08. Trigonometric Ratios and Identities-2 - Copy

7/31/2019 08. Trigonometric Ratios and Identities-2 - Copy

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Parul Institute of Engineering & Technology

Subject Code : 3300001

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Name Of Subject : basic mathematics

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Name of Unit :Trigonometry__________________________________

Topic : basic mathematics

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Slide Title (Calibri 44 pts)

Main Heading (Calibri 36 pts) Sub Heading (Calibri 30 pts)

Text (Calibiri 28 pts)

(Use Images, Graphs, Diagrams where

ever necessary)

Sub: basic mathematics Topic: ____ basic mathematics

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Trigonometric Ratio and Identity 2

Session 2


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Topics

Multiple Angles

Sub Multiple Angles

Conditional Identities

Relation between sides and interior

angle of polygon


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Multiple Angles

An angle of the form of 2A, 3A, 4A

etc are called multiple angles of A

sin 2A = 2sinAcosA

cos2A = cos2A sin2A

cos2A = 1-2sin2A 1 - cos2A = 2sin2Aand

2

tanA2tan A =2

-tan2

A

Trigonometric ratios of 2A in terms of A

1+ cos2A = 2cos2Acos2A = 2cos2A - 1Q

2cot A-2cot A =2

cotA2

J009


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Multiple Angles

sin2A and cos2A in terms of tanA

sin 2A = 2sinAcosA

=

+2 2

sin cos2

cos sin

A A

A A

Dividing Nr and Dr by cos2A we get

=+ 2tanA2

sin A2tan A2

2

2

tan A2cos A2

tan A2

=+

Similarly

J009


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Illustrative Problem

Solution

= o o

2 2

sin cos22 22

( )

=

o o

o o

2 2cos sin2 22 22

2 2

2sin cos2 22 22

2

cos sin2 22 22

sin cos22 22

=

o o

o o

Show that cosec sec2 22 22 2 =o o

L.H.S cosec sec2 22 22= o o

( )=

o o o o

o

sin cos cos sin2 22 22 22 22

sin 22

( )= =

o o

o

sin2 22 22. proved2

sin 22

J009


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Prove that

2 2 2 2

cos cos cos cos2 2 2 22 2 2 2 2 + + + + =

Solution

2

cos cos cos2 2 2

= = Q

2 2 2and cos cos cos

2 2 2

= =

J009


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2 2 2 2

cos cos cos cos2 2 2 22 2 2 2 2

+ + + + =

Prove that

2 2L.H.S cos cos cos cos2 2 2 2

2 2 2 2

= + +

Solution

2 2 2cos cos2 22 2

=

2 2 2sin sin2 2

=

2 22 2

sin sin2 2

2 2 2

=

2 2

cos cos2 2

2 2 2

=

2 2 2 2 2 22 2 2

2 2 2 22 2

= + = = Ans.

J009


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Multiple Angles

Trigonometric ratios of 3A in terms of A

sin3A = 3sinA 4 sin3A

Proof sin3A = sin(2A+A)

= sin2AcosA + cos2AsinA

= 2sinAcosAcosA +(12sin2A)sinA

= 2sinAcos2A + sinA 2sin3A

= 2sinA(1-sin2A) + sinA 2sin3A

= 3sinA 4 sin3ASimilarly

cos3A = 4cos3A 3cosA2

2

tanA tan A2tan A2

tan A2 2

=


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Prove that

sinA.sin(60oA).sin(60o+A) 2sin A22

=

Solution

L.H.S = sinA.sin(60oA).sin(60o+A)

= sinA [sin260osin2A]

2sinA sin A2 2 2sin A2

2 2

= =

22sinA sin A2

=

Proved

sin(A+B).sin(A-B) = sin2Asin2B

J009


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Prove thatsin5 = 5sin 20sin3 + 16sin5

L.H.S = sin 5 =sin(3 +2)

