Test: Trigonometry (त्रिकोणमिति) - 2 - SSC MCQ

tan (45^° + A/2) = tan 45° + tan A / 2 / 1 – tan 45° tan A / 2
= 1 + tan A/2 / 1 – 1 – tan A/2 = cos A/2 + sin A/2 / cos A/2 – sin A/2
= (cos A/2 + sin A/2)² / cos²A/2 – sin² A/2
= cos² A/2 + sin² A/2 + 2 sin A/2 cos A/2 / cos A
= 1 + sin A / cos A

1 + cos x = 2 cos² x / 2
⟹ 1 + cos 45⁰ = 2 cos² 22 1⁰ / 2
⟹1 + 1 / √2 = 2cos² 22 1⁰ / 2
⟹√2 + 1 / √2 = 2 cos² 22 1⁰/2
⟹cos 22 1⁰ / 2 = √2 + 1 / 2√2

a² – b² = (cos a + sin a)² – (cos a – sin a)²
= 4 cos a sin a
= ab = (cos a + sin a) (cos a – sin a
= cos² – sin²a = cos 2a
= a² – b² / 2ab = 2 sin 2a / 2 cos 2a = tan 2a

1 + sin 2x / 1 – sin ²x = sin2x + cos² x + 2 sin x cos x / sin² x + cos² x-2 sin x cos x
= (tan x + 1 / tan x – 1)²
अतः √1 + sin 2x / 1 – sin 2x = tan x + 1 / tan x – 1

cos 20⁰ cos 40⁰ cos 60⁰ cos 80⁰
= ½ 2 cos 20⁰ cos 40⁰ cos 60° cos 80°
= ½ (cos 60° + cos 20°) cos 60° cos 80°
= ¼ (cos 60° cos 80° + cos 20° cos 80°)
= ⅛ cos 80° + ⅛. 2 . cos 20° cos 80°
= ⅛ cos 80° + ⅛ (cos 100° + cos 60°)
= ⅛ (cos 80° + cos 100°) + 1/16 = 0 + 1 / 16 = 1 / 16

sin 85° – sin 25° – cos 55°
= 2 cos 85° + 25° / 2 sin 85° – 25° / 2 – cos 55°
= 2 cos 55° sin 30° – cos 55°
2 cos 55° ⨯ ½ – cos 55° = 0

sin 75° cos 15° = ½ (2sin 75° cos 15°)
= ½ [sin (75° + 15°) + sin (75° – 15°)]
= ½ (sin 90° + sin 60°)
= ½ (1 + √3 / 2) = 2 + √3 / 4

cos A + tan / cot A + tan A = cos A / sin A – sin A / cos A / cos A / sin A + sin A / cos A
= cos² A – sin² A / cos² A + sin² A = cos 2A