2 + 𝐵𝐶2 and 𝑋𝑍 = 𝑋𝑌2 + 𝑌𝑍2

So , 𝐴𝐶

𝑋𝑍=

𝐴𝐵

𝑋𝑌=

𝐵𝐶

𝑌𝑍

∵ ∠𝐴 = ∠𝑋.

7. Given cot 𝜃 =7

8, then evaluate (i)

(1+𝑠𝑖𝑛𝜃 )(1−𝑠𝑖𝑛𝜃 )

(1+𝑐𝑜𝑠𝜃 )(1−𝑐𝑜𝑠𝜃 )

(ii) 1+𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃

Sol: Given cot 𝜃 =7

8 A

by Pythagoreans theorm ∴ 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 8 𝐴𝐶2 = 82 + 72 𝜃 𝐴𝐶2=64+49 B 7 C 𝐴𝐶2=113

AC= 113.

We have sin𝜃=𝐴𝐵

𝐵𝐶=

8

113 , cos𝜃=

𝐵𝐶

𝐴𝐶=

7

113

(i) (1+𝑠𝑖𝑛𝜃 )(1−𝑠𝑖𝑛𝜃 )

(1+𝑐𝑜𝑠𝜃 )(1−𝑐𝑜𝑠𝜃 )

=(1)2−sin 2𝜃

(1)2−cos 2𝜃

=(1−sin 2𝜃)

(1−cos 2𝜃) ∴ cos2𝜃=1 − sin2𝜃 ∴ sin2𝜃=1 −

cos2𝜃

=cos 2𝜃

sin 2𝜃

=cot2𝜃

=(7

8)2

=49

64.

(ii) 1+𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃



=1+

8

1137

113

= 113 +8

1137

113

= 113+8

113. 113

7

= 113+8

7.

8. In a right angle triangle ABC, right angle is at B, if

tan A = 3 then find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C - sin A sin C Sol: Given that a right angle triangle ABC, right angle is at B, tan A

= 3

so B=900 , tan A = 3

tan A = tan600

A=600

We know in a triangle

A+B+C=1800

600+900+C=1800

1500+C=1800

C=1800-1500=300

(i) sin A cos C + cos A sin C

= sin600 cos300 + cos600 sin300

= 3

2 . 3

2 +

1

2.

1

2

=3

4+

1

4

=4

4= 1.

(ii) cos A cos C - sin A sin C =cos600 cos300 – sin600 sin300

= 1

2. 3

2- 3

2.

1

2

= 3

4- 3

4=0.

Exercise: 11.2

1. Evaluate the following

(i) sin450+cos450

Sol: Given

= sin450+cos450=1

2+

1

2=

1+1

2=

2

2=

2. 2

2= 2.

(ii) 𝑐𝑜𝑠 450

𝑠𝑒𝑐300+𝑐𝑜𝑠𝑒𝑐 600

Sol: Given

=𝑐𝑜𝑠 450

𝑠𝑒𝑐300+𝑐𝑜𝑠𝑒𝑐 600=

1

22

3+

2

3

=

1

22+2

3

=

1

24

3

=1

2×

3

4=

3

4 2

(iii) 𝑠𝑖𝑛300+𝑡𝑎𝑛 450−𝑐𝑜𝑠𝑒𝑐600

𝑐𝑜𝑡450+𝑐𝑜𝑠600−𝑠𝑒𝑐300

Sol: Given

=𝑠𝑖𝑛300+𝑡𝑎𝑛 450−𝑐𝑜𝑠𝑒𝑐600

𝑐𝑜𝑡 450+𝑐𝑜𝑠600−𝑠𝑒𝑐300 =

1

2+

1

1−

2

31

1+

1

2−

2

1

=1

1=1.

(iv) 2 𝑡𝑎𝑛2450+𝑐𝑜𝑠2300 − 𝑠𝑖𝑛2600

Sol: Given

=2

𝑡𝑎𝑛2450+𝑐𝑜𝑠2300 −

𝑠𝑖𝑛2600=2. (1)2+( 3

2)2 −(

3

2)2=2.(1)=2.

(v) 𝑠𝑒𝑐 2600−𝑡𝑎𝑛 2600

𝑠𝑖𝑛2300+𝑐𝑜𝑠 2300

Sol: Given

= 𝑠𝑒𝑐 2600−𝑡𝑎𝑛 2600

𝑠𝑖𝑛2300+𝑐𝑜𝑠 2300 =(2)2−( 3)2

(1

2)2+(

3

2)2

= 4−31

4+

3

4

=1

1+3

4

=14

4

=1

1=1.

