Lesson: Introduction to Trigonometry

Exercise 8.1 (11)

Question: 1

In ABC right-angled at B, AB 24 cm, BC 7 cm.





Determine:

(i)

sinA,

cosA

(ii)

sinC,

cosC

Solution

By the Pythagoras theorem in triangle ABC,

     

 

2 2 2

AC AB BC

     

 

2 2 2

AC 24 7

 

 

AC 576 49

 



AC 625

AC 625

Fig. Exc_8.1_1

(i)



 

sideoppositeto A BC

sin A

hypotenuse AC

(ii)



sin A



 

sideadjacent to A AB

cosA

Hypotenuse AC



cosA

(iii)



 

sideoppositeto C AB

sin C

hypotenuse AC



sin C



 

sideadjacent to C BC

cosC

Hypotenuse AC



cosC

Question: 2

In Fig. 8.13, find .

tanP cotR

Fig. Exc_8.1_2

Solution

By the Pythagoras theorem in ,

PQR

     

 

2 2 2

PR PQ QR

     

 

2 2 2

13 12 QR

 

 

QR 169 144

 



QR 25

QR 5



 



sideoppositeto P QR

tanP

sideadjacent to P PQ



tanP



 



sideadjacent to R QR

cotR

sideoppositeto R PQ



  

5 5

tanP cotR

12 12

 tanP cotR 0

Question: 3

If , calculate and .



sin A

cosA

tanA

Solution

Let us consider a triangle ABC, right-angled at B.

Fig. Exc_8.1_3

 

BC 3

sin A

AC 4

Let .

BC 3a

Thus, .

AC 4a

By the Pythagoras theorem in triangle ABC,

     

 

2 2 2

AC AB BC

     

 

2 2 2

4a AB 3a

 

 

2 2

16a 9a AB

 



AB 7a



cosA



tanA



Hence, and .



cosA



tanA

Question: 4

Given , find and .

15cot A 8

sinA

secA

Solution

Let us consider a triangle ABC, right-angled at B.

 

AB 8

cot A

BC 15

Let and .

AB 8a

BC 15a

By the Pythagoras theorem

     

 

2 2 2

AC AB BC

     

 

2 2 2

AC 8a 15a

 

 

2 2

AC 64a 225a

 



AC 289a

AC 17a

Fig. Exc_8.1_4



sinA



15a

17a



secA



Hence, and .



sinA



secA

Question: 5

Given , calculate all other trigonometric



secθ

ratios.

Solution

Let us consider a triangle ABC, right-angled at B.

 

AC 13

secθ

AB 12

Fig. Exc_8.1_5

Let and .

AC 13a

AB 12a

By the Pythagoras theorem,

     

 

2 2 2

AC AB BC

     

 

2 2 2

13a 12a BC

 

 

2 2

BC 169a 144a

 



BC 25a



BC 25a

BC 5a

  

BC 5a 5

sinθ

AC 13a 13

  

AB 12a 12

cosθ

AC 13a 13

  

BC 5a 5

tanθ

AB 12a 12

  

AB 12a 12

cot θ

BC 5a 5

  

AC 13a 13

cosecθ

BC 5a 5

Question: 6

If and are acute angles such that

A

B

, then show that .

cosA cosB

  A B

Solution

Let us consider a triangle ABC such that CD AB.



cosA cosB



AD BD

AC BC



AD AC

BD BC

Let .

 

AD AC

BD BC

AD aBD ......(1)

AC aBC ......(2)

Fig. Exc_8.1_6

In triangles CAD and CBD, by the Pythagoras

theorem,

     

 

2 2 2

CD AC AD ......(3)

     

 

2 2 2

CD BC BD ......(4)

From equations (3) and (4).

       

  

2 2 2 2

AC AD BC BD

       

  

2 2 2 2

aBC aBD BC BD

       

 

  

 

2 2 2 2

a BC BD BC BD



a 1

a 1

Put the value of in equation (2).

