Rd Sharma 2019 Solutions for Class 10 Math Chapter 11 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2019 Book of Class 10 Math Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 11.43:

Question 1:

Prove the following trigonometric identities.

(1 − cos2 A) cosec2 A = 1

Answer:

We know that,

So,

 

Page No 11.43:

Question 2:

Prove the following trigonometric identities.

(1 + cot2 A) sin2 A = 1

Answer:

We know that,

So,

Page No 11.43:

Question 3:

Prove the following trigonometric identities.

tan2θ cos2θ = 1 − cos2θ

Answer:

We know that, .

So,

Page No 11.43:

Question 4:

Prove the following trigonometric identities.

cosec θ1-cos2 θ=1

Answer:

We know that,

So,

 

 

Page No 11.43:

Question 5:

Prove the following trigonometric identities.

(sec2 θ − 1) (cosec2 θ − 1) = 1

Answer:

We know that,

So,

Page No 11.43:

Question 6:

Prove the following trigonometric identities.

tan θ+1tan θ=sec θ cosec θ

Answer:

We know that,

So,

 

                     

 

Page No 11.43:

Question 7:

Prove the following trigonometric identities.

cos θ1-sin θ=1+sin θcos θ

Answer:

We know that,

Multiplying both numerator and the denominator by, we have

 

 

Page No 11.43:

Question 8:

Prove the following trigonometric identities.

cos θ1+sin θ=1-sin θcos θ

Answer:

We know that,

Multiplying the both numerator and the denominator by, we have


 

Page No 11.43:

Question 9:

Prove the following trigonometric identities.

cos2 A+11+cot2 A=1

Answer:

We know that,

So,

Page No 11.43:

Question 10:

Prove the following trigonometric identities.

sin2 A+11+tan2 A=1

Answer:

We know that,

So,

 

 

Page No 11.43:

Question 11:

Prove the following trigonometric identities.

1-cos θ1+cos θ=cosec θ-cot θ

Answer:

We know that,

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

Page No 11.43:

Question 12:

Prove the following trigonometric identities.

1-cos θsin θ=sin θ1+cos θ

Answer:

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have



Page No 11.44:

Question 13:

Prove the following trigonometric identities.

sin θ1-cos θ=cosec θ+cot θ

Answer:

We have to prove .

We know that,

Multiplying both numerator and denominator by , we have

Page No 11.44:

Question 14:

Prove the following trigonometric identities.

1-sin θ1+sin θ=secθ-tanθ2

Answer:

We have to prove

We know that, .

Multiplying both numerator and denominator by , we have

 

              

Page No 11.44:

Question 15:

Prove the following trigonometric identities.

1+cot2θ tanθsec2θ=cotθ

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 16:

Prove the following trigonometric identities.

tan2θ − sin2θ = tan2θ sin2θ

Answer:

We have to prove

We know that,

So,

                       

Page No 11.44:

Question 17:

Prove the following trigonometric identities.

(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 18:

Prove the following trigonometric identities.

(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ

Answer:

We have to prove

We know that,

 

Page No 11.44:

Question 19:

Prove the following trigonometric identities.

secA (1 − sinA) (secA + tanA) = 1

Answer:

We have to prove

We know that,

So,

                                                

Page No 11.44:

Question 20:

Prove the following trigonometric identities.

(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 21:

Prove the following trigonometric identities.

(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1

Answer:

We have to prove

We know that,

So,

                                                

Page No 11.44:

Question 22:

Prove the following trigonometric identities.

sin2 A cot2 A + cos2 A tan2 A = 1

Answer:

We have to prove

We know that,

So,

Page No 11.44:

Question 23:

Prove the following trigonometric identities.

(i) cot θ-tan θ=2 cos2 θ-1sin θ cos θ

(ii) tan θ-cot θ=2 sin2 θ-1sin θ cos θ

(iii) sin A-2sin3 A2cos3 A-cos A=tan A

Answer:

(i) We have to prove

We know that,

So,

                     
 

(ii) We have to prove

We know that,

So,

                     

(iii)
Given
sinA-2sin3A2cos3 A-cosA=tanA

LHS=sinA-2sin3A2cos3A-cosA=sinA 1-2sin2AcosA 2cos2A-1                    =tanAsin2A+cos2A-2sin2A2cos2A-sin2A-cos2A            sin2A+cos2A=1=tanAcos2A- sin2Acos2A- sin2A=tanA=RHS
Hence proved.

Page No 11.44:

Question 24:

Prove the following trigonometric identities.

cos2 θsin θ-cosec θ+sin θ=0

Answer:

We have to prove

We know that,

So,

                                      

Page No 11.44:

Question 25:

Prove the following trigonometric identities.