2. Choose the right opinion and justify your choice.

(i) 2𝑡𝑎𝑛 300

1+𝑡𝑎𝑛 2450

(a) sin600 (b)cos600 (c) tan300 (d) sin300

Sol: Given that

= 2𝑡𝑎𝑛 300

1+𝑡𝑎𝑛 2450=2.

1

3

1+(1)2=

2

3

1+1=

2

3

2=

2

3.

1

2=

1

3

= tan300

(ii) 1−𝑡𝑎𝑛 2450

1+𝑡𝑎𝑛 2450

(a) tan900 (b)1 (c) sin450 (d) 0

Sol: Given that

=1−𝑡𝑎𝑛 2450

1+𝑡𝑎𝑛 2450=1−(1)2

1+(1)2=1−1

1+1=

0

2=0.



(iii) 2𝑡𝑎𝑛 300

1−𝑡𝑎𝑛 2300

(a) cos600 (b)sin600 (c) tan600 (d) sin300

Sol: Given that

=2𝑡𝑎𝑛 300

1−𝑡𝑎𝑛 2300=2.

1

3

1−(1

3)2

=

2

3

1−1

3

=

2

33−1

3

=

2

32

3

=2

3.

3

2=

3

3=

3. 3

3=

3

= tan600

3. Evaluate sin600 .cos300+sin300 .cos600 . what is

the value of sin(600 + 300), what can you conclude.

Sol: Given that

= sin600 .cos300+sin300 .cos600

= 3

2. 3

2+

1

2.

1

2=

3

4+

1

4=

3+1

4=

4

4=1.

=sin(600 + 300)

= sin900=1.

∴sin600 .cos300+sin300 .cos600= sin(600 + 300)

We conclude that

∴ sinA.cosB+sinA.cosB= sin(𝐴 + 𝐵).

4.Is it right to say cos(600 + 300)=

cos600 .cos300+sin600 .sin300

Sol: Given that

cos(600 + 300)= cos600 .cos300+sin600 .sin300

cos900=1

2. 3

2−

3

2.

1

2

0= 3

4−

3

4

0=0

Yes , it is right to say cos(600 + 300)=

cos600 .cos300+sin600 .sin300 .

5. In right angle triangle PQR right angle is at Q and

PQ=6cm ∠𝑅𝑃𝑄 = 600 . Determine the length of QR

and PR.

Sol: given PQ=6cm and ∠𝑅𝑃𝑄 = 600

R

We know that in ∆𝑃𝑄𝑅

300

tan600 =𝑄𝑅

𝑃𝑄

3 =𝑄𝑅

6 Q 6cm P

6 3=QR

sin600 =𝑄𝑅

𝑃𝑅

3

2=

6 3

𝑃𝑅

PR=12cm.

6. In ∆𝑋𝑌𝑍, right angle at Y, YX=x and XZ=2x then

determine ∠𝑌𝑋𝑍 and ∠𝑌𝑍𝑋.

Sol: Given

In ∆𝑋𝑌𝑍, right angle at Y, YZ=x and XY=2x

By Pythagoreans theorem

XZ2=XY2+YZ2 Z

(2x)2=(x)2+YZ2 2x

4x2-x2=YZ2 Y x X

3x2=YZ2

3𝑥2=YZ

3𝑥=YZ

We know

In ∆𝑋𝑌𝑍

tanX=𝑌𝑍

𝑋𝑌=

3𝑥

𝑥= 3=tan600

X=600 .

tanZ=𝑋𝑌

𝑌𝑍=

𝑥

3𝑥=

1

3=tan300

Z=300.

Hence ∠𝑌𝑋𝑍=600, ∠𝑌𝑍𝑋=300.

7. Is it right to say that sin(A+B)=sinA+sinB justify

your answer.

Sol: Given sin(A+B)=sinA+sinB



Let A=300 , B=600

Sin(300 + 600) = 𝑠𝑖𝑛300 + 𝑠𝑖𝑛600

sin900=1

2+

3

2

1= 1+ 3

2

1≠1+ 3

2

Isn’t right to say sin(A+B)=sinA+sinB.

Exercise: 11.3

1. Evaluate

(i) 𝑡𝑎𝑛 360

𝑐𝑜𝑡540

Sol: Given that

=𝑡𝑎𝑛 360

𝑐𝑜𝑡540

∴ cotA=tan(900 −A)

∵ cot360=tan(900 − 540)=tan360

= 𝑡𝑎𝑛 360

𝑐𝑜𝑡 360

=1.

(ii) cos120 − 𝑠𝑖𝑛780

Sol: Given that

= cos120 − 𝑠𝑖𝑛780

∴ cosA=sin(900 −A)

∵ cos120=sin(900 − 120)=sin780

= sin780 − 𝑠𝑖𝑛780

=1.

