Introduction to Trigonometry
Mathematics Class Tenth
NCERT Exercise 8.2 Solution part2
Question (2) Choose the correct option and justify your choice:
(i) 2 tan 30o/1 + tan2 30o
(A) sin 60 o
(B) cos 60 o
(C) tan 60o
(D) sin 30o
Answer : (A) sin 60 o
Explanation:
Given, 2 tan 30o/1 + tan2 30o
After substituting the values of trigonometric ratios of angles
= 2 × 1/3 / 1 + (1/3)2
= 2/3/1 + 1/3
= 2/3/3 + 1/3
= 2/3/4/3
= 2/3 × 3/4
= 3/23
Now, after multiplying with 3/3, we get
3/23 × 3/3
= 33/2 × 3
= 3/2
Since, among the given option, only sin 60o has the value equal to 3/2
Thus, option (A) sin 60o is the right answer.
(ii) 1 – tan2 45o/1 + tan2 45o
(A) tan 90o
(B) 1
(C) sin 45o
(D) 0
Answer: (D) 0
Explanation:
Given, 1 – tan2 45o/1 + tan2 45o
After substituting trigonometric ratios for given angles,
= 1 – 12/1 + 12
= 1 – 1/1 + 1
= 0/2 = 0
Thus, option (D) 0 is the correct answer.
(iii) sin 2 A = 2 sin A is true when A =
(A) 0o
(B) 30o
(C) 45o
(D) 60o
Answer: (A) 0o
Explanation:
Since, sin 0o = 0
Thus, only for the value of A = 0o,
sin 2 A = 2 sin A is true.
As given, sin 2A
By substituting value of A = 0o, we have
sin 2 × 0o = sin 0o = 0
And, 2sin A
= 2sin 0o = 2 × 0 = 0
Thus, option (A) 0o is correct answer.
(iv) 2tan 30o/1 – tan2 30o
(A) cos 60o
(B) sin 60o
(C) tan 60o
(D) sin 30o
Answer: (C) tan 60o
Explanation:
Given, 2tan 30o/1 – tan2 30o
After substituting trigonometric ratios of given angles
= 2 × 1/3 / 1 – (1/3)2
= 2/3 / 1 – 1/3
= 2/3 / 3 – 1/3
= 2/3 / 2/3
= 2/3 × 3/2 = 3/3
Now after multiplying with 3/3 we get
3/3 × 3/3
= 33/3
= 3
Now since tan 60o = 3
Thus, option (C) tan 60o is the correct answer.
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