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Page No 6.42:

Answer:1

We know that,

So,

 



Page No 6.42:

Answer:2

We know that,

So,



Page No 6.42:

Answer:3

We know that, .

So,



Page No 6.42:

Answer:4

We know that,

So,

 

 



Page No 6.42:

Answer:5

We know that,

So,



Page No 6.42:

Answer:6

We know that,

So,

 

                     

 



Page No 6.42:

Answer:7

We know that,

Multiplying both numerator and the denominator by, we have

 

 



Page No 6.42:

Answer:8

We know that,

Multiplying the both numerator and the denominator by, we have


 



Page No 6.42:

Answer:9

We know that,

So,



Page No 6.42:

Answer:10

We know that,

So,

 

 



Page No 6.42:

Answer:11

We know that,

Multiplying numerator and denominator under the square root by , we have



Page No 6.42:

Answer:12

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have



Page No 6.42:

Answer:13

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have



Page No 6.42:

Answer:14

We have to prove

We know that, .

Multiplying both numerator and denominator by , we have

 

              



Page No 6.43:

Answer:15

We have to prove

We know that,

So,



Page No 6.43:

Answer:16

We have to prove

We know that,

So,



Page No 6.43:

Answer:17

We have to prove

We know that,

 



Page No 6.43:

Answer:18

We have to prove

We know that,

So,

                                                



Page No 6.43:

Answer:19

We have to prove

We know that,

So,



Page No 6.43:

Answer:20

We have to prove

We know that,

So,

                       



Page No 6.43:

Answer:21

We have to prove

We know that,

So,

                                                



Page No 6.43:

Answer:22

We have to prove

We know that,

So,



Page No 6.43:

Answer:23

(i) We have to prove

We know that,

So,

                     
 

(ii) We have to prove

We know that,

So,

                     



Page No 6.43:

Answer:24

We have to prove

We know that,

So,

                                      



Page No 6.43:

Answer:25

We have to prove

We know that,

So,


                              



Page No 6.43:

Answer:26

We have to prove

We know that,

Multiplying the denominator and numerator of the second term by , we have

                              



Page No 6.43:

Answer:27

We have to prove that .

We know that,

So,

                                      



Page No 6.43:

Answer:28

We have to prove

Consider the expression

                

Again, we have

                    



Page No 6.43:

Answer:29

We have to prove

We know that,

Multiplying the numerator and denominator by , we have



Page No 6.43:

Answer:30

We need to prove

Now, using in the L.H.S, we get


                             

Further using the identity, we get

Hence



Page No 6.43:

Answer:31

We need to prove

Solving the L.H.S, we get

         

Further using the identity, we get

Hence proved.



Page No 6.43:

Answer:32

We need to prove

Solving the L.H.S, we get

                               ()

Further using the identity, we get

Hence proved.



Page No 6.43:

Answer:33

We need to prove

Solving the L.H.S, we get

Using , we get

                     

Hence proved.



Page No 6.43:

Answer:34

We need to prove

Using the property , we get

LHS =

Further using the identity, , we get


                 = RHS

Hence proved.



Page No 6.43:

Answer:35

We need to prove

Here, we will first solve the LHS.

Now, using , we get

Further, multiplying both numerator and denominator by , we get

Now, using the property, we get

So,

= RHS

Hence proved.



Page No 6.43:

Answer:36

We need to prove

Now, multiplying the numerator and denominator of LHS by , we get

Further using the identity, , we get

Hence proved.



Page No 6.43:

Answer:37

(i) We need to prove

Here, rationalising the L.H.S, we get

Further using the property, , we get

So,

Hence proved.

(ii) We need to prove 

Here, rationaliaing the L.H.S, we get

Further using the property, , we get

So,

Hence proved.



Page No 6.43:

Answer:38

We need to prove

Here, rationaliaing the L.H.S, we get

Further using the property, , we get

So,

Hence proved.



Page No 6.43:

Answer:39

We need to prove

Here, we will first solve the L.H.S.

Now, using , we get

Further using the property , we get

So,

Hence proved.

 



Page No 6.43:

Answer:40

We need to prove

Now, rationalising the L.H.S, we get

Using and, we get

Hence proved.



Page No 6.43:

Answer:41

We need to prove

Solving the L.H.S, we get

Further using the property , we get

So,

                

Hence proved.



