Simplify tan 75° + cot 75°
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tan75° + cot75°
= sin75°/cos75° + cos75°/sin75°
= (sin75° . sin75° + cos75° . cos75°)/(cos75° . sin75°)
= (sin²75° + cos²75°)/(cos75° . sin75°)
we know, sin²x + cos²x = 1
so, sin²75° + cos²75° = 1
= 1/(cos75° . sin75°)
= 2/(2sin75°. cos75°)
we know, 2sinA.cosA = sin2A
so, 2sin75°. cos75° = sin2(75°) = sin150°
= 2/sin150°
= 2/sin(180° - 30°)
= 2/sin30°
= 2/(1/2)
= 4
hence, answer is 4
= sin75°/cos75° + cos75°/sin75°
= (sin75° . sin75° + cos75° . cos75°)/(cos75° . sin75°)
= (sin²75° + cos²75°)/(cos75° . sin75°)
we know, sin²x + cos²x = 1
so, sin²75° + cos²75° = 1
= 1/(cos75° . sin75°)
= 2/(2sin75°. cos75°)
we know, 2sinA.cosA = sin2A
so, 2sin75°. cos75° = sin2(75°) = sin150°
= 2/sin150°
= 2/sin(180° - 30°)
= 2/sin30°
= 2/(1/2)
= 4
hence, answer is 4
Answered by
4
GIVEN:-
tan75° + cot75°
=> sin75°/cos75° + cos75°/sin75°
=> (sin75° . sin75° + cos75° . cos75°)/(cos75° . sin75°)
=> (sin²75° + cos²75°)/(cos75° . sin75°)
we know, sin²x + cos²x = 1
so, sin²75° + cos²75° = 1
2sinA.cosA = sin2A
so, 2sin75°. cos75° = sin2(75°) = sin150°
=> 1/(cos75° . sin75°)
=> 2/(2sin75°. cos75°)
=> 2/sin150°
=> 2/sin(180° - 30°)
=> 2/sin30°
=> 2/(1/2)
=> 4
tan75° + cot75°
=> sin75°/cos75° + cos75°/sin75°
=> (sin75° . sin75° + cos75° . cos75°)/(cos75° . sin75°)
=> (sin²75° + cos²75°)/(cos75° . sin75°)
we know, sin²x + cos²x = 1
so, sin²75° + cos²75° = 1
2sinA.cosA = sin2A
so, 2sin75°. cos75° = sin2(75°) = sin150°
=> 1/(cos75° . sin75°)
=> 2/(2sin75°. cos75°)
=> 2/sin150°
=> 2/sin(180° - 30°)
=> 2/sin30°
=> 2/(1/2)
=> 4
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