11+sinA+11-sinA=2 sec2A

Answer:

We have to prove

We know that,

So,


                              

Page No 11.44:

Question 26:

Prove the following trigonometric identities.

1+sin θcos θ+cos θ1+sin θ=2 sec θ

Answer:

We have to prove

We know that,

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

                              

Page No 11.44:

Question 27:

Prove the following trigonometric identities.

1+sinθ2+1- sinθ22 cos2θ=1+sin2θ1-sin2θ

Answer:

We have to prove that .

We know that,

So,

                                      

Page No 11.44:

Question 28:

Prove the following trigonometric identities.

1+tan2θ1+cot2θ=1-tanθ1-cotθ2=tan2θ

Answer:

We have to prove

Consider the expression

                

Again, we have

                    

Page No 11.44:

Question 29:

Prove the following trigonometric identities.

1+secθsecθ=sin2θ1-cosθ

Answer:

We have to prove

We know that,

Multiplying the numerator and denominator by , we have

Page No 11.44:

Question 30:

Prove the following trigonometric identities.

tanθ1-cotθ+cotθ1-tanθ=1+tanθ+cotθ

Answer:

We need to prove

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


                             

Further using the identity, we get

Hence

Page No 11.44:

Question 31:

Prove the following trigonometric identities.

sec6θ = tan6θ + 3 tan2θ sec2θ + 1

Answer:

We need to prove

Solving the L.H.S, we get

         

Further using the identity, we get

Hence proved.

Page No 11.44:

Question 32:

Prove the following trigonometric identities

cosec6θ = cot6θ + 3 cot2θ cosec2θ + 1

Answer:

We need to prove

Solving the L.H.S, we get

                               ()

Further using the identity, we get

Hence proved.

Page No 11.44:

Question 33:

Prove the following trigonometric identities.

1+tan2θ cotθcosec2θ=tanθ

Answer:

We need to prove

Solving the L.H.S, we get

Using , we get

                     

Hence proved.

Page No 11.44:

Question 34:

Prove the following trigonometric identities.

1+cosAsin2A=11-cosA

Answer:

We need to prove

Using the property , we get

LHS =

Further using the identity, , we get


                 = RHS

Hence proved.

Page No 11.44:

Question 35:

Prove the following trigonometric identities.

secA-tanAsecA+tanA=cos2A1+sinA2

Answer:

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 11.44:

Question 36:

Prove the following trigonometric identities.

1+cos Asin A=sin A1-cos A

Answer:

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 11.44:

Question 37:

Prove the following trigonometric identities.

(i) 1+sin A1-sin A=sec A+tan A

(ii) 

Answer:

(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 11.44:

Question 38:

Prove the following trigonometric identities.

(i) sec θ-1sec θ+1+sec θ+1sec θ-1=2 cosec θ
(ii) 1+sin θ1-sin θ+1-sin θ1+sin θ=2 sec θ
(iii) 1+cos θ1-cos θ+1-cos θ1+cos θ=2 cosec θ
(iv) sec θ-1sec θ+1=sin θ1+cos θ2
(v) sinθ+1-cosθcosθ-1+sinθ=1+sinθcosθ

Answer:

(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 11.45:

Question 39:

Prove the following trigonometric identities.

sec A-tan A2=1-sin A1+sin A

Answer:

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 11.45:

Question 40:

Prove the following trigonometric identities.

1-cos A1+cos A=cot A-cosec A2

Answer:

We need to prove

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

Using and, we get

Hence proved.

Page No 11.45:

Question 41:

Prove the following trigonometric identities.

1secA-1+1secA+1=2 cosecA cotA

Answer:

We need to prove

Solving the L.H.S, we get

Further using the property , we get

So,

                

Hence proved.

Page No 11.45:

Question 42:

Prove the following trigonometric identities.

cos A1-tan A+sin A1-cot A=sin A+cos A

Answer:

We need to prove

Solving the L.H.S, we get

                              
                               = RHS

Hence proved.

 

Page No 11.45:

Question 43:

Prove the following trigonometric identities.

cosecAcosecA-1+cosecAcosecA +1=2 sec2A

Answer:

We need to prove

Using the identity, we get



Further, using the property , we get

So,

                                                   

Hence proved.

Page No 11.45:

Question 44:

Prove the following trigonometric identities.

tan2 A1+tan2 A+cot2 A1+cot2 A=1

Answer:

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 11.45:

Question 45:

Prove the following trigonometric identities.

cot A-cos Acot A+cos A=cosec A-1cosec A+1

Answer:

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 11.45:

Question 46:

Prove the following trigonometric identities.

1+cos θ-sin2 θsin θ (1+cos θ)=cot θ

Answer:

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 11.45:

Question 47:

Prove the following trigonometric identities.