AC BC

Since the angles opposite to equal sides in a triangle

are equal.

Thus,

  A B

Question: 7

If , evaluate:



cot θ

(i)

  

 

1 sinθ 1 sinθ

1 cosθ 1 cosθ

(ii)

cot θ

Solution

Let us consider a triangle ABC, right-angled at B.

 

BC 7

cot θ

AB 8

Fig. Exc_8.1_7

Let and .

BC 7a

AB 8a

In triangle ABC, by the Pythagoras theorem,

     

 

2 2 2

AC AB BC

     

 

2 2 2

AC 8a 7a

 

 

2 2

AC 64a 49a

 



AC 113a

AC 113a



sinθ



113a



113



cosθ



113a



113

(i)

  

 

1 sinθ 1 sinθ

1 cosθ 1 cosθ

  

 



 



 



1 sin θ

1 sinθ 1 sinθ

1 cosθ 1 cosθ

1 cos θ

 



 

 



 



 

 

113







113



113



Thus,

  

 



 

1 sinθ 1 sinθ

1 cosθ 1 cosθ 64

(ii)

cot θ

 



cot θ cot θ

 



 

 



Thus, .



cot θ

Question: 8

If , check whether

3cot A 4

or not.



 



2 2

1 tan A

cos A sin A

1 tan A

Solution

Given:

3cot A 4



cot A

Let us consider a triangle ABC, right-angled at B.

Fig. Exc_8.1_8

 

AB 4

cot A

BC 3

Let and .

AB 4a

BC 3a

By the Pythagoras theorem,

     

 

2 2 2

AC AB BC

     

 

2 2 2

AC 4a 3a

 

 

2 2

AC 16a 9a

 



AC 25a



AC 25a

AC 5a



cosA



sinA



tanA



Now, find the value of .





1 tan A

 



 



 





 



 

 

1 tan A













16 9



Now, find the value of .



2 2

cos A sin A

   

  

   

   

2 2

4 3

cos A sin A

5 5

 

16 9

25 25



Hence, .



 



2 2

1 tan A

cos A sin A

1 tan A

Question: 9

In triangle ABC, right-angled at B, if , find



tanA

the value of:

(i)

sinAcosC cosAsinC

(ii)

cosAcosC sinAsinC

Solution

In the figure, ABC is a triangle, right-angled at B.

Fig. Exc_8.1_9

 

BC 1

tanA

Let and .

BC a

AB 3a

By the Pythagoras theorem,

     

 

2 2 2

AC AB BC

 

 

AC 3a a

 

 

2 2

AC 3a a

 



AC 4a



AC 4a

AC 2a



sinA



cosA



sinC



cosC



(i) Given:

sinAcosC cosAsinC

    

1 1 3 3

sinAcosC cosAsinC

2 2 2 2

 

1 3

4 4

1

Hence, .

 sinAcosC cosAsinC 1

(ii) Given:

cosAcosC sinAsinC

    

3 1 1 3

cosAcosC sinAsinC

2 2 2 2

 

3 3

4 4

 0

Hence, .

 cosAcosC sinAsinC 0

Question: 10

In PQR, right-angled at Q, PR QR 25 cm and







PQ 5 cm. Determine the values of and



sinP,cosP

tanP

Solution

Given:

PR QR 25 cm and PQ 5 cm.





Let .

PR x

Thus, .

 QR 25 x

Fig. Exc_8.1_10

In , by the Pythagoras theorem,

PQR

     

 

2 2 2

PR PQ QR

   

  

2 2

x 5 25 x

     

2 2

x 25 625 x 2 25 x

 50x 25 625



650

x 13

Thus, .

PR 13cm

  QR 25 13 12cm

 

QR 12

sinP

PR 13

 

PQ 5

cosP

PR 13

 

QR 12

tanP

PQ 5

Question: 11

State whether the following are true or false. Justify

your answer.

(i) The value of is always less than 1.

tanA

(ii) for some value of angle A.



secA

(iii) is the abbreviation used for the cosecant

cosA

of angle A.