Solution

= sin 3cos2 + cos3sin2

= (3sin 4sin3) (1 2sin2)

+ (4cos3 3cos) (2sincos)

= (3sin 4sin3) (1 2sin2)

+ cos(4cos2

3) (2sincos)= (3sin 4sin3) (1 2sin2)

+ 2sin (1sin2) {4(1sin2) 3}

J009


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Illustrative ProblemProve that

sin5 = 5sin 20sin3 + 16sin5

Solution

L.H.S. = (3sin 4sin3) (1 2sin2)

+ 2sin (1sin2) {4(1sin2) 3}

= (3sin 4sin3

) (1 2sin2

)+ (2sin 2sin3) (14sin2)

= 3sin 6sin3 4sin3 + 8 sin5

+ 2sin 8sin3

2sin3

+ 8sin5

= 5sin 20sin3 + 16sin5 Proved

J009


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Sub - Multiple Angle

An angle of the form of A/2, A/3,A/4 etc are called sub - multiple

angles of A

Trigonometric ratios of A in terms of A/2

A Asin A sin cos2

2 2

=

2 2A AcosA cos sin2 2

=

2 2A Acos A cos cos cos A2 22 2

2 2

= = +Q

2 2A Acos A sin sin cos A22 2 22 2

= = Q

J0010


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Trigonometric ratios of A in terms of A/2

2

Atan22tanAA

tan22

=

2 Acot 22cot A

Acot2

2

=

AsinA and cosA in terms of tan2

2

Atan2

2sinA

Atan22

=+

2

2

Atan2

2cosA

Atan22

=

+

J0010


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Trigonometric ratios of A interms of A/3

2A AsinA sin sin2 22 2

=

2 A Acos A cos cos2 22 2 =

2

2

A Atan tan2

2 2tanAA

tan2 2 2

=

J0010


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Find the value of

(a) sin18o (b) cos18o (c) tan18o

Solution

(a) Let = 18o 5 = 90o

2

+ 3

= 90o

2

= 90o 3

sin2 sin(90o 3 ) sin2 = cos3

2sincos = 4cos3 3cos

2sin = 4cos2 3

2sin = 4(1-sin2) 3

2sin = 4 4sin2 3 4sin2 + 2sin + 1 = 0

J0010


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Illustrative ProblemFind the value of


This is a quadratic equationin sin

2 2 22

sin2

+

=2 2 2 2 2

2 2

= =

= ,22oQ lies in the first quadrant

sin is positive 2 2sin2

=

(b) cos218o = 1-sin218o2

2 22

2

=

22 2 2 2 2

22

+=

4sin2 + 2sin + 1 = 0

Why ??

J0010


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Illustrative ProblemFind the value of


o2 22 2 2cos 2222

+=

22 2 2cos22

2

+ = o

Again = 18o lies in the first quadrant

cos18o > 0 22 2 2cos222

+ =o

sin22(c) tan22

cos22=

oo

o

2 2

2 22

22 2 2 22 2 2

2

= =

+ +

J0010


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Show that

sin59o sin13o + sin49o sin23o = cos 5o

Solution L.H.S

( ) ( )sin sin sin sin22 22 22 22= + o o o o

22 22 22 22 22 22 22 22cos sin cos sin2 22 2 2 2+ + = +

cos sin cos sin2 22 22 2 22 22= +o o o o

J0010


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Show that

sin59o sin13o + sin49o sin23o = cos5o

Solution

( )cos sin sin2 22 22 22= +o o o

22 22 22 22cos sin cos2 22 2

2 2

+ =

o

cos sin cos2 22 22 2= o o o

2 2 2 2.cos2 2

2 2

+ = o

2 2 cos22= o cos R.H.S2= =o

J0010


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cos cosIf cos

cos cos2

=

then prove that one of the

values of tan is tan cot2 2 2

Solution 2cos sin .....( )2 2 22 =Q

2cos cos .....( )2 2 22

+ =Q

from (1) and (2)