Page No 6.43:

Answer:42

We need to prove

Solving the L.H.S, we get

                              
                               = RHS

Hence proved.

 



Page No 6.43:

Answer:43

We need to prove

Using the identity, we get



Further, using the property , we get

So,

                                                   

Hence proved.



Page No 6.44:

Answer:44

We need to prove .

Using the property , we get

Now, using , we get

Further, using the property, , we get

Hence proved.



Page No 6.44:

Answer:45

In the given question, we need to prove .

Here, we will first solve the LHS.

Now, using , we get

On further solving by taking the reciprocal of the denominator, we get,

Hence proved.



Page No 6.44:

Answer:46

In the given question, we need to prove

Here, we will first solve the LHS.

Now, using, we get

On further solving by taking the reciprocal of the denominator, we get,

Now, taking common from both the numerator and the denominator, we get

Hence proved.

 



Page No 6.44:

Answer:47

(i) We have to prove the following identity-

Consider the LHS.

1+cosθ+sinθ1+cosθ-sinθ

  

= RHS

Hence proved.

(ii) We have to prove the following identity-

Consider the LHS.

sinθ-cosθ+1sinθ+cosθ-1

   
 

              (Divide numerator and denominator by)

RHS

Hence proved.

(iii) We have to prove the following identity-

cos θ - sin θ + 1cos θ + sin θ - 1 = cosec θ + cot θ

Consider the LHS.

cos θ - sin θ + 1cos θ + sin θ - 1

=cos θ - sin θ + 1cos θ + sin θ - 1×cos θ + sin θ + 1cos θ + sin θ + 1=cos θ + 12 - sinθ2cos θ + sin θ2 - 12=cos2θ+ 1 + 2 cos θ - sin2θcos2θ + sin2θ + 2 cos θ sin θ - 1=cos2θ+ 1 + 2 cos θ - 1 - cos2θ1 + 2 cos θ sin θ - 1 =2 cos2θ + 2 cos θ2 cos θ sin θ=2 cos θ cos θ + 12 cos θ sin θ=cos θ + 1sin θ=cos θsin θ + 1sin θ=cot θ + cosec θ

= RHS

Hence proved.

(iv)
Consider the LHS.
sinθ+cosθtanθ+cotθ=sinθ+cosθsinθcosθ+cosθsinθ=sinθ+cosθsin2θ+cos2θsinθ×cosθ=sinθ+cosθsinθ×cosθ                       sin2θ+cos2θ=1=1cosθ+1sinθ=secθ+cosecθ
= RHS
Hence proved.



Page No 6.44:

Answer:48

In the given question, we need to prove .

Here, we will first solve the L.H.S.

Now, using, we get

On further solving, we get

                               

Similarly we solve the R.H.S.

Now, using, we get

On further solving, we get

                               

So, L.H.S = R.H.S

Hence proved.



Page No 6.44:

Answer:49

In the given question, we need to prove

Now, using and in L.H.S, we get

Further, using the identity, we get

Since L.H.S = R.H.S

Hence proved.



Page No 6.44:

Answer:50

In the given question, we need to prove

Now, using and in the L.H.S, we get

Solving further, we get

Hence proved.



Page No 6.44:

Answer:51

In the given question, we need to prove

Using and, we get

Further, using the property , we get

Hence proved.



Page No 6.44:

Answer:52

In the given question, we need to prove

Using the identity, we get

Further, using the property , we get

                                              

Hence proved.



Page No 6.44:

Answer:53

In the given question, we need to prove .

Using the property , we get

So,

1+cosθ-sin2θsinθ1+cosθ=1+cosθ-1-cos2θsinθ1+cosθ=cosθ+cos2θsinθ1+cosθ

Solving further, we get

Hence proved.

 



Page No 6.44:

Answer:54

In the given question, we need to prove

Using the property and, we get

Taking the reciprocal of the denominator, we get

Further, using the identity, we get

Hence proved.



Page No 6.44:

Answer:55

In the given question, we are given

We need to prove

Here L.H.S is

Now, solving the L.H.S, we get

Further using the property, we get

So,

Now, solving the R.H.S, we get

So,

Further using the property, we get,

So,

Hence proved.



Page No 6.44:

Answer:56

In the given question, we need to prove

Now, using the identity in L.H.S, we get

Further using, we get

Also, from the identity , we get

Hence proved.



