Find all the angles between 0° and 90° which satisfy the equation
sec²Θcosec²Θ + 2cosec²Θ = 8
multiply by sin^2 cos^2 to get
1 + 2 cos^2 = 8 sin^2 cos^2
8cos^4 - 6cos^2 + 1 = 0
(4cos^2 - 1)(2 cos^2 - 1)
so,
cos^2 = 1/4 or 1/2
cos = 1/2 or -1/2 or 1/√2 or -1/√2
skip the negative values, since we want 1st quadrant angles only
Θ = 45° or 60°
To find all the angles between 0° and 90° that satisfy the equation sec²Θcosec²Θ + 2cosec²Θ = 8, we can start by simplifying the equation.
1. Recall the trigonometric identities:
sec²Θ = 1 + tan²Θ
cosec²Θ = 1 + cot²Θ
2. Substitute these identities into the equation:
(1 + tan²Θ)(1 + cot²Θ) + 2(1 + cot²Θ) = 8
3. Simplify the equation:
(1 + tan²Θ)(1 + cot²Θ) + 2(1 + cot²Θ) = 8
(1 + tan²Θ)(1 + cot²Θ) + 2cot²Θ + 2 = 8
(1 + tan²Θ)(1 + cot²Θ) + 2cot²Θ = 6
4. Expand the brackets:
(1 + tan²Θ + cot²Θ + tan²Θcot²Θ) + 2cot²Θ = 6
1 + tan²Θ + cot²Θ + tan²Θcot²Θ + 2cot²Θ = 6
5. Combine the terms:
1 + 3tan²Θ + 3cot²Θ + tan²Θcot²Θ = 6
6. Rearrange the terms:
tan²Θcot²Θ + 4tan²Θ + 4cot²Θ = 5
At this point, we notice that the equation is quadratic in nature. We can make a substitution to simplify it further.
Let x = tan²Θ. Then, we can rewrite the equation as:
x² + 4x + 4/x = 5
Now, we have a quadratic equation that we can solve.
7. Multiply both sides of the equation by x to get rid of the denominator:
x³ + 4x² + 4 = 5x
8. Rearrange the equation to have it equal to zero:
x³ + 4x² - 5x + 4 = 0
9. Solve the equation x³ + 4x² - 5x + 4 = 0 using numerical or algebraic methods. Once you find the value(s) of x, substitute them back into the equation x = tan²Θ.
10. Solve for Θ using the equation tan²Θ = x. Take the square root of x to find the positive and negative values of tanΘ. Then, use the inverse tangent (tan⁻¹) function to find the angle Θ.
11. Check that the angles you find are between 0° and 90°.
Following these steps will help you find all the angles between 0° and 90° that satisfy the given equation.