(i) 1+cos θ+sin θ1+cos θ-sin θ=1+sin θcos θ
(ii) sin θ-cos θ+1sin θ+cos θ-1=1sec θ-tan θ
(iii) cos θ-sin θ+1cos θ+sin θ-1=cosec θ+cot θ
(iv) sinθ+cosθtanθ+cotθ=secθ+cosecθ

Answer:

(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 11.45:

Question 48:

Prove the following trigonometric identities.

1sec A+tan A-1cos A=1cos A-1sec A-tan A

Answer:

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 11.45:

Question 49:

Prove the following trigonometric identities.

tan2 A + cot2 A = sec2 A cosec2 A − 2

Answer:

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 11.45:

Question 50:

tanA1+secA-tanA1-secA=2cosecA

Answer:

Consider the LHS.
tanA1+secA-tanA1-secA=tanA1-secA-tanA1+secA1+secA1-secA=tanA-tanAsecA-tanA-tanAsecA1-sec2A=-2tanAsecA1-sec2A=-2tanAsecA-tan2A=2secAtanA=2cosecA
= RHS
Hence proved.

Page No 11.45:

Question 51:

Prove the following trigonometric identities.

1+cot2 θ1+cosec θ=cosec θ

Answer:

In the given question, we need to prove

Using and, we get

Further, using the property , we get

Hence proved.

Page No 11.45:

Question 52:

Prove the following trigonometric identities.

cos θcosec θ+1+cos θcosec θ-1=2 tan θ

Answer:

In the given question, we need to prove

Using the identity, we get

Further, using the property , we get

                                              

Hence proved.

Page No 11.45:

Question 53:

Prove the following trigonometric identities.

1+tan2A+1+1tan2A=1sin2A-sin4A

Answer:

We need to prove .

Using the property , we get

Now, using , we get

Further, using the property, , we get

Hence proved.

Page No 11.45:

Question 54:

Prove the following trigonometric identities.

sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B

Answer:

We know that,

So have,

Hence proved.

Page No 11.45:

Question 55:

Prove the following trigonometric identities.

(i) cotA+tanBcotB+tanA=cotA tanB

(ii) tanA+tanBcotA+cotB=tanA tanB

Answer:

(i) We have to prove

Now,
 
                      

Hence proved.

(ii) We have to prove 

Now,



Hence proved.

Page No 11.45:

Question 56:

Prove the following trigonometric identities.

cot2 A cosec2 B − cot2 B cosec2 A = cot2 A − cot2 B

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 11.45:

Question 57:

Prove the following trigonometric identities.

tan2 A sec2 B − sec2 A tan2 B = tan2 A − tan2 B

Answer:

We have to prove

We know that,

So,

Hence proved.

Page No 11.45:

Question 58:

Prove the following trigonometric identities.

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2y2 = a2b2

Answer:

Given that,

We have to prove

We know that,

So,

Hence proved. 

Page No 11.45:

Question 59:

Prove the following trigonometric identities.

If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ − 3 cos θ = ± 3.

Answer:

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 11.46:

Question 60:

Prove the following trigonometric identities.

If cosec θ + cot θ = m and cosec θ − cot θ = n, prove that mn = 1

Answer:

Given:

We have to prove

We know that,

Multiplying the two equations, we have

 


Hence proved.

Page No 11.46:

Question 61:

Prove the following trigonometric identities.

tan3 θ1+tan2 θ+cot3 θ1+cot2 θ=sec θ cosec θ-2 sin θ cos θ

Answer:

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 11.46:

Question 62:

Prove the following trigonometric identities.

If Tn=sinn θ+cosn θ, prove that T3-T5T1=T5-T7T3

Answer:

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 11.46:

Question 63:

Prove the following trigonometric identities.

tan θ+1cos θ2+tan θ-1cos θ2=21+sin2 θ1-sin2 θ

Answer:

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 11.46:

Question 64:

Prove the following trigonometric identities.

1sec2 θ-cos2 θ+1cosec2 θ-sin2 θ sin2 θ cos2 θ=1-sin2 θ cos2 θ2+sin2 θ cos2 θ

Answer:

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 11.46:

Question 65:

Prove the following trigonometric identities.

(i) 1+sin θ-cos θ1+sin θ+cos θ2=1-cos θ1+cos θ
(ii)  1+secθ-tanθ1+secθ+tanθ=1-sinθcosθ

Answer:

(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 11.46:

Question 66:

Prove the following trigonometric identities.

(sec A + tan A − 1) (sec A − tan A + 1) = 2 tan A

Answer:

We have to prove

We know that,

So, we have

So, we have

Hence proved.

Page No 11.46:

Question 67:

Prove the following trigonometric identities.

