Important Questions for CBSE Class 10 Maths Chapter 8 - Introduction to Trigonometry 4 Mark Question

1. Express the trigonometric ratios   and   in terms of 
Ans. For 
By using identity 
 
 
 
 
For 
By using identity 
 
  = 
 
 
For 

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2. Write the other trigonometric ratios of A in terms of 
Ans.  For 
By using identity, 
 
  = 
 
For 

For 
By using identity 
 
 
For 
 = 
 
For 

 

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3. Evaluate:
(i) 
(ii) 
Ans. (i) 
= 
= 

=  
(ii) 
= 
= 

=  = 1

=  = 1

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4. Choose the correct option. Justify your choice:
(i)   =
(A) 1 (B) 9 (C) 8 (D) 0
(ii)   =
(A) 0 (B) 1 (C) 2 (D) none of these
(iii)   =
(A)   (B)   (C)   (D) 
(iv)   =
(A)   (B)   (C)   (D) none of these
Ans. (i) (B)
=  = 9 x 1 = 9
(ii) (C)
= 
= 
= 
= 
= 

=  = 2
(iii) (D)
= 
= 
=  =  = 

(iv) (D)
=  = 
=  =  = 

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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 
(vi) 
(vii) 
(viii) 
(ix) 
(x) 
Ans.  Proof:
(i) L.H.S. 
= 
= 
= 
= 
= 

= 
= 
=  = R.H.S.
(ii) L.H.S. 
= 
= 
= 

=  = 
=  =  = R.H.S
(iii) L.H.S. 
= 
= 
= 
= 
= 
= 

=  
=  = 
= 
(iv) L.H.S. 
=  = 
= 
= 
=  = R.H.S.
(v) L.H.S. 
Dividing all terms by 
=  = 
= 
= 
= 
=  = R.H.S.
(vi) L.H.S. 
= 
= 

=  
=  = 
=  = R.H.S.
(vii) L.H.S.  = 
=  
= 
=  =  =  = R.H.S
(viii) L.H.S. 
= 
= 
= 
= 
= 
= 

= 
= R.H.S.
(ix) L.H.S. 
= 
= 
=  = 
=  
Dividing all the terms by ,
=  = 
=  = R.H.S.
(x) L.H.S. 
=  
=  =  = R.H.S.
Now, Middle side =  = 
= 
=  =  =  = R.H.S.

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6. Use Euclid’s division algorithm to find the HCFof:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Ans. (i) 135 and 225
We have 225 > 135,
So, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Here remainder 90 ≠ 0, we apply the division lemma again to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45≠ 0, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since that time the remainder is zero, the process get stops.
The divisor at this stage is 45
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
We have 38220 > 196,
So, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since we get the remainder is zero, the process stops.
The divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
We have 867 > 255,
So, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Here remainder 102 ≠ 0, we apply the division lemma again to 255 and 102 to obtain
255 = 102 × 2 + 51
Here remainder 51 ≠ 0, we apply the division lemma again to 102 and 51 to obtain
102 = 51 × 2 + 0
Since we get the remainder is zero, the process stops.
The divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.

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7. Evaluate:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
Ans. (i)  = 
=  =1
(ii)  = 
=  =2
(iii) 
=  = 
=  = 
= 
= 
= 
= 
(iv)  = 
= 
= 
= 
=  = 
(v) 
= 
= 
= 
= 

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8. 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 
(vi) 
(vii) 
(viii) 
(ix) 
(x) 
Ans. (i) L.H.S. 
= 
= 
= 
= 
=  
= 
=  =  = R.H.S.
(ii) L.H.S. 
= 
= 
=  
=  =  = 
=  = R.H.S
(iii) L.H.S. 
= 
= 
= 
= 
= 
= 

=  
= 
= 
= 
(iv) L.H.S. 
=  = 
= 
= 
=  = R.H.S.
(v) L.H.S. 
Dividing all terms by 
=  = 
= 
= 
= 
=  = R.H.S.
(vi) L.H.S. 
= 
=  
=  
=  =  =  = R.H.S.
(vii) L.H.S.  = 
=  
= 
=  =  =  = R.H.S
(viii) L.H.S. 
= 
= 
= 
= 
= 
= 

= 
= R.H.S.
(ix) L.H.S. 
= 
= 
=  = 
= 

Dividing all the terms by ,
=  = 
=  = R.H.S.
(x) L.H.S.  = 

=  =  = R.H.S.
Now, Middle side = 
= 
= 
=  = 
=  = R.H.S.