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2. Choose the correct option and justify your choice : (i) 2tan 30°/1+tan230° = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° (ii) 1-tan245°/1+tan245° = (A) tan 90° (B) 1 (C) sin 45° (D) 0 (iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60° (iv) 2tan30°/1-tan230° = (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan245°/1+tan245° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°

(iv) 2tan30°/1-tan230° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8741 Points
3 months ago
(i) (A) is correct. Substitute the of tan 30° in the given equation tan 30° = 1/√3 2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2 = (2/√3)/(1+1/3) = (2/√3)/(4/3) = 6/4√3 = √3/2 = sin 60° The obtained solution is equivalent to the trigonometric ratio sin 60° (ii) (D) is correct. Substitute the of tan 45° in the given equation tan 45° = 1 1-tan245°/1+tan245° = (1-12)/(1+12) = 0/2 = 0 The solution of the above equation is 0. (iii) (A) is correct. To find the value of A, substitute the degree given in the options one by one sin 2A = 2 sin A is true when A = 0° As sin 2A = sin 0° = 0 2 sin A = 2 sin 0° = 2 × 0 = 0 or, Apply the sin 2A formula, to find the degree value sin 2A = 2sin A cos A ⇒2sin A cos A = 2 sin A ⇒ 2cos A = 2 ⇒ cos A = 1 Now, we have to check, to get the solution as 1, which degree value has to be applied. When 0 degree is applied to cos value, i.e., cos 0 =1 Therefore, ⇒ A = 0° (iv) (C) is correct. Substitute the of tan 30° in the given equation tan 30° = 1/√3 2tan30°/1-tan230° = 2(1/√3)/1-(1/√3)2 = (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60° The value of the given equation is equivalent to tan 60°.

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