(1 + cot A − cosec A) (1 + tan A + sec A) = 2

Answer:

We have to prove

We know that, .

So,

                                                           

                                                         

Hence proved.

Page No 11.46:

Question 68:

Prove the following trigonometric identities.

(cosec θ − sec θ) (cot θ − tan θ) = (cosec θ + sec θ) ( sec θ cosec θ − 2)

Answer:

We have to prove

We know that,

Consider the LHS.

Now, consider the RHS.

∴ LHS = RHS

Hence proved.

Page No 11.46:

Question 69:

Prove the following trigonometric identities.

(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A

Answer:

We have to prove

We know that,

So,

                                                      

                                                      

                                                      

Hence proved.

Page No 11.46:

Question 70:

Prove the following trigonometric identities.

(i) cos A cosec A-sin A sec Acos A+sin A=cosec A-sec A

(ii) sin Asec A+tan A-1+cos Acosec A+cot A-1=1

Answer:

(i) We have to prove

So,

                                         

                                         

Hence proved.

(ii) We have to prove 

We know that, 

So,

            

            

            

Hence proved.

Page No 11.46:

Question 71:

Prove the following trigonometric identities.

sin Asec A+tan A-1+cos Acosec A+cot A-1=1

Answer:

We have to prove

We know that,

So,

            

            

            

Hence proved.

Page No 11.46:

Question 72:

Prove the following trigonometric identities.

tan A1+tan2 A2+cot A1+cot2 A=sin A cos A

Answer:

We have to prove

We know that, .

So,



  

 
 

Hence proved.

Page No 11.46:

Question 73:

Prove the following trigonometric identities.

sec4 A(1 − sin4 A) − 2 tan2 A = 1

Answer:

We have to prove

We know that,

So,

                                             

                                             

Hence proved.

Page No 11.46:

Question 74:

Prove the following trigonometric identities.

cot2 Asec A-11+sin A=sec2 A1-sin A1+sec A

Answer:

We have to prove .

We know that,

So,

                           

                           

Multiplying both the numerator and denominator by , we have

Hence proved.

Page No 11.46:

Question 75:

Prove the following trigonometric identities.

1+cot A+tan A sin A-cos A=sec Acosec2 A-cosec Asec2 A=sin A tan A-cot A cos A

Answer:

We have prove that



We know that,

So,

             

             

             

             

Now,

                              

Hence proved.

Page No 11.46:

Question 76:

Prove the following trigonometric identities.

If xacos θ+ybsin θ=1 and xasin θ-ybcos θ=1, prove that x2a2+y2b2=2

Answer:

Given that,



We have to prove

We know that,

Squaring and then adding the above two equations, we have

Page No 11.46:

Question 77:

Prove the following trigonometric identities.

If cosec θ − sin θ = a3, sec θ − cos θ = b3, prove that a2 b2 (a2 + b2) = 1

Answer:

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 11.47:

Question 78:

Prove the following trigonometric identities.

If a cos3 θ + 3 a cos θ sin2 θ = m, a sin3 θ + 3 a cos2 θ sin θ = n, prove that (m + n)2/3 + (mn)2/3 = 2a2/3

Answer:

Given that,

We have to prove

Adding both the equations, we get

Also.

Therefore, we have

Hence proved.

Page No 11.47:

Question 79:

Prove the following trigonometric identities.

If x = a cos3 θ, y = b sin3 θ, prove that xa2/3+yb2/3=1

Answer:

Given:

We have to prove

We know that

So, we have



Hence proved.

Page No 11.47:

Question 80:

Prove the following trigonometric identities.

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that a2 + b2 = m2 + n2

Answer:

Given:



We have to prove

We know that,

Now, squaring and adding the two equations, we get



Hence proved.

Page No 11.47:

Question 81:

Prove the following trigonometric identities.

if cos A + cos2 A = 1, prove that sin2 A + sin4 A = 1

Answer:

Given:

We have to prove

Now,

Therefore, we have



Hence proved.

Page No 11.47:

Question 82:

If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1

Answer:

Given:

We have to prove

From the given equation, we have

 

Therefore, we have

 

Hence proved.

Page No 11.47:

Question 83:

Given that:
(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)

Show that one of the values of each member of this equality is sin α sin β sin γ

Answer:

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 11.47:

Question 84:

If sin θ + cos θ = x, prove that sin6θ+cos6θ=4-3 x2-124.

Answer:

Given:

Squaring the given equation, we have

Squaring the last equation, we have

Therefore, we have


                       

Hence proved.

Page No 11.47:

Question 85:

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that x2a2+y2b2-z2c2=1.

Answer:

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 11.47:

Question 86:

If sinθ+2cosθ=1 prove that 2sinθ-cosθ=2.