(iv) is the product of cot and A.

cot A

(v) for some angle .



sinθ

Solution

(i) False.

Let us consider a triangle ABC, which is right-

angled at B.

Let and .

BC 12

AB 5

Fig. Exc_8.1_11_(i)

 

BC 12

tanA

AB 5

 

tanA 1

Hence, the value of is not always less

tanA

than 1.

(ii) True.

Let us consider a triangle ABC, which is right-

angled at B.

Fig. Exc_8.1_11_(ii)

 

AC 12

secA

AB 5

Let and .

AC 12a

AB 5a

By the Pythagoras theorem.

     

 

2 2 2

AC AB BC

     

 

2 2 2

12a 5a BC

 

 

2 2

BC 144a 25a

 



BC 119a

BC 10.9a ......(1)

But, BC should satisfy the condition,

   AC AB BC AC AB

   12a 5a BC 12a 5a

       12a 5a BC 12a 5a12a 5a BC 12a 5a

From equation (1).

BC 10.9a

Hence, this type of triangle can be drawn and

the given statement is true.

(iii) False.

is the abbreviation used for cosecant of

cosec A

angle A and the abbreviation used for cosine of

angle A is .

cosA

(iv) False.

cot A is the cotangent of .

A

(v) False.





sideoppositeto θ

sinθ

hypotenuse

We know that in a right-angled triangle,

hypotenuse is greater than the other two sides.

Thus, such value of is impossible.

sinθ

Exercise 8.2 (4)

Question: 1

Evaluate the following:

(i)

    sin60 cos30 sin30 cos60

(ii)

    

2 2 2

2tan 45 cos 30 sin 60

(iii)



  

cos45

sec30 cosec30

(iv)

    

    

sin30 tan45 cosec60

sec30 cos60 cot 45

(v)

    

  

2 2 2

2 2

5cos 60 4sec 30 tan 45

sin 30 cos 30

Solution

(i) Given:

    sin60 cos30 sin30 cos60

  

      

  

  

3 3 1 1

sin60 cos30 sin30 cos60

2 2 2 2

 

3 1

4 4



1

Thus,

     sin60 cos30 sin30 cos60 1

(ii)

    

2 2 2

2tan 45 cos 30 sin 60

 

   

       

   

   

2 2

2 2 2

3 3

2tan 45 cos 30 sin 60 2 1

2 2

  

3 3

4 4

 2

Thus,

     

2 2 2

2tan 45 cos 30 sin 60 2

(iii) Given:



  

cos45

sec30 cosec30





  



cos45

sec30 cosec30





2 2 3

 





2 2 2 3





2 2 2 6

Multiply numerator and denominator by

2 6 2 2

 

 

  

 

cos45 3 2 6 2 2

sec30 cosec30

2 2 2 6 2 6 2 2

 

   







2 2

2 3 6 2

2 6 2 2

 







2 3 6 2

24 8

 





2 3 6 2





18 6





3 2 6

Thus, .

 



  

cos45 3 2 6

sec30 cosec30 8

(iv) Given:

    

    

sin30 tan45 cosec60

sec30 cos60 cot 45

 

    



    

 

1 2

sin30 tan45 cosec60

2 1

sec30 cos60 cot 45







3 2







3 3 4

2 3

3 3 4

2 3







3 3 4

Multiply both numerator and denominator by

3 3 4

      

 

    

 

sin30 tan45 cosec60 3 3 4 3 3 4

sec30 cos60 cot 45

3 3 4 3 3 4

 







3 3 4

 





27 16 24 3

27 16





43 24 3

Thus,

     



    

sin30 tan45 cosec60 43 24 3

sec30 cos60 cot 45 11

(v) Given:

    

  

2 2 2

2 2

5cos 60 4sec 30 tan 45

sin 30 cos 30

 

 

 

 

    

 



  

 



 

 

2 2 2

2 2

1 2

5 4 1

5cos 60 4sec 30 tan 45

sin 30 cos 30

1 3

2 2

 

