sect + tan t= (sqr 3)
To solve the equation sect + tan t = √3, we can use the trigonometric identities:
1. sec t = 1 / cos t
2. tan t = sin t / cos t
Let's substitute these values into the equation:
1 / cos t + sin t / cos t = √3
Now, let's find a common denominator for the left side:
(sin t + cos t) / cos t = √3
To eliminate the fraction, we can multiply both sides of the equation by cos t:
(sin t + cos t) = √3 * cos t
Expanding the equation:
sin t + cos t = √3 * cos t
Next, let's use the Pythagorean identity:
sin^2 t + cos^2 t = 1
Rearranging the equation:
cos^2 t = 1 - sin^2 t
Substituting this into the previous equation:
sin t + √(1 - sin^2 t) = √3 * √(1 - sin^2 t)
Squaring both sides of the equation to eliminate the square root:
(sin t)^2 + 2sin t√(1 - sin^2 t) + (1 - sin^2 t) = 3(1 - sin^2 t)
Simplifying the equation:
(sin t)^2 + 2sin t√(1 - sin^2 t) + 1 - sin^2 t = 3 - 3(sin^2 t)
Collecting like terms:
2sin t√(1 - sin^2 t) - 2(sin^2 t) = 2 - 2(sin^2 t)
Dividing both sides by 2:
sin t√(1 - sin^2 t) - sin^2 t = 1 - sin^2 t
Rearranging the equation:
sin t√(1 - sin^2 t) = 1
Square both sides of the equation to eliminate the square root:
(sin t)^2 * (1 - sin^2 t) = 1
Expanding the equation:
(sin^3 t) - (sin^5 t) = 1
Rearranging the equation:
(sin^5 t) - (sin^3 t) + 1 = 0
This is a quintic equation in terms of sin t. Unfortunately, solving a general quintic equation algebraically is not always possible. In this case, we can use numerical methods such as graphical methods or approximation techniques like the Newton-Raphson method to find an approximate solution.
Therefore, the solution to the equation sect + tan t = √3 involves finding the roots of the quintic equation (sin^5 t) - (sin^3 t) + 1 = 0 using numerical methods.