Page No 6.44:

Answer:57

In the given question, we need to prove

Now, using and in L.H.S, we get

Further using the identity, we get

Further using the identity , we get

Now, from the identity, we get

So,

Hence proved.



Page No 6.44:

Answer:58

(i) In the given question, we need to prove

Taking common from the numerator and the denominator of the L.H.S, we get

Now, using the property , we get

Using , we get

Taking common from the numerator, we get

Using and , we get

Now, using the property , we get

Hence proved.

(ii)
Consider the LHS.
1+secθ-tanθ1+secθ+tanθ=secθ-tanθ+11+secθ+tanθ=secθ-tanθ+sec2θ-tan2θ1+secθ+tanθ                           sec2θ-tan2θ=1=secθ-tanθ1+secθ+tanθ1+secθ+tanθ=secθ-tanθ=1cosθ-sinθcosθ=1-sinθcosθ
= RHS
Hence proved.



Page No 6.44:

Answer:59

We have to prove

We know that,

So, we have

So, we have

Hence proved.



Page No 6.44:

Answer:60

We have to prove

We know that, .

So,

                                                           

                                                         

Hence proved.



Page No 6.44:

Answer:61

We have to prove

We know that,

Consider the LHS.

Now, consider the RHS.

∴ LHS = RHS

Hence proved.



Page No 6.44:

Answer:62

We have to prove

We know that,

So,

                                                      

                                                      

                                                      

Hence proved.



Page No 6.45:

Answer:63

(i) We have to prove

So,

                                         

                                         

Hence proved.

(ii) We have to prove 

We know that, 

So,

            

            

            

Hence proved.



Page No 6.45:

Answer:64

We have to prove

We know that,

So,

            

            

            

Hence proved.



Page No 6.45:

Answer:65

We have to prove

We know that, .

So,



  

 
 

Hence proved.



Page No 6.45:

Answer:66

We have to prove

We know that,

So,

                                             

                                             

Hence proved.



Page No 6.45:

Answer:67

We have to prove .

We know that,

So,

                           

                           

Multiplying both the numerator and denominator by , we have

Hence proved.



Page No 6.45:

Answer:68

We have prove that



We know that,

So,

             

             

             

             

Now,

                              

Hence proved.



Page No 6.45:

Answer:69

We know that,

So have,

Hence proved.



Page No 6.45:

Answer:70

(i) We have to prove

Now,
 
                      

Hence proved.

(ii) We have to prove 

Now,



Hence proved.



Page No 6.45:

Answer:71

We have to prove

Now,



Hence proved.



Page No 6.45:

Answer:72

We have to prove

We know that,

So,

Hence proved.



Page No 6.45:

Answer:73

We have to prove

We know that,

So,

Hence proved.



Page No 6.45:

Answer:74

Given that,

We have to prove

We know that,

So,

Hence proved. 



Page No 6.45:

Answer:75

Given that,



We have to prove

We know that,

Squaring and then adding the above two equations, we have



Page No 6.45:

Answer:76

Given that,



We have to prove

We know that

Now from the first equation, we have

 

Again from the second equation, we have

Therefore, we have


                      

Hence proved.



Page No 6.45:

Answer:77

Given that,

We have to prove

Adding both the equations, we get

Also.

Therefore, we have

Hence proved.



Page No 6.45:

Answer:78

Given:

We have to prove

We know that

So, we have



Hence proved.



Page No 6.45:

Answer:79

Given:

We have to prove that 5sin θ – 3cos θ = ±3.

We know that,

Squaring the given equation, we have



⇒                                                   (5sin θ – 3cos θ)2 = 9
⇒                                                       5sin θ – 3cos θ = ±3

Hence proved.

 



Page No 6.45:

Answer:80

Given:



We have to prove

We know that,

Now, squaring and adding the two equations, we get



Hence proved.



Page No 6.45:

Answer:81

Given:

We have to prove

We know that,

Multiplying the two equations, we have

 


Hence proved.



Page No 6.45:

Answer:82

Given:

We have to prove

Now,

Therefore, we have



Hence proved.