(iii) cos𝑒𝑐310 − 𝑠𝑒𝑐590

Sol: Given that

= cos𝑒𝑐310 − 𝑠𝑒𝑐590

∴ cosecA=sec(900 −A)

∵ cos𝑒𝑐310=sec(900 − 310)=sec590

= s𝑒𝑐590 − 𝑠𝑒𝑐590

=1.

(iv) sin150 . 𝑠𝑒𝑐750

Sol: Given that

= sin150 . 𝑠𝑒𝑐750

= 𝑠𝑖𝑛150

𝑐𝑜𝑠750


∵ cos750=sin(900 − 750)=sin150

= 𝑠𝑖𝑛150

𝑠𝑖𝑛150

=1.

(v) tan260 . tan640

Sol: Given that

= tan260 . tan640

∴ tanA=cot(900 −A)

∵ tan640=cot(900 − 640)=cot260

= tan260 . cot260

= tan260 .1

tan 260

=1.

2. (i) Show that tan480 . tan160 . tan420 . tan740=1

Sol: LHS. = tan480 . tan160 . tan420 . tan740


∵ tan420=cot(900 − 420)=cot480

∵ tan740=cot(900 − 740)=cot160

= tan480 . tan160 . cot480 . cot160

= tan480 . tan160 . 1

tan 480 .1

tan 160

=1. RHS

(ii) cos360 .cos540-sin360 .sin540=0

Sol: LHS

= cos360 .cos540 −sin360 .sin540



∵ cot360=tan(900 − 540)=tan360



∵ cos360=sin(900 − 360)=sin540

∵ cos540=sin(900 − 540)=sin360

= sin540 .sin360 − sin360 . sin540

=1. RHS

3. If tan2A=cot(𝐴 − 180), where 2A is an acute

angle. Find the value of ‘A’.

Sol: Given that

=tan2A=cot(𝐴 − 180)


∵tan2A=cot(900 − 2A)

cot(900 − 2A)= cot(𝐴 − 180)

900 − 2A= 𝐴 − 180

900+180=A+2A

1080=3A

360=A.

4. If tanA = cotB where A and B are acute angles

prove that A+B=900

Sol: Given that

tanA = cotB


∵ cotB=tan(900 − 𝐵)

tanA= tan(900 − 𝐵)

A=900 − 𝐵

A+B=900 .

5. If A,B and C are the interior angles of a triangle

ABC then show that tan(𝐴+𝐵

2) = 𝑐𝑜𝑡

𝐶

2 .

Sol: We know that in triangle ABC

A+B+C= 1800 (dividing the equation by ‘2’)

𝐴+𝐵+𝐶

2=

1800

2

𝐴+𝐵

2+

𝐶

2=900

𝐴 + 𝐵

2= 900 −

𝐶

2

LHS. =tan(𝐴+𝐵

2) ∴ tanA=cot(900 −A)=cotA

=tan(900 −𝐶

2)

= 𝑐𝑜𝑡𝐶

2 RHS.

6. Express sin750 +cos650 in terms of

trigonometric ratios of angle between 00 and 450 .

Sol: Given that

= sin750 +cos650

∴ sinA=cos(900 −A)

∵ cos750=sin(900 − 750)=sin150


∵ cos650=sin(900 − 650)=sin250

= cos150 +sin250 .

Exercise: 11.4

1.Evaluate the following

(i) (1+tan 𝜃 -sec 𝜃)(1+cot 𝜃 -cosec 𝜃)

Sol: Given that (1+tan 𝜃 -sec 𝜃)(1+cot 𝜃 -

cosec 𝜃)

=(1+𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃−

1

𝑐𝑜𝑠𝜃 )(1+

𝑐𝑜𝑠𝜃

𝑠𝑖𝑛𝜃−

1

𝑠𝑖𝑛𝜃 )

= 𝑐𝑜𝑠𝜃+𝑠𝑖𝑛𝜃−1

𝑐𝑜𝑠𝜃

𝑠𝑖𝑛𝜃+𝑐𝑜𝑠𝜃−1

𝑠𝑖𝑛𝜃

= (𝑐𝑜𝑠𝜃+sin θ)2− (1)2

𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃

= 𝑐𝑜𝑠 2𝜃+𝑠𝑖𝑛2𝜃+2𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃−1

𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃 ∴ 𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2𝜃=1

= ∤+2𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃−∤


=2𝑐𝑜𝑠𝜃 .𝑠𝑖𝑛𝜃


=2.

