NCERT Solutions for Class 10 Maths Ex 8.3

NCERT Solutions for Class 10 Maths Ex 8.3

In this post, you will find the NCERT solutions for class 10 maths ex 8.3. These solutions are based on the latest syllabus of NCERT Maths class 10.


NCERT Solutions for Class 10 Maths Ex 8.3


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, cosecA – cotA = 1

 cosecA = 1 + cotA

 

 

 

For sec A,

Let us use the identity, secA – tanA = 1

 secA = 1 + tanA

  

  

  

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, sinA + cosA = 1

 sinA = 1 – cosA

 

 

For cos A,


For tan A,

Let us use the identity, secA – tanA = 1                 

⇒ tanA = secA – 1

 

For cot A,

 

For cosec A,

 

3. 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 – sinA = cos2 A]

(iv) (D) 

                                = 

                                = tan2 A

4. 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 cosec2A = 1 + cot2A.

(vi) 

(vii) 

(viii) (sin A + cosec A)2 + (cos A + sec A)7 + tanA + cotA

(ix) (cosec A – sin A) (sec A – cos A)

(x) 

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

                                 = cosecθ + cotθ – 2 cosec θ cot θ

                                 = 

                                 = 

                                 = 

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

                                 = 

                                 = 

                                  = 

                                  = R.H.S.

(ii) L.H.S. = 

                  = 

                  = 

                  =                        [Since, sinθ + cosθ = 1]

                  = 

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

(iii) L.H.S. =

                  =

                  =

                  =

                  =

                  =

                  =

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

                  =     [Since, sinθ + cosθ = 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 – sinθ = cosθ]

= 

= 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 + cosecA + secA

= 5 + 1 + cotA + 1 + tanA

[Since, cosecθ = 1 + cotθ and secθ = 1 + tanθ]

= 7 + tanA + cotA

= R.H.S.

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

= 

= 

= 

= sin A cos A

=                      [Since, sinθ + cosθ = 1]

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

= 

= 

=  

= R.H.S.

(x) L.H.S. = 

[Since, 1 + tanθ = secθ and 1 + cotθ = cosecθ]

=  

= tanA = R.H.S.

Now, the middle term = 

=  

= (– tan A)2

= tan

= R.H.S.

Please do not enter any spam link in the comment box.

Post a Comment (0)
Previous Post Next Post