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secondary 3 | A Maths
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Date Posted: 4 years ago
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tan1° tan2° tan 3° ... tan 88° tan89°
= tan(90 - 89)° tan(90 - 88)° tan (90 - 87)° ... tan 88° tan89°
= (1/tan89°)(1/tan88°)(1/tan87°) ... tan88° tan89°
This is done using the property tan(90 - θ)° = 1/tanθ
So tan1° cancels out tan 89°, tan2° cancels out tan88°, tan3° cancels out tan87° and so on, leaving you with just 1 multiplied by itself many times.
Eventually we reach tan44° cancels out tan46°. We have tan45° left, which isn't cancelled out.
But tan45° = 1.
So,
tan1° tan2° tan 3° ... tan 88° tan89°
= (1/tan89°)(1/tan88°)(1/tan87°) ... tan88° tan89°
= (tan89°/tan89°)(tan88°/tan88°)(tan87°/tan87°)...(tan47°/tan47)(tan46°/tan46°)(tan45°)
= (1)(1)(1)...(1)(1)(1)
= 1
= tan(90 - 89)° tan(90 - 88)° tan (90 - 87)° ... tan 88° tan89°
= (1/tan89°)(1/tan88°)(1/tan87°) ... tan88° tan89°
This is done using the property tan(90 - θ)° = 1/tanθ
So tan1° cancels out tan 89°, tan2° cancels out tan88°, tan3° cancels out tan87° and so on, leaving you with just 1 multiplied by itself many times.
Eventually we reach tan44° cancels out tan46°. We have tan45° left, which isn't cancelled out.
But tan45° = 1.
So,
tan1° tan2° tan 3° ... tan 88° tan89°
= (1/tan89°)(1/tan88°)(1/tan87°) ... tan88° tan89°
= (tan89°/tan89°)(tan88°/tan88°)(tan87°/tan87°)...(tan47°/tan47)(tan46°/tan46°)(tan45°)
= (1)(1)(1)...(1)(1)(1)
= 1
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