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Question
Without using trigonometric tables, find the value of the expression:
`2((cos 58^circ)/(sin 32^circ)) - sqrt(3)((cos 38^circ "cosec" 52^circ)/(tan 15^circ tan 60^circ tan 75^circ))`
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Solution
Given: `2((cos 58^circ)/(sin 32^circ)) - sqrt(3)((cos 38^circ "cosec" 52^circ)/(tan 15^circ tan 60^circ tan 75^circ))`
Step-wise calculation:
1. sin 32° = cos 58° ...(Since sin(90° – θ) = cos θ)
So `(cos 58^circ)/(sin 32^circ) = 1`
⇒ 2(...) = 2
2. `cos 38^circ · "cosec" 52^circ = (cos 38^circ)/(sin 52^circ)`.
But sin 52° = cos 38° ...(52° = 90° – 38°)
So numerator = 1.
3. `tan 60^circ = sqrt(3)` and `tan 75^circ = cot 15^circ = 1/(tan 15^circ)`.
Thus tan 15° · tan 60° · tan 75°
= `tan 15° · sqrt(3) · (1/tan 15^circ)`
= `sqrt(3)`
4. Therefore the second term is `sqrt(3) · (1/sqrt(3)) = 1`.