(𝑖𝑖) (𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃)2 + (𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃)2

𝑠𝑜𝑙: (𝑠𝑖𝑛𝜃 + 𝑐𝑜𝑠𝜃)2 + (𝑠𝑖𝑛𝜃 − 𝑐𝑜𝑠𝜃)2

=𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2𝜃+2sin 𝜃.cos 𝜃 +𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠2 𝜃 -2sin 𝜃.cos 𝜃



=1+1

=2.

(iii) ( 𝑠𝑒𝑐2𝜃 − 1)( 𝑐𝑜𝑠𝑒𝑐2𝜃-1)

Sol: ( 𝑠𝑒𝑐2𝜃 − 1)( 𝑐𝑜𝑠𝑒𝑐2𝜃-1)

∴ 𝑡𝑎𝑛2= 𝑠𝑒𝑐2-1 ∴ 𝑐𝑜𝑡2=𝑐𝑜𝑠𝑒𝑐2-1

=𝑡𝑎𝑛2𝜃. 𝑐𝑜𝑡2 𝜃

=𝑡𝑎𝑛2. 1

𝑡𝑎𝑛 2

=1.

2. Show that ( 𝑐𝑜𝑠𝑒𝑐𝜃 − cot𝜃)2= 1−𝑐𝑜𝑠𝜃

1+𝑐𝑜𝑠𝜃

Sol: LHS.

=( 𝑐𝑜𝑠𝑒𝑐𝜃 − cot𝜃)2

= 1

𝑠𝑖𝑛𝜃−

𝑐𝑜𝑠𝜃

𝑠𝑖𝑛𝜃

2

= 1−𝑐𝑜𝑠𝜃

𝑠𝑖𝑛𝜃

2

= (1−𝑐𝑜𝑠𝜃)2

𝑠𝑖𝑛2 𝜃

= (1−𝑐𝑜𝑠𝜃)2

12−𝑐𝑜𝑠 2𝜃

= (1−cos 𝜃)(1−cos 𝜃)

(1−cos 𝜃)(1+cos 𝜃)

= (1−cos 𝜃)

(1+cos 𝜃). RHS

3.Show that 1+𝑠𝑖𝑛𝐴

1−𝑠𝑖𝑛𝐴 =secA+tanA

Sol: LHS. 1+𝑠𝑖𝑛𝐴

1−𝑠𝑖𝑛𝐴

Multiply and divided by (1+sinA)

= 1+𝑠𝑖𝑛𝐴

1−𝑠𝑖𝑛𝐴×

1+𝑠𝑖𝑛𝐴

1+𝑠𝑖𝑛𝐴

= (1+𝑠𝑖𝑛𝐴 )2

12−𝑠𝑖𝑛𝐴2

= (1+𝑠𝑖𝑛𝐴 )2

𝑐𝑜𝑠𝐴2

=1+𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴

=1

𝑐𝑜𝑠𝐴+

𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴 ∴ 𝑠𝑒𝑐𝐴 =

1

𝑐𝑜𝑠𝐴 ∴tanA=

𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴

=secA+tanA. RHS

4.Show that 1−𝑡𝑎𝑛 2𝐴

𝑐𝑜𝑡 2𝐴−1= 𝑡𝑎𝑛2𝐴.

Sol: LHS. = 1−𝑡𝑎𝑛 2𝐴

𝑐𝑜𝑡 2𝐴−1

=1−𝑡𝑎𝑛 2𝐴

1

𝑡𝑎𝑛 2𝐴−

1

1

=1−𝑡𝑎𝑛 2𝐴

1

1−𝑡𝑎𝑛 2𝐴

𝑡𝑎𝑛 2𝐴

= 1−𝑡𝑎𝑛 2𝐴

1×

𝑡𝑎𝑛 2𝐴

1−𝑡𝑎𝑛 2𝐴

= 𝑡𝑎𝑛2𝐴. RHS.

5. Show that 1

𝑐𝑜𝑠𝜃− 𝑐𝑜𝑠𝜃= tan 𝜃.sin 𝜃

Sol: LHS. = 1

𝑐𝑜𝑠𝜃− 𝑐𝑜𝑠𝜃

= 1−𝑐𝑜𝑠 2𝜃

𝑐𝑜𝑠𝜃 ∴ 𝑠𝑖𝑛2𝜃 = 1 − 𝑐𝑜𝑠2𝜃

= 𝑠𝑖𝑛2𝜃

𝑐𝑜𝑠𝜃

= 𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃 .sin 𝜃

= tan 𝜃.sin 𝜃 RHS.

