Given that the tan thetha = p/2 where then angle is between 0 and 90, determine the following in terms of p:
sec thetha
cos thetha
Since tan is opposite/ adajcent.. 2 would be the adjacent. p would be the opposit. Then in order to determine the hypotenuse you would use theroem of pythagoras(it is a right angled triangle)
So heres my question:
The hypotenuse would be equal to the sqaure root of p^2 + 4, but can you simplify that to p+2, or should it remain in square root form ?
sec thetha
cos thetha
Since tan is opposite/ adajcent.. 2 would be the adjacent. p would be the opposit. Then in order to determine the hypotenuse you would use theroem of pythagoras(it is a right angled triangle)
So heres my question:
The hypotenuse would be equal to the sqaure root of p^2 + 4, but can you simplify that to p+2, or should it remain in square root form ?
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Angle is between 0 and 90
=> Angle Θ is in quadrant One
=> All Trig. Ratios are positive
Given, tan Θ = p/2
sec Θ = √(1 + tan² Θ)
= √(1 + p²/4)
= √(p² + 4) / 2
sec Θ = 1 / cosΘ = 2 /√(p² + 4)
Your line of thinking is almost correct
Opp. side = p*m and Adj, side = 2m, where m is any real number
=> Hypot. = m*√(p² + 4)
But you CAN NOT write √(p² + 4) as equal to p + 2 , because
(p + 2)² = p² + 4 + 4p
=> Angle Θ is in quadrant One
=> All Trig. Ratios are positive
Given, tan Θ = p/2
sec Θ = √(1 + tan² Θ)
= √(1 + p²/4)
= √(p² + 4) / 2
sec Θ = 1 / cosΘ = 2 /√(p² + 4)
Your line of thinking is almost correct
Opp. side = p*m and Adj, side = 2m, where m is any real number
=> Hypot. = m*√(p² + 4)
But you CAN NOT write √(p² + 4) as equal to p + 2 , because
(p + 2)² = p² + 4 + 4p
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