Answer:

It is given that,
sinθ+2cosθ=12cosθ=1-sinθ                 .....iSquaring both sides, we get2cosθ2=1-sinθ24cos2θ=1+sin2θ-2sinθ4cos2θ=1+sin2θ-2sinθ-1+14cos2θ=2-2sinθ-1-sin2θ4cos2θ=2-2sinθ-cos2θ5cos2θ=21-sinθ 5cos2θ=4cosθ               Using (i)cosθ5cosθ-4=0cosθ=0, cosθ=45 Putting cosθ=4 5 in sinθ+2cosθ=1, we getsinθ=1-2cosθ=1-245=-35This is not possible. Putting cosθ=0  in sinθ+2cosθ=1, we getsinθ=1Thus, the value of 2sinθ-cosθ=21-0=2.



Page No 11.54:

Question 1:

If cos θ=45, find all other trigonometric ratios of angle θ.

Answer:

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 11.54:

Question 2:

If sin θ=12, find all other trigonometric ratios of angle θ.

Answer:

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 11.54:

Question 3:

If tan θ=12, find the value of cosec2 θ-sec2 θcosec2 θ+cot2 θ.

Answer:

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 11.54:

Question 4:

If 4tanθ = 3, evaluate 4sin θ-cos θ+14sin θ+cos θ-1

Answer:

Given: 4tanθ = 3 ⇒ tan θ = 34
Let us suppose a right angle triangle ABC right angled at B, with one of the acute angle θ. Let the sides be BC = 3and AB = 4k, where k is a positive number.

By Pythagoras theorem, we get
AC2=BC2+AB2AC2=3k2+4k2AC2=9k2+16k2AC=25k2AC=±5k
Ignoring AC = − 5k , as k is a positive number, we get
AC = 5k
If tanθ=BCAB=34, then sinθ=BCAC=35 and cosθ=ABAC=45
Putting the values in 4sinθ-cosθ+14sinθ+cosθ-1, we get
4×35-45+14×35+45-1=12-4+5512+4-55=1311

Page No 11.54:

Question 5:

If tan θ=125, find the value of 1+sin θ1-sin θ.

Answer:

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 11.54:

Question 6:

If cot θ=13, find the value of 1-cos2 θ2-sin2 θ.

Answer:

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 11.54:

Question 7:

If cosec A=2, find the value of 2 sin2 A+3 cot2 A4tan2 A-cos2 A.

Answer:

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 11.54:

Question 8:

If cot θ=3, find the value of cosec2 θ+cot2 θcosec2 θ-sec2 θ.

Answer:

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 11.54:

Question 9:

If 3cosθ = 1, find the value of 6 sin2 θ+tan2 θ4 cos θ

Answer:

Given:

We have to find the value of the expression .

We have,




Therefore,

Hence, the value of the expression is 10.

Page No 11.54:

Question 10:

If 3 tan θ=3 sin θ, find the value of sin2θ − cos2θ.

Answer:

Given:

We have to find the value of .

Therefore,

Hence, the value of the expression is 13.

Page No 11.54:

Question 11:

If cosec θ=1312, find the value of 2 sin θ-3 cos θ4 sin θ-9 cos θ.

Answer:

Given:

We have to find the value of the expression .

Now,

Therefore,

Hence, the value of the expression is 3.

Page No 11.54:

Question 12:

If sin θ + cos θ = 2 cos 90°-θ, find cot θ.

Answer:

Given:

We have to find the value of .

Now,

Hence,

Page No 11.54:

Question 13:

If 2 sin2θ - cos2θ = 2  , then find the value of θ . 

Answer:

It is given that,
2sin2θ-cos2θ=22sin2θ-2=cos2θ-21-sin2θ=cos2θ -2cos2θ=cos2θ                        1-sin2θ=cos2θ3cos2θ=0cosθ=0θ=90°



Page No 11.55:

Question 1:

Define an identity.

Answer:

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 11.55:

Question 2:

What is the value of (1 − cos2 θ) cosec2 θ?

Answer:

We have,

Page No 11.55:

Question 3:

What is the value of (1 + cot2 θ) sin2 θ?

Answer:

We have,

Page No 11.55:

Question 4:

What is the value of sin2 θ+11+tan2 θ?

Answer:

We have,

Page No 11.55:

Question 5:

If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.

Answer:

Given:

We know that,

Therefore,


Hence,

Page No 11.55:

Question 6:

If cosec θ − cot θ = α, write the value of cosec θ + cot α.

Answer:

Given:

We know that,

Therefore,

 
Hence,

Page No 11.55:

Question 7:

Write the value of cosec2 (90° − θ) − tan2 θ.