Page No 6.45:

Answer:83

(i) We have,

                                     

                                     

(ii) We have,

                                    

(iii) We have,

                                      

(iv) We have,

Multiplying both the numerator and the denominator by, we have

(v) We have,

Multiplying both the numerator and the denominator by , we have

                          

                          

                          



Page No 6.46:

Answer:84

Given:

We have to prove

From the given equation, we have

 

Therefore, we have

 

Hence proved.



Page No 6.46:

Answer:85

Given:

Let us assume that

We know that,

Then, we have

Therefore, we have

Taking the expression with the positive sign, we have



Page No 6.46:

Answer:86

Given:

Squaring the given equation, we have

Squaring the last equation, we have

Therefore, we have


                       

Hence proved.



Page No 6.46:

Answer:87

Given:


We have to prove that .

Squaring the above equations and then subtracting the third from the sum of the first two, we have


 


Hence proved.



Page No 6.53:

Answer:1

Given:

Now, we have to find all the other trigonometric ratios.

We have the following right angle triangle.

From the above figure,

Perpendicular=Hypotenuse2-Base2


Therefore,



Page No 6.53:

Answer:2

Given:

We have to find all the trigonometric ratios.

We have the following right angle triangle.

From the above figure,

Base=Hypotenuse2-Perpendicular2BC=AC2-AB2BC=22-12BC=1



Page No 6.53:

Answer:3

Given:
We have to find the value of the expression

We know that,

Therefore, the given expression can be written as

Hence, the value of the given expression is .



Page No 6.53:

Answer:4

Given:
We have to find the value of the expression .

From the above figure, we have


Therefore,

Hence, the value of the given expression is 19.



Page No 6.53:

Answer:5

Given:
We have to find the value of the expression .

From the above figure, we have

Therefore,

Hence, the value of the given expression is 25.



Page No 6.53:

Answer:6

Given:
We have to find the value of the expression

We know that,

Using the identity , we have

Hence, the value of the given expression is 35.



Page No 6.53:

Answer:7

Given:
We have to find the value of the expression

We know that,

Therefore,

Hence, the value of the given expression is 2.



Page No 6.53:

Answer:8

Given:

We have to find the value of the expression .

We know that,

Therefore,

Hence, the value of the given expression is 218.



Page No 6.53:

Answer:9

Given:

We have to find the value of the expression .

We have,




Therefore,

Hence, the value of the expression is 10.



Page No 6.53:

Answer:10

Given:

We have to find the value of .

Therefore,

Hence, the value of the expression is 13.



Page No 6.53:

Answer:11

Given:

We have to find the value of the expression .

Now,

Therefore,

Hence, the value of the expression is 3.



Page No 6.53:

Answer:12

Given:

We have to find the value of .

Now,

Hence,



Page No 6.54:

Answer:1

An identity is an equation which is true for all values of the variable (s).

For example,

Any number of variables may involve in an identity.

An example of an identity containing two variables is

The above are all about algebraic identities. Now, we define the trigonometric identities.

An equation involving trigonometric ratios of an angle (say) is said to be a trigonometric identity if it is satisfied for all valued of for which the trigonometric ratios are defined.

For examples,

sin2θ+cos2θ=11+tan2θ=sec2θ1+cot2θ=cosec2θ



Page No 6.54:

Answer:2

We have,



Page No 6.54:

Answer:3

We have,



Page No 6.54:

Answer:4

We have,



Page No 6.54:

Answer:5

Given:

We know that,

Therefore,


Hence,



Page No 6.54:

Answer:6

Given:

We know that,

Therefore,

 
Hence,



Page No 6.54:

Answer:7

We have,

We know that,

Therefore, cosec290°-θ-tan2θ=1



Page No 6.54:

Answer:8

We have,

We know that,

Therefore, sinAcos90°-A+cosAsin90°-A=1



Page No 6.54:

Answer:9

We have,

We know that,

Therefore, cot2θ-1sin2θ=-1



Page No 6.54:

Answer:10

Given:

x = a sinθ and y = b cosθ

So,

b2x2+a2y2=b2asinθ2+a2bcosθ2=a2b2sin2θ+a2b2cos2θ=a2b2sin2θ+cos2θ

We know that,

Therefore, b2x2+a2y2=a2b2



Page No 6.54:

Answer:11

Given:

We know that,


We have,

Hence, the value of cotθ + cosecθ is 2.