 

 

 





1 16

5 1

4 3

1 3

4 4

 



15 64 12



Thus,

    



  

2 2 2

2 2

5cos 60 4sec 30 tan 45 67

sin 30 cos 30 12

Question: 2

Choose the correct option and justify your choice:

(i)





 

2tan30

1 tan 30

(A)

sin60

(B)

cos60

(C)

tan60

(D)

sin30

(ii)

 



 

1 tan 45

(A)

tan90

(B) 1

(C)

sin45

(D) 0

(iii) is true when

sin2A 2sinA

A

(A)

0

(B)

30

(C)

45

(D)

60

(iv)





 

2tan30

1 tan 30

(A)

cos60

(B)

sin60

(C)

tan60

(D)

sin30

Solution

(i)



 

2tan30

1 tan 30

(A)

 

 



 



 

 



 

 

2tan30

1 tan 30











3 2

4 3



We know that .

 

sin60

(ii)

 

 

1 tan 45

(D)

 



 



 



1 1

1 tan 45

1 1







1 1

 0

(iii)

sin2A 2sinA

(A)

Put .

 A 0

 

 sin2A sin2 0

 sin0

 0

 2sinA 2sin0

 2 0

 0

is true when .

sin2A 2sinA

 A 0

(iv)



 

2tan30

1 tan 30

(C)

 

 



 



 

 



 

 

2tan30

1 tan 30









3 1



 3

We know that .

 tan60 3

Thus, .



 

 

2tan30

tan60

1 tan 30

Question: 3

 

 tan A B 3

 

 

tan A B

; , find A and B.

    0 A B 90

A B

Solution

 

 tan A B 3

We know that .

 tan60 3

 

  tan A B tan60

Thus,

  A B 60 ......(1)

 

 

tan A B

 

  tan A B tan30

Thus,

  A B 30 ......(2)

Add equations (1) and (2).

    A B A B 60 30

2A 90



A 45

Put the value of A in equation (1).

 A B 60

 45 B 60

 B 60 45

B 15

Thus, and .

 A 45

 B 15

Question: 4

State whether the following are true or false. Justify

your answer.

(i)

 

  sin A B sinA sinB

(ii) The value of increases as increases.

sinθ

(iii) The value of increases as increases.

cosθ

(iv) for all values of .

sinθ cosθ

(v) is not defined for

cot A

 A 0

Solution

(i) False.

 

  sin A B sinA sinB

Let and .

 A 30

 B 60

   

    sin A B sin 30 60

 sin90

1

    sinA sinB sin30 sin60

 

1 3

2 2





1 3

Hence, .

 

  sin A B sinA sinB

(ii) True.

Find the value of for increasing values of .

sinθ

Let and .

 θ 30 ,

45

60

  

sin30 0.5

  

sin45 0.707

  

sin60 0.866

 sin90 1

Thus, the value of increases as increases.

sinθ

(iii) False.

Find the value of for increasing values of

cosθ

Let and .

 θ 30 ,

45

60

  

cos30 0.866

  

cos45 0.707

  

cos60 0.5

  

cos30 0.866

  

cos45 0.707

  

cos60 0.5

Hence, the value of does not increase as

cosθ

increases.

(iv) False.

Find and for some values of .

sinθ

cosθ

Let and .

 θ 45

30

  θ 45 and30

and

 

sin45

 

cos45

and

 

sin30

 

cos30

Hence, for all values of .

sinθ cosθ

(v) True.



cosA

cot A

sinA

Let .

 A 0



 



cos0

cot 0

sin0

 

undefined

Hence, is not defined for

cot A

 A 0

Exercise 8.3 (7)

Question: 1

Evaluate:

(i)



sin18

cos72

(ii)



tan26

cot 64

(iii)

  cos48 sin42

(iv)

  cosec31 sec59

Solution

(i) Given:



sin18

cos72

 

  





 

sin 90 72

sin18

cos72 cos72

 



     

 

 



cos72

sin 90 72 cos72

cos72



1

Hence, .







sin18

cos72

(ii) Given:



tan26

cot 64

 

  





 

tan 90 64

tan26

cot 64 cot 64

 



     

 

 



cot 64

tan 90 64 cot 64

cot 64



Hence, .

