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secondary 3 | A Maths
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please help me with this question :)
since cos is adjcacent/hypotenuse,
We can let the adjacent be k and hypotenuse be 1.
Using Pythagoras' theorem,
k² + opposite² = 1²
opposite² = 1 - k²
opposite = √(1 - k²)
sin35° = √(1 - k²) / 1 = √(1 - k²)
tan35° + tan45° + tan55°
= sin35°/cos35° + 1 + tan(90° - 35°)
(Recall tan 45° = 1 and tanx = sinx/cosx)
= sin35°/cos35° + 1 + 1/tan35°
(Recall tan(90° - θ) = 1/tanθ for degrees,
or (tan(½π) - θ) = 1/tanθ for radians)
= sin35°/cos35° + 1 + cos35°/sin35°
= (sin²35° + cos²35°)/(sin35°cos35°) + 1
= 1/ksin35° + 1
= 1/k(√(1 - k²)) + 1
= 1/(k√(1-k)(1+k)) + 1
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