2 cos2tancos2 2

=

+

cos cos2 cos cos2

cos cos2

cos cos2

=

+

J0010


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Illustrative Problemcos cos

If coscos cos2

=

then prove that one of thevalues of tan is tan cot

2 2 2

Solution

cos cos cos cos2

cos cos cos cos2

+

= +

( ) ( )

( ) ( )

cos cos cos2 2

cos cos cos2 2

+

= + +2 2

2 2

sin . cos2 22 2

cos . sin2 22 2

= 2 2tan tan

2 2

=

2 2tan tan tan tan tan2 2 2 2 2

= = Proved.

J0010


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Identities with certain givencondition.

Four Types of Problems

(1) Identities involving sines or cosines of multiples andsubmultiples of the angle

(2) Identities involving squares of sines or cosines ofangle

Conditions :A+B+C = , /2, 2

J0011


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(3) Identities involving tangents or co-tangents of angles

(4) Identities which can be solved by transforming the

given identities in trigonometric form.

J0011


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(1) Identities involving sines orcosines of multiples and sub

multiples of the angle involved.

If A+B+C = , prove that

cos2A + cos2B cos2C = 1-4sinAsinBcosC

Solution L.H.S = (cos2A + cos2B) cos2C

A B A B2 2 2 2cos cos cos C2 2

2 2

+ =

= 2cos(A+B) cos(AB) (2cos2C 1)

= 2cos( C) cos(AB) 2cos2C +1

J0011


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cos2A + cos2B cos2C = 1-4sinAsinBcosC

Solution

= 1-2cosC {cos(A B) + cos ((A+B))}

= 1-2cosC {cos(A B) cos (A+B)}

= 1 4sinAsinBcosC Proved.

A B A B A B A BcosC sin sin2 2 2 2 2

+ + + + =

= 2cosC cos(AB) 2cos2C +1

J0011


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If A + B + C = , prove that

A B Ccos cos cos2 2 2+ +

A B Ccos cos cos2

2 2 2

=

Solution

A B CL.H.S cos cos cos

2 2 2

= + +

A B A BC2 2 2 2

cos cos cos2

2 2 2

+

= +A B A B C

cos cos sin22 2 2 2

+ = +

J0011


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Illustrative ProblemIf A + B + C = , prove that

A B Ccos cos cos

2 2 2+ +

A B Ccos cos cos2

2 2 2

=

Solution

C A B C Ccos cos sin cos2 2

2 2 2 2

= + A B C

A B C2 2 2 2

+ + = + + =Q

C A B Ccos cos sin2

2 2 2

= +

C A B Ccos cos cos2

2 2 2 2

= +

J0011


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Illustrative ProblemIf A + B + C = , prove that

A B Ccos cos cos

2 2 2+ +

A B Ccos cos cos2

2 2 2

=

Solution

C A B Ccos cos cos2

2 2 2

+ = +

C A B C C A Bcos cos cos2 2

2 2 2

+ + + + =

A B C A C B and B C A + + = + = + = Q

C B B A Acos cos cos2 2

2 2 2

+ + = A B C

cos cos cos22 2 2

= Proved.

J0011


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(2) Identities involving squares of sines orcosines of angle


cos2

A+cos2

B+cos2

C+2cosAcosBcosC = 1

Solution L.H.S = cos2A+cos2B+cos2C+2cosAcosBcosC

= cos2A+(1sin2B)+cos2C+2cosAcosBcosC

= 1+(cos2Asin2B)+cos2C+2cosAcosBcosC

= 1+cos(A+B)cos(AB)+cos2C+2cosAcosBcosC

= 1+cos(C)cos(AB)+cos2C+2cosAcosBcosC

J0011


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cos2A+cos2B+cos2C+2cosAcosBcosC = 1

Solution

= 1+cos(C)cos(AB)+cos2C+2cosAcosBcosC

= 1cosCcos(AB)+cos2C+2cosAcosBcosC

= 1cosC{cos(AB)+cos(A+B)+2cosAcosBcosC

= 12cosAcosBcosC+ 2cosAcosBcosC

= 1 Proved.