1







tan26

cot 64

(iii) Given:

  cos48 sin42

 

        cos48 sin42 cos 90 42 sin42

 

        

 

 

sin42 sin42 cos 90 42 sin42

 0

Hence, .

   cos48 sin42 0

(iv) Given:

  cosec31 sec59

 

        cosec31 sec59 cosec 90 59 sec59

 

        

 

 

sec59 sec59 cosec 90 59 sec59

 0

Hence, .

   cosec31 sec59 0

Question: 2

Show that:

(i)

    tan48 tan23 tan42 tan67 1

(ii)

     cos38 cos52 sin38 sin52 0

Solution

(i) Given:

    tan48 tan23 tan42 tan67 1

We know that .

 

 tan 90 θ cot θ

 

  

 

 

       

 

 

 

tan 90 42

tan48 tan23 tan42 tan67 tan 90 67

tan42 tan67

    cot 42 cot 67 tan42 tan67

    cot 42 tan42 cot 67 tan67

   

  

   

 

   

1 1

cot 42 cot 67

1

Thus, .

    tan48 tan23 tan42 tan67 1

(ii) Given:

     cos38 cos52 sin38 sin52 0

We know that .

 

 cos 90 θ sinθ

 

  

 

 

        

 

  

 

cos 90 52

cos38 cos52 sin38 sin52 cos 90 38

sin38 sin52

     sin52 sin38 sin38 sin52

 0

Thus, .

     cos38 cos52 sin38 sin52 0

Question: 3

If ,where is an acute angle,

 

  tan2A cot A 18

find the value of A.

Solution

Given:

 

  tan2A cot A 18

 

  tan2A cot A 18

   

    cot 90 2A cot A 18

    90 2A A 18

    2A A 90 18

 3A 108





108

 A 36

Hence, A is .

36

Question: 4

If , prove that .

tanA cotB

  A B 90

Solution

Given:

 

   

 

 

tanA cotB tan 90 B cotB

 

  tanA tan 90 B

  A 90 B

  A B 90

Hence proved.

Question: 5

If , where 4A is an acute

 

  sec4A cosec A 20

angle, find the value of A.

Solution

Given:

 

  sec4A cosec A 20

 

  sec4A cosec A 20

     

       

 

 

cosec 90 4A cosec A 20 cosec 90 4A sec4A

    90 4A A 20

    4A A 90 20

 5A 110





110

 A 22

Hence, .

 A 22

Question: 6

If A, B and C are interior angles of a triangle ABC,

then show that



 



 

 

B C A

sin cos

2 2

Solution

By the angle sum property of a triangle,

      A B C 180

      B C 180 A

Divide both sides by 2.

     



B C 180 A

2 2

   

  

B C A

2 2

Take sin of angles on both sides.



   

  

   

   

B C A

sin sin 90

2 2

 



 

 

cos

Hence, .



 



 

 

B C A

sin cos

2 2

Question: 7

Express in terms of trigonometric

  sin67 cos75

ratios of angles between and .

0

45

Solution

Given:

  sin67 cos75

   

          sin67 cos75 sin 90 23 cos 90 15

   cos23 sin15

Hence,

      sin67 cos75 cos23 sin15

Exercise 8.4 (5)

Question: 1

Express the trigonometric ratios sin A, sec A and tan

A in terms of cot A.

Solution

We know that,

 

2 2

cosec A 1 cot A





2 2

1 1

cosec A 1 cot A

 

 

 



 

1 1

sin A sinA

1 cot A cosec A



Thus,





sinA

1 cot A

We know that,

and



sinA

tanA

cosA



cosA

cot A

sinA

Thus,



tanA

cot A

We know that,

 

2 2

sec A 1 tan A

 

sec A 1

cot A





cot A 1

sec A

cot A





cot A 1

secA

cot A

Thus,





sinA

1 cot A

and .





cot A 1

secA

cot A



tanA

cot A

Question: 2

Write all the other trigonometric ratios of in

A

terms of sec A.

