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4. Choose the correct option. Justify your choice. (i) 9 sec2A – 9 tan2A = (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) 1+tan2A/1+cot2A = (A) sec2 A (B) -1 (C) cot2A (D) tan2A

4. Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(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) 1+tan2A/1+cot2A =

(A) sec2 A (B) -1 (C) cot2A (D) tan2A

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 60787 Points
2 years ago
(i) (B) is correct. Justification: Take 9 outside, and it becomes 9 sec2A – 9 tan2A = 9 (sec2A – tan2A) = 9×1 = 9 (∵ sec2 A – tan2 A = 1) Therefore, 9 sec2A – 9 tan2A = 9 (ii) (C) is correct Justification: (1 + tan θ + sec θ) (1 + cot θ – cosec θ) We know that, tan θ = sin θ/cos θ sec θ = 1/ cos θ cot θ = cos θ/sin θ cosec θ = 1/sin θ Now, substitute the above values in the given problem, we get = (1 + sin θ/cos θ + 1/ cos θ) (1 + cos θ/sin θ – 1/sin θ) Simplify the above equation, = (cos θ +sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ = (cos θ+sin θ)2-12/(cos θ sin θ) = (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ) = (1+ 2cos θ sin θ -1)/(cos θ sin θ) (Since cos2θ + sin2θ = 1) = (2cos θ sin θ)/(cos θ sin θ) = 2 Therefore, (1 + tan θ + sec θ) (1 + cot θ – cosec θ) =2 (iii) (D) is correct. Justification: We know that, Sec A= 1/cos A Tan A = sin A / cos A Now, substitute the above values in the given problem, we get (secA + tanA) (1 – sinA) = (1/cos A + sin A/cos A) (1 – sinA) = (1+sin A/cos A) (1 – sinA) = (1 – sin2A)/cos A = cos2A/cos A = cos A Therefore, (secA + tanA) (1 – sinA) = cos A (iv) (D) is correct. Justification: We know that, tan2A =1/cot2A Now, substitute this in the given problem, we get 1+tan2A/1+cot2A = (1+1/cot2A)/1+cot2A = (cot2A+1/cot2A)×(1/1+cot2A) = 1/cot2A = tan2A So, 1+tan2A/1+cot2A = tan2A

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