J0011


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Illustrative Problem(3) Identities involving tangents orco-tangents of angles

If A+B+C =, prove thatcotBcotC+cotCcotA+cotAcotB = 1

Solution A B C + + =Q

A B C + =

( ) ( )cot A B cot C + =

cot A cotB 2cotC

cot A cotB

=

+cot A cotB cot A cotC cotBcotC2 =

cotA cotB cotBcotC cotCcot A 2 + + = Proved.

J0011


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(4) Identities which can be solved bytransforming the given identities in

trigonometric form.

If xy+yz+zx = 1, prove that

( ) ( ) ( )2 2 2 2 2 2

x y z 2

x y z2 2 2 x y z2 2 2

+ + =

Solution Let x=tanA, y=tanB, z=tanC

xy yz zx 2+ + =Q

tanAtanB+tanBtanC+tanCtanA = 1

tanB(tanA+tanC) = 1 tanAtanC

J0011


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( ) ( ) ( )2 2 2 2 2 2x y z 2

x y z2 2 2 x y z2 2 2+ + =

Solution

tanA tanC 2cotB

tanA tanC tanB2

+ = =

tan(A C) tan B

2

+ =

A C B2

+ =

A B C2

+ + = A B C .....( )2 2 2 2 + + =

2 2 2

x y zL.H.S

x y z2 2 2= + +

2 2 2tan A tanB tanC

tan A tan B tan C2 2 2= + +

J0011


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( ) ( ) ( )2 2 2 2 2 2x y z 2

x y z2 2 2 x y z2 2 2+ + =

Solution

2 2 2

tanA tanB tanC2 2 2 2

tan A tan B tan C2 2 2 2

= + +

2 tan A tan B tan C2 2 22

= + +

2tan A tan B tan C2 2 2

2=

2 2 2tanA tanB tanC2 2 2 2. .tan A tan B tan C2 2 2 2

= ( ) ( ) ( )2 2 2xyz2

x y z2 2 2=

A B C2 2 2 + + = Q

[putting the values of tanA, tanB, tanC] Proved.

J0011


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Relation Between side and interiorangles of a polygon

No of Sides = n

Let O be a point inside the polygon

Polygon = n triangles

sum of all the angle of the triangle n .....( )222 2= o

A N

M

DC

BO

J0011


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Sum of all the angle of the trianglen .222= o

A N

M

DC

BO

Sum of all the angle of the triangle

= Sum of all the angles at O +

Sum of all interior angle of polygon

Sum of all interior angle of polygon + 3600= nx1800

Sum of all interior angle of polygon

= nx1800 - 3600

J0011


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Sum of interior angle of polygon

n 222222= o o

( n )2 2 2 2= o

(n )2 222= o

If the polygon is regular

( )n 2222

n

=

o

Each interior angle

Sum of all the angles at O= nx1800 - 3600

A N

M

DC

BO

J0011


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The angle in one regular polygon is tothat in another as 3:2; also the

number of sides in the first is twicethat in second ; how many sides thepolygons have ?

Solution

Let no. of sides of one regular polygon = x

No. of sides of other polygon = 2x

x2 2222

2x2Given x 2 2

222x

=

o

ox

22

x 22 = x x2 2 2 2 = x 2 =

No. of sides = 4 ,8

J0011


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Class Exercise - 1

If cotA tanA = 4, the value ofcot2A is

(a) 4

(b) 2

(c) 1

(d) 5

cot A tan A 2 =Q

Solution :-

2tanA 2

tanA =

2tan A22

tanA

=

2 2

tan A tan A22 2

2 2tan A tan A2 2 = =

2tan A2

2 = cot2A = 2


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Class Exercise - 2

Solution :-

Provethat

sin12 sin48 sin54 = .2

2

LHS = sin12 sin48 sin54

( )2

sin sin sin2 22 22 222

=

( ) ( )2 cos cos sin2222 2222 222

= +

[ ] ( )2

cos cos sin22 22 22222

=

[ ]

