1. Chapter 8 Class 10 Introduction to Trignometry
2. Serial order wise

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Ex 8.2, 2 Choose the correct option and justify your choice : (i) "2 tan 30°" /"1 + tan2 30°" = sin 60° (B) cos 60° (C) tan 60° (D) sin 30° We know that , tan "30°" = 1/√3 So, (2 tan⁡〖30°〗)/(1+𝑡𝑎𝑛2 30°) = (2 × (1/√3))/(1 + (1/√3)^2 ) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = 2/√3×3/4 = 2/√3×(√3 × √3)/4 = √3/2 Options are (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° Since sin "60°" = √3/2 Hence, option (A) is correct Ex 8.2, 2 Choose the correct option and justify your choice : (ii) "1− tan2 45°" /"1 + tan2 45°" = tan 90° (B) 1 (C) sin 45° (D) 0 We know that, tan 45° = 1 So, (1 −〖 tan〗^2⁡〖45°〗)/(1 +〖 tan〗^2⁡〖45°〗 ) = (1 − (1)^2)/(1 + (1)^2 ) = (1 − 1)/(1 + 1) = 0/2 = 0 Hence, Option (D) is correct Ex 8.2, 2 Choose the correct option and justify your choice : (iii) sin 2A = 2 sin A is true when A = 0° (B) 30° (C) 45° (D) 60° sin 2A = 2 sin A Here, we substitute the value of option in question and whichever satisfies the question would be solution . Option (A) = 0° Lets’s check it sin 2A = 2sin A sin 2(0) = 2 sin(0) sin 0° = 2 sin 0° 0 = 0 L.H.S = R.H.S Hence option A is correct Option (B) = 30° sin 2A = 2 sin A sin 2(30°) = 2(sin 30°) sin 60° = 2 sin 30° √3/2=2×1/2 √3/2=1 But √3/2 " ≠ " 1 So, Option (B) is not possible Option (c) = 45° sin 2A = 2 sin A sin 2 (45°) = 2 sin 45° sin 90° = 2 sin45° 1 = 2 ×1/√2 1 = √2 × √2 ×1/√2 1 = √2 But 1 ≠√2 So, Option (C) cannot be possible Option (D) = 60° sin 2A = 2 sin A sin 2(60°) = 2 sin(60°) sin 120° = 2 sin 60° sin (90 + 30) = 2 sin 60° cos 30° = 2sin 60° √3/2=2×√3/2 √3/2=√3 But √3/2≠√3 Hence, option (D) is incorrect So, only option (A) is correct