NCERT Solutions Class 10 Mathematics Introduction To Trigonometry Exercise 8.4

Exercise 8.4

Question : 1:Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer :

(i) sin A = √(sin² A)
               = √(1/coses² A)     
               = √{1/(1 + cot² A)}                   [coses² A = 1 + cot² A]

(ii) sec A = √(sec² A)     
                = √(1 + tan² A)                        [sec² A = 1 + tan² A]    
                = √(1 + 1/cot² A)
                = √{(1 + cot² A)/ cot² A}
                = √{(1 + cot² A)/ cot A        

(iii) tan A = 1/cot A

Question : 2:Write all the other trigonometric ratios of ∠ A in terms of sec A.
(i) sin A = √(sin² A)
(ii) cos A = 1/sec A

(iii) tan A = √(tan² A)
(iv) cot A = √(cot² A)
(v) cosec A = √(cosec² A)

Answer :

(i) sin A = √(sin² A)
              = √(1 – cos² A)
              = √(1 – 1/sec² A)
              = √{( sec² A – 1)/sec² A}
              = √( sec² A – 1)/sec A

(ii) cos A = 1/sec A

(iii) tan A = √(tan² A)
              = √(sec² A – 1)

(iv) cot A = √(cot² A)
              = √(1/tan² A)
              = √{1/(sec² A – 1)}
              = 1/√ (sec² A – 1)

(v) cosec A = √(cosec² A)
                    = √(1 + cot² A)
                    = √(1 + 1/tan² A)
                    = √{(1 + 1/(sec² A – 1)}
                    = √{(sec² A – 1 + 1)/(sec² A – 1)}
                    = √{sec² A/(sec² A – 1)}
                    = sec A/√(sec² A – 1)  

Question : 3:Evaluate :
(i) (sin² 63^o + sin² 27^o)/(cos² 17^o + cos² 73^o)
(ii) sin 25^o cos 65^o + cos 25^o sin 65^o

Answer :

(i) (sin² 63^o + sin² 27^o)/(cos² 17^o + cos² 73^o)
= {sin² 63^o + cos² (90 – 27^o)}/{cos² 17^o + sin² (90 – 73^o)}          
    [sin θ = cos(90^o – θ) and cos θ = sin(90^o – θ)]
= (sin² 63^o + cos² 63^o)/(cos² 17^o + sin² 17^o)
= 1
(ii) sin 25^o cos 65^o + cos 25^o sin 65^o
= cos(90 – 25^o)cos 65^o + sin(90 – 25^o)sin 65^o
= cos 65^o cos 65^o + sin 65^o sin 65⁰                  [sin θ = cos(90^o – θ) and cos θ = sin(90^o – θ)]
= cos² 65^o + sin² 65^o
= 1                                       [sin² θ + cos² θ = 1]

Question : 4 :
Choose the correct option. Justify your choice.
(i) 9 sec² A – 9 tan²  A =
(A) 1                      (B) 9                    (C) 8                                  (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
(A) 0                      (B) 1                    (C) 2                                  (D) -1
(iii) (sec A + tan A) (1 – sin A) =
(A) sec A               (B) sin A              (C) cosec A                      (D) cos A
(iv) (1 + tan² A)/(1 + cot² A) =
(A) sec² A             (B) –1                  (C) cot²A                        (D) tan² A

Answer :

(i) 9 sec² A – 9 tan² A =
Given, 9 sec² A – 9 tan² A
= 9(sec² A – tan² A)
= 9*1                      {since sec² A – tan² A = 1}
= 9
Hence, option (B) is the correct answer.

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =
= 1 + cot θ – cosec θ + tan θ + tan θ cot θ – tan θ cosec θ + sec θ + sec θ cot θ – sec θ cosec θ
= 1 + cos θ/sin θ – 1/sin θ + sin θ/cos θ + 1 – sin θ/cos θ * 1/sin θ + 1/cos θ
+ 1/cos θ * cos θ/sin θ – 1/cos θ * 1/sin θ                  [since tan θ cot θ = 1]
= 2 + cos θ/sin θ – 1/sin θ + sin θ/cos θ – 1/cos θ + 1/cos θ + 1/sin θ – 1/(cos θ * sin θ)
= 2 + cos θ/sin θ + sin θ/cos θ – 1/(cos θ * sin θ)
= 2 + (cos² θ + sin² θ – 1)/ (cos θ * sin θ)
= 2 + (1 – 1)/ (cos θ * sin θ)                           [sin² θ + cos² θ = 1]
= 2 + 0
= 2
Hence, option (C) is the correct answer.

(iii) (sec A + tan A) (1 – sin A) = (1/cos A + sin A/cos A) (1 – sin A)
                                                    = {(1 + sin A)/cos A}(1 – sin A)
                                                    = (1 – sin² A)/cos A
                                                    = cos² A/cos A                      [sin² θ + cos² θ = 1]
                                                    = cos A
Hence, option (D) is the correct answer.   

(iv) (1 + tan² A)/(1 + cot² A) = sec² A/cosec² A
[sec² A = 1 + cot² A and cosec² A = 1 + tan² A]
                                                 = (1/cos² A)/(1/sin² A)
                                                 = sin² A/ cos² A
                                                 = tan² A
Hence, option (D) is the correct answer.   
 

Question : 5:Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ)² = (1 – cos θ)/(1 + cos θ)        (ii) cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A
(iii) tan θ/(1 – cot θ) + cot θ/(1 – tan θ) = 1 + sec θ + cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin² A/(1 – cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A – sin A + 1)/ (cos A + sin A – 1) = cosec A, using the identity cosec² A = 1 + cot² A.
(vi) √{(1 + sin A)/(1 – sin A)} = sec A + tan A
(vii) (sin θ – 2 sin³ θ)/(2cos³ θ – cos θ) = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A + cot A)
[Hint : Simplify LHS and RHS separately]
(x) (1 + tan² A)/(1 + cot² A) = (1 – tan² A)/(1 – cot² A) = tan² A

Answer :

(i) LHS:
(cosec θ – cot θ)² = (1/sin θ – cos θ/sin θ)²      
                                = {(1 – cos θ)/sin θ}²
                                = (1 – cos θ)²/sin² θ    
                                = (1 – cos θ)²/(1 – cos² θ)                           [sin² θ + cos² θ = 1]
                                = (1 – cos θ)²/{(1 – cos θ) (1 + cos θ)}      [a² – b² = (a – b)(a + b)]
                                = (1 – cos θ)/{(1 + cos θ) 
                                = RHS

(ii) LHS:
cos A/(1 + sin A) + (1 + sin A)/cos A = {cos² A + (1 + sin A)²}/{(1 + sin A) * cos A}
                                                               = {cos² A + 1 + sin² A + 2sin A}/{(1 + sin A) * cos A}     
                                                              = {1 + 1 + 2sin A}/{(1 + sin A) * cos A}       [sin² θ + cos² θ = 1]
                                                              = {2 + 2sin A}/{(1 + sin A) * cos A}
                                                              = 2(1 + sin A)/{(1 + sin A) * cos A}
                                                              = 2/cos A
                                                              = 2 sec A
                                                              = RHS