2

cos cos cos22 22 222=

2 2 2 2 2 2

2 2 2 2

+ +

= 2 2 2 2 2

2 2 2

+=

2 2 2RHS

22 2

= = =


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Class Exercise - 3

Solution :-

If sinx + siny = a and cosx + cosy

= b, show that sin(x + y) = . 2 2ab2

a b+

2ab = 2(sinx + siny)(cosx + cosy)

= 2(sinx cosx + sinx cosy + siny cosx + siny cosx)

= 2 sinx cosx + 2 siny cosy + 2(sinx cosy + cosx siny)

= sin2x + sin2y + 2 sin(x + y)

= 2 sin(x + y) . cos(x y) + 2 sin(x + y)2ab = 2 sin(x + y){1+ cos(x y)} ... (i)

and a2 + b2 = (sinx + siny)2 + (cosx + cosy)2


44/59

Class Exercise - 3

Solution :-

If sinx + siny = a and cosx + cosy

= b, show that sin(x + y) = . 2 2

ab2

a b+

= sin2x + sin2y + 2 sinx siny + cos2x+ cos2y + 2 cosx cosy

= 2 + 2 cos(x y) = 2(1 + cos(x y) ... (ii)

{ }

{ }2 2sin(x y) cos(x y)2ab2

cos(x y)2a b

+ + =

+ +RHS =

= sin(x + y) = LHS

[ From (i) and (ii)]


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Class Exercise - 4If a cos2 and bsin2 = c has and as its solution, then prove that

b2tan tanc a

+ =+

(i) c atan tanc a

= +(ii)

Solution :-

... (i)acos b sin c2 2 + =Q

( )22 2

a tan2 b tan2c

tan tan2 2

+ =

+ +

( ) ( )2 2a tan b tan c tan2 2 2 + = +

( ) 2a c tan b tan (c a)2 2 + + + = ... (ii)


46/59

Class Exercise - 4

If a cos2 and bsin2 = c has and as its solution, then prove that

Solution :-

and are the roots of equation (i),

tan and tan are the roots of equation (ii).

c atan tan

c a

=

+(Product of roots)

b2ta n ta n

a c + =

+

(Sum of roots)


47/59

Class Exercise - 5

tan x 2

+

If ,, and satisfy the equation

= 3 tan3x, then

tan tan tan tan + + +

(a) 1

(c) 0 (d) 2

2

2(b)


48/59

Class Exercise - 5


= 3 tan3x, then

tan x2

+

tan tan tan tan + + + Solution :-

tan x tan x2 22

+ = Q . (i)

( )22

tan x tan tan x tan x2 22

tan x2 2tanx. tan22

+ =

( )2

2

tan x tan x2 2tanx2

tanx2 tan x2 2

+ =

(1 + tanx)(1 3 tan2x) = (9 tanx 3 tan3x)(1 tanx)


49/59

Class Exercise - 5


= 3 tan3x, then

tan x2

+

tan tan tan tan + + +

Solution :-

3 tan4x 6 tan2x + 8 tanx 1 = 0 ... (ii)

,, and are the roots of equation (i),

are the roots of equation (ii).tan tan tan tan + + +

2

2

coefficient of tan xtan tan tan tan

coefficient tan x + + + = = 0


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Class Exercise - 6Prove that

cos3A cos3A + sin3A sin3A = cos3 2A.

