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secondary 3 | A Maths
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Rachel
Rachel

secondary 3 chevron_right A Maths chevron_right Singapore

please help me with this question :)

Date Posted: 4 years ago
Views: 333
J
J
4 years ago
cos35° = k = k/1

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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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
We must draw out the triangle and use Pythagoras theorem and Toa cah Soh to obtain the expressions for Tan 35 and Tan 55. As for Tan 45, it’s a special angle case.
Rachel
Rachel
4 years ago
thank you so much!!