Answer:

We have,

We know that,

Therefore, cosec290°-θ-tan2θ=1

Page No 11.55:

Question 8:

Write the value of sin A cos (90° − A) + cos A sin (90° − A).

Answer:

We have,

We know that,

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

Page No 11.55:

Question 9:

Write the value of cot2 θ-1sin2 θ.

Answer:

We have,

We know that,

Therefore, cot2θ-1sin2θ=-1

Page No 11.55:

Question 10:

If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?

Answer:

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 11.55:

Question 11:

If sin θ=45, what is the value of cotθ + cosecθ?

Answer:

Given:

We know that,


We have,

Hence, the value of cotθ + cosecθ is 2.

Page No 11.55:

Question 12:

What is the value of 9cot2 θ − 9cosec2 θ?

Answer:

We have,

We know that,

Therefore, 9cot2θ-9cosec2θ=-9

Page No 11.55:

Question 13:

What is the value of 6 tan2θ-6cos2θ.

Answer:

We have,

We know that,

Therefore, 6 tan2θ-6cos2θ=-6

Page No 11.55:

Question 14:

What is the value of tan2 θ-sec2 θcot2 θ-cosec2 θ.

Answer:

We have,

We know that,

Therefore,

Page No 11.55:

Question 15:

What is the value of (1 + tan2 θ) (1 − sin θ) (1 + sin θ)?

Answer:

We have,

We know that,

Therefore,

Page No 11.55:

Question 16:

If cos A=725, find the value of tan A + cot A.

Answer:

Given:

We know that,

 

Therefore,

 
                    

Page No 11.55:

Question 17:

If sin θ=13, then find the value of 2cot2 θ + 2.

Answer:

Given:

We know that,

Therefore,

Page No 11.55:

Question 18:

If cos θ=34, then find the value of 9tan2 θ + 9.

Answer:

Given:

We know that,

Therefore,

Page No 11.55:

Question 19:

If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.

Answer:

Given:

Hence, the value of k is 1.

Page No 11.55:

Question 20:

If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.

Answer:

Given:

Thus, the value of λ is 1.

Page No 11.55:

Question 21:

If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.

Answer:

Given:


λ=1×1=1

Hence, the value of λ is 1.

Page No 11.55:

Question 22:

If 5x = sec θ and 5x=tan θ, find the value of 5x2-1x2.

Answer:

Given:

We know that,

 

Page No 11.55:

Question 23:

If cosec θ = 2x and cot θ=2x, find the value of 2x2-1x2.

Answer:

Given:

We know that,

 



Page No 11.56:

Question 24:

 Write True' or False' and justify your answer in each of the following :
(i) The value of sin θ is x + 1x, where 'x'  is a positive real number .
(ii) cos θ = a2 + b22ab, where a and b are two distinct numbers such that ab > 0.

(iii) The value of cos223 - sin267 is positive .

(iv) The value of the expression sin 80° - cos 80°  is negative .

(v) The value of sin θ + cos θ  is always greater than 1 .

Answer:

(i)
sinθ=x+1x-1x+1x1x+1x1x2+1xx2+1-x0Take x= 1,1+1-1010Which is false, so x is not always a positive real number.The given statement is false.

(ii)
It is given that,
cosθ=a2+b22abSince, a-b2=a2+b2-2ab>0  a-b2>0, a and b are distint numbersa2+b2>2aba2+b22ab>1cosθ>1But, -1cosθ1Therefore, the given statement is false.

(iii)
cos223°-sin267°=sin290°-23°-sin267°=sin267°-sin267°=0
Which is not positive, the given statement is false.

(iv)
Consider the table.

θ 30º 45º 60º 90º
sinθ 0 12 12 32 1
cosθ 1 32 12 12 0

Here,
sin60°-cos60°=32-12>0sin90°-cos90°=1-0>0So, sin80°-cos80°>0          sinθ-cosθ0 45°θ90°
Therefore, the given statement is false.

(v)
Consider the table.
θ 30º 45º 60º 90º
sinθ 0 12 12 32 1
cosθ 1 32 12 12 0
Here,
sin90°+cos90°=1+0=1, which is not greater than 1Therefore, the given statement is false.

Page No 11.56:

Question 25:

What is the value of cos2 67° – sin2 23°?

Answer:

cos267°-sin223°=cos267°-sin290°-67°=cos267°-cos267°                    sin290-θ=cos2θ=0

Page No 11.56:

Question 1:

If sec θ + tan θ = x, then sec θ =

(a) x2+1x
(b) x2+12x
(c) x2-12x
(d) x2-1x

Answer:

Given:

We know that,

Now,

Adding the two equations, we get

Therefore, the correct choice is (b).