Page No 6.54:

Answer:12

We have,

We know that,

Therefore, 9cot2θ-9cosec2θ=-9



Page No 6.54:

Answer:13

We have,

We know that,

Therefore, 6 tan2θ-6cos2θ=-6



Page No 6.54:

Answer:14

We have,

We know that,

Therefore,



Page No 6.54:

Answer:15

We have,

We know that,

Therefore,



Page No 6.54:

Answer:16

Given:

We know that,

 

Therefore,

 
                    



Page No 6.54:

Answer:17

Given:

We know that,

Therefore,



Page No 6.54:

Answer:18

Given:

We know that,

Therefore,



Page No 6.54:

Answer:19

Given:

Hence, the value of k is 1.



Page No 6.54:

Answer:20

Given:

Thus, the value of λ is 1.



Page No 6.54:

Answer:21

Given:


λ=1×1=1

Hence, the value of λ is 1.



Page No 6.54:

Answer:22

Given:

We know that,

 



Page No 6.55:

Answer:23

Given:

We know that,

 



Page No 6.55:

Answer:1

Given:

We know that,

Now,

Adding the two equations, we get

Therefore, the correct choice is (b).



Page No 6.55:

Answer:2

Given:

We know that,

Now,

Subtracting the second equation from the first equation, we get

Therefore, the correct choice is (d).



Page No 6.55:

Answer:3

The given expression is .

Multiplying both the numerator and denominator under the root by , we have


Therefore, the correct option is (a).



Page No 6.55:

Answer:4

The given expression is .

Multiplying both the numerator and denominator under the root by, we have

Therefore, the correct choice is (b).



Page No 6.55:

Answer:5

The given expression is .

Taking common from both the terms, we have



Disclaimer: The options given in (c) and (d) are same by the commutative property of addition.

Therefore, the correct options are (c) or (d).



Page No 6.55:

Answer:6

The given expression is .

Factorising the given expression, we have

Therefore, the correct option is (b).



Page No 6.55:

Answer:7

The given expression is .

Multiplying both the numerator and denominator under the root by , we have

Therefore, the correct option is (c).



Page No 6.56:

Answer:8

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).



Page No 6.56:

Answer:9

The given expression is

Simplifying the given expression, we have



Therefore, the correct option is (b).



Page No 6.56:

Answer:10

The given expression is .

Simplifying the given expression, we have



Therefore, the correct option is (c).



Page No 6.56:

Answer:11

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).



Page No 6.56:

Answer:12

Given:

So,

We know that,

Therefore,

Hence, the correct option is (a).



Page No 6.56:

Answer:13

Given:

So,

We know that,

Therefore,

Hence, the correct option is (d).



Page No 6.56:

Answer:14

The given expression is .

Simplifying the given expression, we have



Therefore, the correct option is (b).



Page No 6.56:

Answer:15

The given expression is .

Simplifying the given expression, we have

 

Therefore, the correct option is (c).



Page No 6.56:

Answer:16

Given:

Squaring and then adding the above two equations, we have


Hence, the correct option is (c).



Page No 6.56:

Answer:17

Given:

Squaring both the equations and then subtracting the second from the first, we have


 
 

Hence, the correct option is (b).



Page No 6.56:

Answer:18

The given expression is .

 

Hence, the correct option is (a).



Page No 6.56:

Answer:19

Given:

Squaring and adding these equations, we get





Hence, the correct option is (a).



Page No 6.56:

Answer:20

Given:

Now,

Hence, the correct option is (b).



Page No 6.56:

Answer:21

Given:

Squaring and adding these equations, we have




Hence, the correct option is (d).



Page No 6.57:

Answer:22

Given:

So,

 
 

Hence, the correct option is (c).



Page No 6.57:

Answer:23

Given:

Now,




Hence, the correct option is (d).



Page No 6.57:

Answer:24

Given:

Squaring on both sides, we have

 



Hence, the correct option is (b).



Page No 6.57:

Answer:25

Given:

We know that,

Therefore,

Hence, the correct option is (b).



Page No 6.57:

Answer:26

The given expression is .

Simplifying the given expression, we have



Disclaimer: None of the given options match with the answer.



Page No 6.57:

Answer:27

The given expression is .

Simplifying the given expression, we have


Therefore, the correct option is (d).



Page No 6.57:

Answer:28

Given:


 

Therefore, the correct option is (d).



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