6. Simplify secA(1-sinA)(secA+tanA).

Sol: secA(1-sinA)(secA+tanA)

= (secA-secA.sinA)(secA+tanA)

= (secA-𝑠𝑖𝑛𝐴

𝑐𝑜𝑠𝐴)(secA+tanA)

= (secA-tanA)(secA+tanA)

= 𝑠𝑒𝑐2𝐴 − 𝑡𝑎𝑛2𝐴 ∴ 𝑠𝑒𝑐2𝐴 − 𝑡𝑎𝑛2𝐴 =1

= 1.

7. Prove that (𝑠𝑖𝑛𝐴 + 𝑐𝑜𝑠𝑒𝑐𝐴)2 + (𝑐𝑜𝑠𝐴 +

𝑠𝑒𝑐𝐴)2=7+𝑡𝑎𝑛2𝐴+𝑐𝑜𝑡2𝐴.

Sol: LHS. = (𝑠𝑖𝑛𝐴 + 𝑐𝑜𝑠𝑒𝑐𝐴)2 + (𝑐𝑜𝑠𝐴 + 𝑠𝑒𝑐𝐴)2

=𝑠𝑖𝑛2𝐴 + 𝑐𝑜𝑠𝑒𝑐𝐴2+2.sinA.cosecA+𝑐𝑜𝑠2𝐴 +

𝑠𝑒𝑐𝐴2+2.cosA.secA

= 𝑠𝑖𝑛2𝐴+𝑐𝑜𝑠2𝐴+𝑐𝑜𝑠𝑒𝑐𝐴2+𝑠𝑒𝑐𝐴2+2+2

= 1+1+ 𝑐𝑜𝑡𝐴2+1 + 𝑡𝑎𝑛𝐴2+4 ∴ 𝑐𝑜𝑠𝑒𝑐𝐴2

=1+ 𝑐𝑜𝑡𝐴2 ∴ 𝑠𝑒𝑐𝐴2 = 1 + 𝑡𝑎𝑛𝐴2



= 7+𝑡𝑎𝑛𝐴2 + 𝑐𝑜𝑡𝐴2. RHS.

8. Simplify (1-cos 𝜃)(1+cos 𝜃)(1+𝑐𝑜𝑡2𝜃)

Sol: Given that

=(1-cos 𝜃)(1+cos 𝜃)(1+𝑐𝑜𝑡2𝜃)

=(12 − 𝑐𝑜𝑠2𝜃)(1 + 𝑐𝑜𝑡2𝜃)

=𝑠𝑖𝑛2𝜃(1+𝑐𝑜𝑠 2𝜃

𝑠𝑖𝑛2𝜃)

=𝑠𝑖𝑛2𝜃(𝑠𝑖𝑛2𝜃+𝑐𝑜𝑠 2𝜃

𝑠𝑖𝑛2𝜃) ∴ 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃=1

= 𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃

=1.

9. sec 𝜃 +tan 𝜃 =p then what is the value of sec 𝜃 -

tan 𝜃 ?

Sol: Given that

sec 𝜃 +tan 𝜃 =p

∴ sec 𝜃 -tan 𝜃 =1

sec 𝜃+tan𝜃

= sec 𝜃 -tan 𝜃 =1

p

10. If cosec 𝜃 +cot 𝜃 =k then prove that cos 𝜃 =𝑘2−1

k2+1

Sol: Given that

cosec 𝜃 +cot 𝜃 =k

∴ 𝑐𝑜sec 𝜃 -cot 𝜃 =1

cosec 𝜃+cot 𝜃

= cosec 𝜃 -cot 𝜃 =1

k

From and

cosec 𝜃 +cot 𝜃 =k

cosec 𝜃 -cot 𝜃 =1

k

2cosec 𝜃 = k+1

k

2cosec 𝜃 = 𝑘2+1

𝑘

cosec 𝜃 = 𝑘2+1

2𝑘

∴ 𝑐𝑜𝑠2𝜃 =1- 𝑠𝑖𝑛2𝜃

cos 𝜃 = 1 − 𝑠𝑖𝑛2𝜃

cos 𝜃 = 1 − 2k

k2+1

2

cos 𝜃 = (k2+1)2 −4k2

(k2+1)2

cos 𝜃 = (k2−1)2

(k2+1)2

cos 𝜃 = (k2−1)2

(k2+1)2 RHS.

1

1

1

2

2

2

1

1

1

A SQUARE SERIES [Pick the date] - Webnodefiles.urdu4medn.webnode.com/200000016-d25d8d3587/11...SinA...

Documents

Transcript of A SQUARE SERIES [Pick the date] - Webnodefiles.urdu4medn.webnode.com/200000016-d25d8d3587/11...SinA...