Solution

We know that,



cosA

secA

Now,

 

2 2

sin A cos A 1

 

2 2

sin A 1 cos A

 

sinA 1 cos A

 

 

 

 

sinA 1

secA





sec A 1

sinA

sec A





sec A 1

sinA ......(1)

secA

Now, we know that,

 

2 2

tan A 1 sec A

 

2 2

tan A sec A 1

Now, for ,

cot A



cosA

cot A

sinA

From equation (1).

 

tanA sec A 1



cosA

cot A

sinA





secA

sec A 1

secA





sec A 1

We know that .



cosec A

sinA

From equation (1).



cosec A

sinA





secA

sec A 1

Question: 3

Evaluate:

(i)

  

2 2

sin 63 sin 27

cos 17 cos 73

(ii)

    sin25 cos65 cos25 sin65

Solution

(i) Given:

  

2 2

sin 63 sin 27

cos 17 cos 73

 

    

 

  

 



  

    

 

 

2 2

sin 90 27 sin 27

sin 63 sin 27

cos 17 cos 73

cos 90 73 cos 73

 

  



  

cos27 sin 27

sin73 cos 73

  



  

2 2

cos 27 sin 27

sin 73 cos 73



1

Hence, .

  



  

2 2

sin 63 sin 27

cos 17 cos 73

(ii) Given:

    sin25 cos65 cos25 sin65

 

   

 

     

 

    

 

 

sin25 cos 90 25

sin25 cos65 cos25 sin65

cos25 sin 90 25

     sin25 sin25 cos25 cos25

   

2 2

sin 25 cos 25

1

Hence, .

     sin25 cos65 cos25 sin65 1

Question: 4

Choose the correct option. Justify your choice.

(i)

 

2 2

9sec A 9tan A

(A) 1

(B) 9

(D) 0

(ii)

  

    1 tanθ secθ 1 cot θ cosecθ

(A) 0

(B) 1

(D)

1

(iii)

  

  secA tanA 1 sinA

(A)

secA

(B)

sinA

(C)

cosec A

(D)

cosA

(iv)







1 tan A

1 cot A

(A)

sec A

(B)

1

(C)

cot A

(D)

tan A

Solution

(i) Given:

 

2 2

9sec A 9tan A

(B)

 

  

2 2 2 2

9sec A 9tan A 9 sec A tan A

 

   

 

2 2

9 1 sec A tan A 1

 9

Thus, .

 

2 2

9sec A 9tan A 9

(ii) Given:

  

   1 tanθ secθ 1 cot θ cosecθ

(C)

  

 

 

 

 

 

 

 

    

 

 

 

 

 

 

 

sinθ 1

cosθ cosθ

1 tanθ secθ 1 cot θ cosecθ

cosθ 1

sinθ sinθ

   

  



  

  

cosθ sinθ 1 sinθ cosθ 1

cosθ sinθ

   

 



2 2

sinθ cosθ 1

sinθcosθ

  



2 2

sin θ cos θ 2sinθcosθ 1

sinθcosθ

We know that .

 

2 2

sin θ cos θ 1

Thus,

  

 

    

1 2sinθcosθ 1

1 tanθ secθ 1 cot θ cosecθ

sinθcosθ



2sinθcosθ

sinθcosθ

 2

Hence,

  

    1 tanθ secθ 1 cot θ cosecθ 2

(iii) Given:

  

 secA tanA 1 sinA

(D)

    

 

    

 

 

1 sinA

secA tanA 1 sinA 1 sinA

cosA cosA

 



 

 

 

 

1 sinA

cosA





1 sin A

cosA

 

  

 

2 2

cos A

1 sin A cos A

cosA



 cosA

Hence,

  

  secA tanA 1 sinA cosA

(iv) Given:



1 tan A

1 cot A

(D)







sin A

1 tan A

cos A

1 cot A

sin A







2 2

cos A sin A

cos A

sin A cos A

sin A



sin A

cos A



tan A

Hence, .







