Use the following to answer questions :
$\alpha$ and $\beta$ are acute angles with tan($\alpha$) = and sec($\beta$) = .
-Find sin($\alpha$ +$\beta$).

Question

Use the following to answer questions :
$\alpha$ and $\beta$ are acute angles with tan($\alpha$) =  and sec($\beta$) =  .
-Find sin($\alpha$ +$\beta$).

Quizplus · Accepted Answer

We can use the trigonometric identity sin(a+b)=sin(a)cos(b)+cos(a)sin(b) to find sin(F1F1F1S1F1F1F10 + F1F1F1S1F1F1F10). 
We know that tan(F1F1F1S1F1F1F10) = opposite/adjacent = x/1, where x is the length of the opposite side of angle F1F1F1S1F1F1F10 and sec(F1F1F1S1F1F1F10) = hypotenuse/adjacent = 1/y, where y is the length of the hypotenuse of angle F1F1F1S1F1F1F10.
Using the Pythagorean theorem, we can find the length of the adjacent side of both angles:
- For angle F1F1F1S1F1F1F10, a^2 + b^2 = c^2 becomes 1^2 + x^2 = (1/tan(F1F1F1S1F1F1F10))^2, which is equivalent to 1 + x^2 = (1/x)^2. Rearranging the terms, we get x^4 + x^2 - 1 = 0. Solving this quadratic equation (by substituting y=x^2), we get y = x^2 = (sqrt(5)-1)/2, which is positive since x is an acute angle. Thus, x = sqrt((sqrt(5)-1)/2).
- For angle F1F1F1S1F1F1F10, a^2 + b^2 = c^2 becomes y^2 - 1^2 = (y/sec(F1F1F1S1F1F1F10))^2, which is equivalent to y^4 - 2y^2 - 1 = 0. Solving this quadratic equation, we get y^2 = 1 + sqrt(2), which is positive since y is an acute angle. Thus, y = sqrt(1 + sqrt(2)).
Now, we can use the trigonometric identities sin(F1F1F1S1F1F1F10) = x/sqrt(x^2+1) and cos(F1F1F1S1F1F1F10) = 1/y to find sin(F1F1F1S1F1F1F10 + F1F1F1S1F1F1F10) as follows:
sin(F1F1F1S1F1F1F10 + F1F1F1S1F1F1F10) = sin(F1F1F1S1F1F1F10)cos(F1F1F1S1F1F1F10) + cos(F1F1F1S1F1F1F10)sin(F1F1F1S1F1F1F10)
= (x/sqrt(x^2+1))(1/y) + sqrt(x^2+1)/sqrt(x^2+1)(1/y)
= (xy + sqrt(x^2+1))/sqrt(x^2+1)y
= {[sqrt((sqrt(5)-1)/2)]*[sqrt(1+sqrt(2)]) + sqrt((sqrt(5)-1)/2+1)}/sqrt((sqrt(5)-1)/2)(sqrt(1+sqrt(2)))
= [sqrt(2) + sqrt(10)]/sqrt(2sqrt(2)+2)
Rationalizing the denominator, we get
[sqrt(2) + sqrt(10)]/sqrt(2sqrt(2)+2) * (sqrt(2sqrt(2)+2)/sqrt(2sqrt(2)+2))
= [sqrt(2)(2+sqrt(2)) + sqrt(10)(2-sqrt(2))]/2
= [2sqrt(2) + 2 + sqrt(20) - sqrt(5)]/2
= sqrt(2) + sqrt(5)/2
= 11ea7f29_4ff9_0d53_9ecd_0f12a260cb03_TB3307_11

Use the Following to Answer Questions α\alphaα And β\betaβ Are Acute Angles with Tan

Use the Following to Answer Questions $\alpha$ And $\beta$ Are Acute Angles with Tan