NCERT Solutions Maths Class 10 Exercise 8.4

NCERT Solutions Maths Class 10 Exercise 8.4

                          Maths Class 10 Exercise 8.4 

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

 

Solution:  For sin A,

Let us use the identity, cosec2 A – cot2 A = 1

cosec2 A = 1 + cot2 A

 

 

 

For sec A,

Let us use the identity, sec2 A – tan2 A = 1

 sec2 A = 1 + tan2 A

  

  

  

For tan A,


2. Write all the other trigonometric ratios of A in terms of sec A.

Solution: For sin A,

Let us use the identity, sin2 A + cos2 A = 1

 sin2 A = 1 – cos2 A

 

 

For cos A,


For tan A,

Let us use the identity, sec2 A – tan2 A = 1                 

tan2 A = sec2 A – 1

 

For cot A,

 

For cosec A,

 

3. Evaluate:

(i) 

(ii) sin 25° cos 65° + cos 25° sin 65°

Solution: (i)

=

= 

[Since, sin (90° – θ) = cos θ and cos (90° – θ) = sin θ]

= 1/1 = 1                   [Since, sin2 θ + cos2 θ = 1 ]

 

(ii) sin 25° cos 65° + cos 25° sin 65°

= sin 25° cos (90° – 25°) + cos 25° sin (90° – 25°)

= sin 25° sin 25° + cos 25° cos 25°

[Since sin (90° – θ) = cos θ and cos (90° – θ) = sin θ]

= sin2 25° + cos2 25° = 1                [ Since, sin2 θ + cos2 θ = 1]

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) =

(A)  sec2 A                   (B)  –1                      (C) cot2 A                  (D) tan2 A

 

Solution: (i) (B) 9 sec2 A  – 9 tan2 A

                                                       = 9(sec2 A  – tan2 A)

                                                       = 9 × 1 = 9

(ii) (C) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

                            = 

                            = 

                            = 

                            = 

                            =           [Since, sin2θ +cos2θ =1]

                            = 

                            = 2

(iii) (D) (sec A + tan A) (1 – sin A)

                               = 

                               = 

                               =     [Since, 1 – sin2 A = cos2 A]

(iv) (D) 

                                = 

                                = tan2 A

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) 

(ii) 

(iii) 

(iv)

(v), using the identity cosec2 A = 1 + cot2 A

(vi) 

(vii) 

(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)

(x) 

Solution: (i) L.H.S. = (cosec θ – cot θ)2

                                 = cosec2 θ + cot2 θ – 2 cosec θ cot θ

                                 = 

                                 = 

                                 = 

  =           [Since, a2 + b2 – 2ab = (a – b)2]

                                 = 

                                 = 

                                  = 

                                  = R.H.S.

(ii) L.H.S. = 

                  = 

                  = 

                  =                        [Since, sin2 θ + cos2 θ = 1]

                  = 

                  = 2/cos A = 2 sec A = R.H.S.

(iii) L.H.S. =

                  =

                  =

                  =

                  =

                  =

                  =

                                  [Since, a3 – b3 = (a –b) (a2 + b2 + ab)]

                  =     [Since, sin2 θ + cos2 θ = 1]

                  =

                = 1 + sec θ cosec θ

(iv) L.H.S. =

 

=  

= 1 + cos A

= 

= 

=  

= R.H.S.

(v) L.H.S. = 

Dividing all the terms by sin A, we get

= 

= 

= 

= 

= cot A + cosec A 

= R.H.S.

(vi) L.H.S. = 

=

=              [Since, (a + b)(a – b) = a2 – b2]

=               [Since, 1 – sin2 θ = cos2 θ]

= 

= sec A + tan A 

= R.H.S.

(vii) L.H.S. = 

= 

=    [Since, 1 – sinθ = cosθ]

= 

= 

= tan θ 

= R.H.S.

(viii) L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2

= 

= 

= 

= 

= 5 + cosec2 A + sec2 A

= 5 + 1 + cot2 A + 1 + tan2 A

[Since, cosec2 θ = 1 + cot2 θ and sec2 θ = 1 + tan2 θ]

= 7 + tan2 A + cot2 A

= R.H.S.

(ix) L.H.S. = (cosec A – sin A) (sec A – cos A)

= 

= 

= 

= sin A cos A

=                      [Since, sin2 θ + cos2 θ = 1]

Dividing the numerator and the denominator by sin A cos A, we get

= 

= 

=  

= R.H.S.

(x) L.H.S. = 

[Since, 1 + tan2 θ = sec2 θ and 1 + cot2 θ = cosec2 θ]

=  

= tan2 A = R.H.S.

Now, the middle term = 

=  

= (– tan A)2

= tan2

= R.H.S.

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