1 tan A

tan A

1 cot A

Question: 5

Prove the following identities, where the angles

involved are acute angles for which the expressions

are defined.

(i)

 



 



1 cosθ

cosecθ cot θ

1 cosθ

(ii)



 



cosA 1 sinA

2secA

1 sinA cosA

(iii)

  

 

tanθ cot θ

1 secθ cosecθ

1 cot θ 1 tanθ

(iv)







1 secA sin A

secA 1 cosA

(v)

 

 

 

cosA sinA 1

cosec A cot A

cosA sinA 1

, using the identity

 

2 2

cosec A 1 cot A

(vi)



 



1 sinA

secA tanA

1 sinA

(vii)







sinθ 2sin θ

tanθ

2cos θ cosθ

(viii)

   

     

2 2

sinA cosec A cosA secA 7 tan A cot A

(ix)

  

  



cosec A sinA secA cosA

tanA cot A

(x)

 

 

 

 

 

 

 

1 tan A 1 tanA

tan A

1 cot A 1 cot A

Solution

(i) Given:

 



 



1 cosθ

cosecθ cot θ

1 cosθ

LHS

 

 

  

 

 

1 cosθ

cosecθ cot θ

sinθ sinθ

 





1 cosθ

sin θ

 



 

  

 



2 2

1 cosθ

1 cos θ sin θ

1 cos θ



  

 



 

1 cosθ 1 cosθ

 







1 cosθ

 RHS

Hence proved.

(ii) Given:



 



cosA 1 sinA

2secA

1 sinA cosA

LHS

     

  

  



 

 

cosA cosA 1 sinA 1 sinA

cosA 1 sinA

1 sinA cosA 1 sinA cosA

  

  





2 2

cos A 1 sin A 2sinA

1 sinA cosA

  

  





2 2

sin A cos A 1 2sinA

1 sinA cosA

  

 

 

  

 



2 2

1 1 2sinA

sin A cos A 1

1 sinA cosA



 

  







2 1 sinA

1 sinA cosA

 



cosA

 2secA

 RHS

Hence proved.

(iii) Given:

  

 

tanθ cot θ

1 secθ cosecθ

1 cot θ 1 tanθ

LHS

  

 

sinθ cosθ

tanθ cot θ

cosθ sinθ

1 cot θ 1 tanθ

1 1

sinθ cosθ

 

 

sinθ cosθ

cosθ sinθ

sinθ cosθ cosθ sinθ

sinθ cosθ

   

 

 

2 2

sin θ cos θ

cosθ sinθ cosθ sinθ sinθ cosθ

 

 

 

 



 

2 2

1 sin θ cos θ

sinθ cosθ cosθ sinθ

 

 





 



 

3 3

1 sin θ cos θ

sinθ cosθ sinθcosθ

 

 

  

 



 



 

2 2

sinθ cosθ sin θ cos θ sinθcosθ

sinθ cosθ sinθcosθ



 

  

 

2 2

1 sinθcosθ

sin θ cos θ 1

sinθcosθ



 

1 sinθcosθ

sinθcosθ sinθcosθ

 secθ cosecθ 1

 RHS

Hence proved.

(iv) Given:







1 secA sin A

secA 1 cosA

LHS





1 secA

cosA

secA

cosA





cosA 1

cosA

 cosA 1

Multiply both numerator and denominator by

 

1 cosA

  

   

 





 

1 cosA 1 cosA

1 cos A

1 cosA 1 cosA

 

 

  

 



2 2

sin A

1 cos A sin A

1 cosA



 RHS

Hence proved.

