Question 3 - Case Based Questions (MCQ) - Chapter 2 Class 12 Inverse Trigonometric Functions

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Question In the school project Sheetal was asked to construct a triangle and name it as ABC. Two angles A and B were given to be equal to tan−1 1/3 and tan−1 1/3 respectively. Question 1 (i) The value of sin A is _______. (a) 1/2 (b) 1/3 (c) 1/√5 (d) 2/√5 Given that A = tan−1 1/2 tan A = 𝟏/𝟐 Now, we know that 1 + tan2 A = sec2 A 1 + tan2 A = 1/cos^2⁡𝐴 1 + tan2 A = 1/(1 − sin^2⁡𝐴 ) Putting tan A = 1/2 1 + (𝟏/𝟐)^𝟐 = 𝟏/(𝟏 − 〖𝒔𝒊𝒏〗^𝟐⁡𝑨 ) 1 + 1/4 = 1/(1 − sin^2⁡𝐴 ) 5/4 = 1/(1 − sin^2⁡𝐴 ) 1 − sin2 A = 4/5 1 − 4/5 = sin2 A 1/5 = sin2 A sin2 A = 1/5 sin A = 𝟏/√𝟓 So, the correct answer is (C) Question 2 cos(A + B + C) = _______. (A) 1 (B) 0 (C) −1 (D) 1/2 Since ABC is a triangle, By Angle sum property of triangle A + B + C = 180° Thus, cos (A + B + C) = cos 180° = –1 So, the correct answer is (C) Question 3 If B = cos–1 x, then x = _______. (a) 1/√5 (b) 3/√10 (C) 1/√10 (d) 2/√5 Given B = tan−1 1/3 tan B = 1/3 Now, we know that 1 + tan2 B = sec2 B 1 + tan2 B = 1/cos^2⁡𝐵 cos B = 3/√10 B = cos−1 3/√10 Thus, x = 3/√10 So, the correct answer is (B) Question 4 The value of A + B = _______. (a) 𝜋/6 (b) 𝜋/4 (C) 𝜋/3 (d) 𝜋/2 Given A = tan−1 1/2 , B = tan−1 1/3 Now, A + B = tan−1 𝟏/𝟐 + tan−1 𝟏/𝟑 = tan−1 ((1/2 + 1/3)/(1 − 1/2 × 1/3)) = tan−1 (((3 + 2)/(2 × 3))/(1 − 1/6)) = tan−1 ((5/6)/(5/6)) = tan−1 (1) = 𝝅/𝟒 So, the correct answer is (B) Question 5 The third angle, ∠C = _______. (A) 𝜋/4 (B) 𝜋/2 (C) 𝜋/3 (D) 3𝜋/4 Now, A + B = 𝝅/𝟒 And, since ABC is a triangle A + B + C = 180° A + B + C = 𝜋 𝝅/𝟒 + C = 𝜋 C = 𝜋 − 𝜋/4 C = 𝟑𝝅/𝟒 So, the correct answer is (D)