Page No 11.56:

Question 2:

If secθ+tanθ=x, then tanθ=

(a) x2+1x
(b) x2-1x
(c) x2+12x
(d) x2-12x

Answer:

Given:

We know that,

Now,

Subtracting the second equation from the first equation, we get

Therefore, the correct choice is (d).

Page No 11.56:

Question 3:

1+sin θ1-sin θ is equal to

(a) sec θ + tan θ
(b) sec θ − tan θ
(c) sec2 θ + tan2 θ
(d) sec2 θ − tan2 θ

Answer:

The given expression is .

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


Therefore, the correct option is (a).

Page No 11.56:

Question 4:

The value of 1+cos θ1-cos θ is

(a) cot θ − cosec θ
(b) cosec θ + cot θ
(c) cosec2 θ + cot2 θ
(d) (cot θ + cosec θ)2

Answer:

The given expression is .

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

Therefore, the correct choice is (b).

Page No 11.56:

Question 5:

sec4 A − sec2 A is equal to

(a) tan2 A − tan4 A
(b) tan4 A − tan2 A
(c) tan4 A + tan2 A
(d) tan2 A + tan4 A

Answer:

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 11.56:

Question 6:

cos4 A − sin4 A is equal to

(a) 2 cos2 A + 1
(b) 2 cos2 A − 1
(c) 2 sin2 A − 1
(d) 2 sin2 A + 1

Answer:

The given expression is .

Factorising the given expression, we have

Therefore, the correct option is (b).



Page No 11.57:

Question 7:

sin θ1+cos θis equal to

(a) 1+cos θsin θ
(b) 1-cos θcos θ
(c) 1-cos θsin θ
(d) 1-sin θcos θ

Answer:

The given expression is .

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

Therefore, the correct option is (c).

Page No 11.57:

Question 8:

sin θ1-cot θ+cos θ1-tan θ is equal to

(a) 0
(b) 1
(c) sin θ + cos θ
(d) sin θ − cos θ

Answer:

The given expression is .

Simplifying the given expression, we have

Therefore, the correct option is (c).

Page No 11.57:

Question 9:

The value of (1 + cot θ − cosec θ) (1 + tan θ + sec θ) is

(a) 1
(b) 2
(c) 4
(d) 0

Answer:

The given expression is

Simplifying the given expression, we have



Therefore, the correct option is (b).

Page No 11.57:

Question 10:

tan θsec θ-1+tan θsec θ+1 is equal to

(a) 2 tan θ
(b) 2 sec θ
(c) 2 cosec θ
(d) 2 tan θ sec θ

Answer:

The given expression is .

Simplifying the given expression, we have



Therefore, the correct option is (c).

Page No 11.57:

Question 11:

(cosec θ − sin θ) (sec θ − cos θ) (tan θ + cot θ) is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

Answer:

The given expression is

Simplifying the given expression, we have

Therefore, the correct option is (b).

Page No 11.57:

Question 12:

If x = a cos θ and y = b sin θ, then b2x2 + a2y2 =

(a) a2 b2
(b) ab
(c) a4 b4
(d) a2 + b2

Answer:

Given:

So,

We know that,

Therefore,

Hence, the correct option is (a).

Page No 11.57:

Question 13:

If x = a sec θ and y = b tan θ, then b2x2a2y2 =

(a) ab
(b) a2b2
(c) a2 + b2
(d) a2 b2

Answer:

Given:

So,

We know that,

Therefore,

Hence, the correct option is (d).

Page No 11.57:

Question 14:

cot θcot θ-cot 3θ+tan θtan θ-tan 3θis equal to

(a) 0
(b) 1
(c) −1
(d) 2

Answer:

The given expression is .

Simplifying the given expression, we have



Therefore, the correct option is (b).

Page No 11.57:

Question 15:

2 (sin6 θ + cos6 θ) − 3 (sin4 θ + cos4 θ) is equal to

(a) 0
(b) 1
(c) −1
(d) None of these

Answer:

The given expression is .

Simplifying the given expression, we have

 

Therefore, the correct option is (c).

Page No 11.57:

Question 16:

If a cos θ + b sin θ = 4 and a sin θ − b sin θ = 3, then a2 + b2 =

(a) 7
(b) 12
(c) 25
(d) None of these

Answer:

Given:

Squaring and then adding the above two equations, we have


Hence, the correct option is (c).

Page No 11.57:

Question 17:

If a cot θ + b cosec θ = p and b cot θ − a cosec θ = q, then p2q2 =

(a) a2 b2
(b) b2a2
(c) a2 + b2
(d) ba

Answer:

Given:

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


 
 

Hence, the correct option is (b).

Page No 11.57:

Question 18:

The value of sin2 29° + sin2 61° is

(a) 1
(b) 0
(c) 2 sin2 29°
(d) 2 cos2 61°

Answer:

The given expression is .