= Q 2cos A cos A cos A2 2 2Solution :-

( ) = +22

cos A cos A cos A2 22

( )= 2 2sin A sin A sin A2 22

Similarly

+2 2cos A cos A sin A sin A2 2L.H.S

( ) ( )= + + 2 2

cos A cos A cos A sin A sin A sin A2 2 2 2 2 22 2

( ) ( )= + + 2 22 2

cos A cos A sin A sin A cos A sin A2 2 2 22 2


51/59

Class Exercise - 6Prove that

cos3A cos3A + sin3A sin3A = cos3 2A.

Solution :-

( )= +2 2

cos A A cos ( A)2 222 2

= +2 2

cos A cos A2 2

2 2

= +2 2

cos A cos ( A)2 222 2

1 = +

22

cos A cos A cos A2 2 2 2 2

2 2

= RHS= cos32A


52/59

Class Exercise - 7

= cos cos .cos

+ =2tan tan tan2 2 2

If then prove that

= Q cos cos . cos

Solution :-

2tan2

LHS

coscoscos

=

(1)

2 2

2 2

sin sin2cos22 2cos2

cos cos22 2

= = =

+


53/59

Class Exercise - 7

= cos cos .cos

+ =

2

tan tan tan2 2 2

If then prove that

Solution :-

cos2

cos coscoscos cos cos

2cos

= = + +

sin . sin22 2 tan . tan

2 2cos . cos22 2

+ +

= = +

= RHS


54/59

Class Exercise - 8If A + B + C = , prove that

( ) ( ) ( )cot B cot C cot C cot A cot A cot B

cos ecA.cosecB.cosecC

+ + +

=

Solution :-

cos B cos Ccot B cot C

sinB sin C+ = +Q

sinC cosB sinB cos C

sinB sinC

+=

( ) ( )sin B C sin A sinAcot B cot C

sinB sinC sinB sinC sinB sinC

+ + = = = . (i)

sin B

sinC sinBSimilarly, cotC + cotA = ... (ii)


55/59

Class Exercise - 8

If A + B + C = , prove that( ) ( ) ( )cot B cot C cot C cot A cot A cot B

cosecA.cosecB.cosecC

+ + +

=

Solution :-

LHS = (cotB + cotC)(cotC + cotA)(cotA + cotB)

and cotA + cotB =sin C

sinA sinB (iii)

= cosecA cosecB cosecC

sin A sinB sinC

sinB sinC sinC sin A sin A sinB=

2

sinA sinB sinC=


56/59

Class Exercise - 9

a btan tan

a b2 2

=

+

a cos bcos

a b cos

+ =

+

IF

than prove that

Solution :-

2

2

tan22cos

tan22

=

+

Q

= +

a b

tan tana b2 2Given .. (i)

2

2

a btan2

a b 2

a b tan2a b 2

+= + +


57/59

Class Exercise - 9

a btan tan

a b2 2

=

+

acos bcosa b cos

+ =+

IF

than prove that

Solution :-

2 2

2 2

(a b) cos (a b) sin2 2

(a b) cos (a b) sin2 2

+ =

+ +

2 2 2 2

2 2 2 2

a cos sin b cos sin

2 2 2 2

a cos sin b cos sin2 2 2 2

+ + =

+ +

acos b L.H.Sa b cos

+= =+


58/59

Class Exercise - 10The number of sides in two regularpolygons are 5 : 4 and the

difference between their angles is9. Find the number of sides in thepolygons.

Solution :-

Let the number of sides of regular polygons are5x and 4x respectively.

Each interior angle of regular polygons is

x x2 2 2 2 2 2and22 22

x x2 2

x x2 2 2 2 2 222 22 2

x x2 2

= Given


59/59

Class Exercise - 10The number of sides in two regular polygonsare 5 : 4 and the difference between theirangles is 9. Find the number of sides in the

polygons.

Solution :-

x x22 2 2 2

x x2 2 22

4 =

x x22 22 22 22 2

x22 22

+ =

2 2x 2

x22 22 = =

Sides are 10 and 8.

08. Trigonometric Ratios and Identities-2 - Copy

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