hi please answer my question now itself tommorow exam
Answers
9)
sin² 37° + sin² 53° + sin² 90°
= sin² 37° + cos² 37° + sin² 90° [sin x = cos (90° - x)]
= 1 + 1 [sin²x + cos²x = 1 ; sin 90° = 1]
= 2
10)
tan 35° tan 45° tan 55°
= tan 35° tan 45° cot 35° [tan x = cot (90° - x)]
= 1 × 1 [tan x · cot x = 1 ; tan 45° = 1]
= 1
11)
(sec 72° sin 18° + tan 72° cot 18°) / cos 60°
= (cosec 18° sin 18° + tan 72° tan 72°) / (1 / 2) [sec x = cosec (90° - x) ; tan x = cot (90° - x)]
= 2 (1 + tan² 72°) [cosec x sin x = 1]
= 2 sec² 72° [1 + tan² x = sec² x]
= 2[6 + 2√5] [cos 72° = 1 / (1 + √5)]
= 12 + 4√5
12)
tan 60° / tan 30°
= √3 / (1 / √3)
= √3 × √3
= 3
Answer:
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