 

Hence, the correct option is (a).

Page No 11.57:

Question 19:

If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then

(a) x2+y2+z2=r2
(b) x2+y2-z2=r2
(c) x2-y2+z2=r2
(d) z2+y2-x2=r2

Answer:

Given:

Squaring and adding these equations, we get





Hence, the correct option is (a).



Page No 11.58:

Question 20:

If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =

(a) −1
(b) 1
(c) 0
(d) None of these

Answer:

Given:

Now,

Hence, the correct option is (b).

Page No 11.58:

Question 21:

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =

(a) m2n2
(b) m2n2
(c) n2m2
(d) m2 + n2

Answer:

Given:

Squaring and adding these equations, we have




Hence, the correct option is (d).

Page No 11.58:

Question 22:

If cos A + cos2 A = 1, then sin2 A + sin4 A =

(a) −1
(b) 0
(c) 1
(d) None of these

Answer:

Given:

So,

 
 

Hence, the correct option is (c).

Page No 11.58:

Question 23:

If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, then x2a2+y2b2=

(a) z2c2
(b) 1-z2c2
(c) z2c2-1
(d) 1+z2c2

Answer:

Given:

Now,




Hence, the correct option is (d).

Page No 11.58:

Question 24:

If a cos θ − b sin θ = c, then a sin θ + b cos θ =

(a) ±a2+b2+c2
(b) ±a2+ b2-c2
(c) ±c2-a2-b2
(d) None of these

Answer:

Given:

Squaring on both sides, we have

 



Hence, the correct option is (b).

Page No 11.58:

Question 25:

9 sec2 A − 9 tan2 A is equal to

(a) 1
(b) 9
(c) 8
(d) 0

Answer:

Given:

We know that,

Therefore,

Hence, the correct option is (b).

Page No 11.58:

Question 26:

(1 + tan θ + sec θ) (1 + cot θ − cosec θ) =

(a) 0
(b) 1
(c) 1
(d) −1

Answer:

The given expression is .

Simplifying the given expression, we have



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

Page No 11.58:

Question 27:

(sec A + tan A) (1 − sin A) =

(a) sec A
(b) sin A
(c) cosec A
(d) cos A

Answer:

The given expression is .

Simplifying the given expression, we have


Therefore, the correct option is (d).

Page No 11.58:

Question 28:

1+tan2 A1+cot2 Ais equal to

(a) sec2 A
(b) −1
(c) cot2 A
(d) tan2 A

Answer:

Given:


 

Therefore, the correct option is (d).

Page No 11.58:

Question 29:

If sin θ - cos θ = 0 , then the value of sin4 θ + cos4 θ is 
(a) 1     (b)  -1      (c)   12      (d)  14

Answer:

It is given that,
sinθ-cosθ=0sinθ=cosθsinθcosθ=1tanθ=1tanθ=tan45°θ=45°
sin4θ+cos4θ=sin445°+cos445°=124+124=14+14=12
Hence, the correct answer is option (c).

Page No 11.58:

Question 30:

The value of sin (45° +θ) - cos (45° - θ) is equal to
(a) 2 cos θ    (b) 0     (c)     2 sin θ    (d) 1

Answer:

We know that, sin90-θ=cosθ.
So, sin45°+θ=cos90-45°+θ=cos45°-θ
sin45°+θ-cos45°-θ=cos45°-θ-cos45°-θ=0
Hence, the correct answer is option (b).

Page No 11.58:

Question 31:

If ABC is right angled at C , then the value of cos ( A + B ) is

Answer:

In a right angled triangle ABC, C is a right angle.
We know that, the sum of angles of a triangle is 180º.
A+B+C=180°A+B+90°=180°A+B=90°
cosA+B=cos90°=0
Hence, the correct answer is option (a).



Page No 11.59:

Question 32:

If cos 9θ = sin θ and  9θ  < 900 , then the value of tan 6 θ is

Answer:

It is given that,
cos9θ=sinθ, 9θ<90°              sin90°-9θ=sinθ       sin90°-θ=cosθ90°-9θ=θ10θ=90°θ=9°Therefore, tan6θ=tan54°.
Disclaimer: Answer of the given question is not matching with the options provided in the textbook.

Page No 11.59:

Question 33:

If  cos (α + β) = 0 , then sin α - β can be reduced to 

Answer:

It is given that,
cosα+β=0cosα+β=cos90°             cos90°=0α+β=90° α=90°-βNow, put α=90°-β in sinα-β.sinα-β=sin90°-β-β=sin90°-2β       =cos2β                       sin90°-θ=cosθ
Hence, the correct answer is option (b).



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