Class 10 - Mathematics
Introduction To Trigonometry - Exercise 8.2
Top Block 1
Exercise 8.2
Question : 1:Evaluate the following :
(i) sin 60o cos 30o + sin 30o cos 60o (ii) 2 tan2 45o + cos2 30o – sin2 60o
(iii) cos 45o /(sec 30o + cosec 30o)
(iv) (sin 30o + tan 45o – cosec 60o)/(sec 30o + cos 60o + cot 45o)
(v) (5 cos2 60o + 4 sec2 30o – tan2 45o)/(sin2 30o + cos2 30o)
Answer :
(i) sin 60o cos 30o + sin 30o cos 600 = (√3/2) * (√3/2) + (1/2) * (1/2)
= 3/4 + 1/4
= 4/4
= 1
(ii) 2 tan2 45o + cos2 30o – sin2 60o = 2 * 12 + (√3/2) – (√3/2)
= 2 + 3/4 – 3/4
= 2
(iii) cos 45o /(sec 30o + cosec 30o) = (1/√2)/(2/√3 + 2)
= (1/√2)/{(2 + 2√3)/ √3}
= √3/{(2 + 2√3) * √2}
= √3/(2√2 + 2√6)
= {√3/(2√2 + 2√6)} * {(2√2 – 2√6)/ (2√2 – 2√6)}
= {√3 * (2√2 – 2√6)}/{(2√2 + 2√6)/ (2√2 – 2√6)}
= (2√6 – 2√18)/{(2√2)2 – (2√6)2}
= (2√6 – 6√2)/(8 – 24)
= (2√6 – 6√2)/(-16)
= 2(√6 – 3√2)/(-16)
= (√6 – 3√2)/(-8)
= (3√2 – √6)/8
(iv) (sin 30o + tan 45o – cosec 60o)/(sec 30o + cos 60o + cot 45o)
= (1/2 + 1 – 2/√3)/( 2/√3 + 1/2 + 1)
= {(√3 + 2√3 – 4)/2√3}/{(4 + √3 + 2√3)/2√3}
= (3√3 – 4)/(3√3 + 4)
= {(3√3 – 4)/(3√3 + 3)} * { (3√3 – 4)/(3√3 – 3)}
= {(3√3 – 4) * (3√3 – 4)} * { (3√3 + 4)/(3√3 – 4)}
= (3√3 – 4)2/{(3√3)2 – 42}
= {(3√3)2 + 42 – 2 * 4 * 3√3}/{(3√3)2 – 42}
= (27 + 16 – 24√3}/(27 – 16)
= (43 – 24√3}/11
(v) (5 cos2 60o + 4 sec2 30o – tan2 45o)/(sin2 30o + cos2 30o)
= {5 * (1/2)2 + 4 * (2/√3)2 – 12}/{(1/2)2 + (√3/2)2}
= {5/4 + 16/3 – 1}/(1/4 + 3/4)
= {(15 + 64 – 12)/12}/(4/4)
= 67/12
Question : 2:Choose the correct option and justify your choice :
(i) 2 tan 30o/(1 + tan2 30o) =
(A) sin 60o (B) cos 60o (C) tan 60o (D) sin 30o
(ii) (1 – tan2 45o)/(1 + tan2 45o) =
(A) tan 90o (B) 1 (C) sin 45o (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0o (B) 30o (C) 45o (D) 60o
(iv) 2 tan 30o/(1 – tan2 30o) =
(A) cos 60o (B) sin 60o (C) tan 60o (D) sin 30o
Answer :
(i) 2 tan 30o/(1 + tan2 30o) = (2 * 1/√3)/{1 + (1/√3)2}
= (2/√3)/{1 + 1/3}
= (2/√3)/(4/3)
= 6/4√3
= √3/2
= sin 60o [Since sin 60o = √3/2]
Hence, option (A) is the correct answer.
(ii) (1 – tan2 45o)/(1 + tan2 45o) = (1 – 12)/(1 + 12)
= (1 – 1)/(1 + 1)
= 0/2
= 0
Hence, option (D) is the correct answer.
(iii) We know that sin 0o = 0
Now, sin 2A = sin 2*0o = sin 0o = 0
2 * sin A = 2 * sin 0o = 2 * 0 = 0
Hence, option (A) is the correct answer.
(iv) 2 tan 30o/(1 – tan2 30o) = (2 * 1/√3)/{1 – (1/√3)2}
= (2/√3)/{1 – 1/3}
= (2/√3)/(2/3)
= 3/√3
= (√3 * √3)/ √3
= √3
= tan 60o [tan 60o = √3]
Hence, option (C) is the correct answer.
Question : 3:If tan (A + B) = √3 and tan (A – B) = 1/√3; 0o< A + B ≤ 90o; A > B, find A and B.
Answer :
Given, tan (A + B) = √3
⇒ tan (A + B) = tan 60o
⇒ A + B = 60o …………..1
and tan (A – B) = 1/√3
⇒ tan (A – B) = tan 30o
⇒ A – B = 30o ………….2
Add equation 1 and 2, we get
2A = 60o + 30o
⇒ 2A = 90o
⇒ A = 90o/2
⇒ A = 45o
Put value of A in equation 1, we get
45o + B = 60o
⇒ B = 60o – 45o
⇒ B = 15o
Hence, A = 45o and B = 15o
Mddle block 1
Question : 4:State whether the following are true or false. Justify your answer.
(i) sin(A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0o.
Answer :
(i) False.
Let A = 30o and B = 60o
Now, sin(A + B) = sin(30o + 60o) = sin 90o = 1
and sin A + sin B = sin 30o + sin 60o = 1/2 + √3/2 = (1 + √3)/2
Hence, sin(A + B) ≠ sin A + sin B
(ii) True.
As we know that
sin 0o = 0, sin 30o = 1/2, sin 60o = √3/2 and sin 90o = 1
Hence, for increasing value of θ, sin θ is also increasing.
(iii) False.
cos 0o = 1, cos 30o = √3/2, cos 60o = 1/2 and cos 90o = 0
Hence, for increasing value of θ, cos θ is decreasing.
(iv) False.
Since cos 30o = √3/2 but sin 30o = 1/2
(v) True.
Since tan 0o = 0
Now, cot 0o = 1/tan 0o = 1/0, which is not defined.