(v) Given:

 

 

 

cosA sinA 1

cosec A cot A

cosA sinA 1

Divide both numerator and denominator in LHS

by .

sinA

LHS

 



 

cosA sinA 1

sinA sinA sinA

cosA sinA 1

sinA sinA sinA

 



 

cot A 1 cosec A

Multiply both numerator and denominator by

 

 cot A 1 cosec A

 

   

   

cot A 1 cosec A cot A 1 cosec A

 

   

 



 

2 2

cot A 1 cosec A

    



  

2 2

cot A 1 cosec A 2cot A 2cosec A 2cot Acosec A

cot A 1 cosec A 2cosec A

   

  



  

2 2

2cosec A cosec A cot A 2 cot A cosec A

cot A cosec A 1 2cosec A

  

 



  

cosec A cot A 2cosec A 2

1 1 2cosec A

  

 

 





cosec A cot A 2cosec A 2

2cosec A 2

 cosec A cot A

Hence proved.

 RHS

(vi) Given:



 



1 sinA

secA tanA

1 sinA

Multiply both numerator and denominator in

LHS by .

1 sinA

LHS

  

 





  

1 sinA 1 sinA

1 sinA

1 sinA 1 sinA 1 sinA







1 sinA

1 sin A



 

  

 

2 2

1 sinA

1 sin A cos A

cos A







1 sinA

cosA

 

1 sinA

cosA cosA

 secA tanA

 RHS

Hence proved.

(vii) Given:







sinθ 2sin θ

tanθ

2cos θ cosθ

LHS

 







sinθ 1 2sin θ

sinθ 2sin θ

2cos θ cosθ

cosθ 2cos θ 1

 



 

  

 

 

2 2

sinθ 1 2sin θ

1 sin θ cos θ

cosθ 2 1 sin θ 1



 







sinθ 1 2sin θ

cosθ 1 2sin θ

 tanθ

 RHS

Hence proved.

(viii) Given:

   

     

2 2

sinA cosec A cosA secA 7 tan A cot A

LHS

   

  

2 2

sinA cosec A cosA secA

 

 



 

  

 

2 2

sin A cosec A 2sinAcosec A

cos A sec A 2cosAsecA

   

 

  

 



 

   

 

   

 

   

 

2 2 2 2

sin A cos A cosec A sec A

1 1

2sinA 2cosA

sinA cosA

      

2 2

1 1 cot A 1 tan A 2 2

  

2 2

7 tan A cot A

 RHS

Hence proved.

(ix) Given:

  

  



cosec A sinA secA cosA

tanA cot A

Given:

  

  

    

  

  

1 1

cosec A sinA secA cosA sinA cosA

sinA cosA

  

 



  

  

2 2

1 sin A 1 cos A

sinA cosA



2 2

cos Asin A

sinAcosA

 sinAcosA

Now take RHS.





1 1

sinA cosA

tanA cot A

cosA sinA





2 2

sin A cos A

sinAcosA





2 2

sinAcosA

sin A cos A

 

  

 

2 2

sinAcosA sin A cos A 1

Thus, LHS RHS.



Hence proved.

(x) Given:

 

 

 

 

 

 

 

1 tan A 1 tanA

tan A

1 cot A 1 cot A

LHS



 





 



 



sin A

1 tan A

cos A

1 cot A

sin A





2 2

cos A+sin A

cos A

sin A cos A

sin A

 

  

 

2 2

cos A

sin A cos A 1

sin A





sin A

cos A



tan A

Now take middle term.

  

 



 

  

 

1 tanA 1 tan A 2tanA

1 cot A 1 cot A 2cot A

 

 





 



 

 

2 2

1 tan A sec A

sec A 2tanA

cosec A 2cot A

1 cot A cosec A









1 2sinA

cos A cosA

1 2cosA

sin A sinA







1 2sinAcosA

cos A

1 2sinAcosA

sin A



sin A

cos A



tan A

Thus, .

 

 

 

 

 

 

 

1 tan A 1 tanA

tan A

1 cot A 1 cot A

Hence proved.