Question 4 - Case Based Questions (MCQ) - Chapter 2 Class 12 Inverse Trigonometric Functions

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Question Architect Taqwa was asked to design an office building for a multi-national company. The fine storied building has five pillars in the lawn, which are congruent and in the shape of triangular prisms. Two of the base angles are given to be tan–1 2 and tan–1 3. Question 1 (i) tan−1 2 + tan−1 3 _______. (a) 𝜋/4 (b) 𝜋/2 (c) 3𝜋/4 (d) 𝜋 Now, tan−1 2 + tan−1 3 = tan−1 ((2 + 3)/(1 − 2 × 3)) = tan−1 (5/(1 − 6)) = tan−1 (5/(−5)) = tan−1 (−1) = 𝟑𝝅/𝟒 So, the correct answer is (C) = tan−1 (5/(−5)) = tan−1 (−1) = 𝟑𝝅/𝟒 So, the correct answer is (C) Question 2 (i) The third angle is _______. (a) 𝜋/4 (b) 𝜋/2 (c) 3𝜋/4 (d) 𝜋 Let the third angle be x Since all three angles are in a triangle Sum of angles = 180° Sum of angles = 𝜋 tan−1 2 + tan−1 3 + x = 𝜋 3𝜋/4 + x = 𝜋 x = 𝜋 − 3𝜋/4 x = 𝝅/𝟒 So, the correct answer is (a) Question 3 (i) If tan–1 x + tan–1 y = 𝜋/4 , than x + y + xy = _______. (a) 1 (b) 0 (c) −1 (d) None of these Given that tan–1 x + tan–1 y = 𝜋/4 tan−1 ((𝒙 + 𝒚)/(𝟏 − 𝒙𝒚)) = 𝝅/𝟒 (𝑥 + 𝑦)/(1 − 𝑥𝑦) = tan 𝜋/4 (𝑥 + 𝑦)/(1 − 𝑥𝑦) = 1 x + y = 1 − xy x + y + xy = 1 So, the correct answer is (a) Question 4 If tan–1 x + tan–1 y + tan–1 z = 𝜋, then x + y + z = (A) 1 (B) 0 (C) xyz (D) xy + yz + zx Given that tan–1 x + tan–1 y + tan–1 z = 𝜋 tan–1 x + tan–1 y = 𝜋 − tan–1 z tan–1 x + tan–1 y = tan–1 0 − tan–1 z tan−1 ((𝑥 + 𝑦)/(1 − 𝑥𝑦)) = tan−1 ((0 − 𝑧)/(1 − 0 × 𝑧)) tan−1 ((𝑥 + 𝑦)/(1 − 𝑥𝑦)) = tan−1 (−𝑧) (𝑥 + 𝑦)/(1 − 𝑥𝑦) = −𝑧 𝑥+𝑦 = −𝑧(1−𝑥𝑦) 𝑥+𝑦 = −𝑧+𝑥𝑦𝑧 𝑥+𝑦+𝑧=𝒙𝒚𝒛 So, the correct answer is (c) Question 5 If tan–1 x + tan–1 y + tan–1 z = 𝜋/2 , then xy + yz + zx = (A) 1 (B) 0 (C) xyz (D) xy + yz + zx Given that tan–1 x + tan–1 y + tan–1 z = 𝜋/2 tan–1 x + tan–1 y = 𝝅/𝟐 − tan–1 z tan–1 x + tan–1 y = cot–1 z tan–1 x + tan–1 y = tan–1 (1/𝑧) tan−1 ((𝒙 + 𝒚)/(𝟏 − 𝒙𝒚)) = tan−1 (𝟏/𝒛) (𝑥 + 𝑦)/(1 − 𝑥𝑦) = 1/z 𝑥𝑧+𝑦𝑧=1−𝑥𝑦 𝒙𝒚+𝒙𝒛+𝒚𝒛=𝟏 So, the correct answer is (a)