Question

find a solution for the equation. assume that all angles are acute angles. sec θ = csc (θ + 42°) a. 21° b. 66° c. 69° d. 24°

Explanation:

Step1: Recall reciprocal and co-function identities

We know that \(\sec\theta=\frac{1}{\cos\theta}\) and \(\csc\alpha = \frac{1}{\sin\alpha}\), also \(\cos\theta=\sin(90^{\circ}-\theta)\) (co - function identity). So the equation \(\sec\theta=\csc(\theta + 42^{\circ})\) can be rewritten as \(\frac{1}{\cos\theta}=\frac{1}{\sin(\theta + 42^{\circ})}\), which implies \(\cos\theta=\sin(\theta + 42^{\circ})\).
Using the co - function identity \(\cos\theta=\sin(90^{\circ}-\theta)\), we get \(\sin(90^{\circ}-\theta)=\sin(\theta + 42^{\circ})\).

Step2: Solve for \(\theta\)

Since the sine function is one - to - one for acute angles (because the sine function is increasing on the interval \([0^{\circ},90^{\circ}]\)), if \(\sin A=\sin B\) and \(A,B\) are acute angles, then \(A = B\).
So we set \(90^{\circ}-\theta=\theta + 42^{\circ}\).
Add \(\theta\) to both sides: \(90^{\circ}=\theta+\theta + 42^{\circ}\).
Simplify: \(90^{\circ}=2\theta + 42^{\circ}\).
Subtract \(42^{\circ}\) from both sides: \(90^{\circ}- 42^{\circ}=2\theta\).
\(48^{\circ}=2\theta\).
Divide both sides by 2: \(\theta=\frac{48^{\circ}}{2}=24^{\circ}\).
We can also check by plugging \(\theta = 24^{\circ}\) into the original equation:
Left - hand side: \(\sec(24^{\circ})=\frac{1}{\cos(24^{\circ})}\approx\frac{1}{0.9135}\approx1.094\)
Right - hand side: \(\csc(24^{\circ}+ 42^{\circ})=\csc(66^{\circ})=\frac{1}{\sin(66^{\circ})}\approx\frac{1}{0.9135}\approx1.094\)

Answer:

D. \(24^{\circ}\)