If $\sec 4A = \csc(3A - 50^\circ) $, where $ 4A $ and $ 3A $ are acute angles, find the value of $\csc(A + 25^

Question

If $\sec 4A = \csc(3A - 50^\circ) $, where $ 4A $ and $ 3A $ are acute angles, find the value of $\csc(A + 25^\circ) $.

Examoogle · Accepted Answer

The correct answer is Option D: $\sqrt{2}$. Use the complementary angle property $\sec 	heta = \csc(90^\circ - 	heta)$ to equate the angles directly without converting to sine and cosine.

Step-by-Step Solution & Concepts:
Shortcut Trick: Use the complementary angle property $\sec 	heta = \csc(90^\circ - 	heta)$ to equate the angles directly without converting to sine and cosine.

Formula Used: $$\sec 	heta = \csc(90^\circ - 	heta)$$

Given: $$\sec 4A = \csc(3A - 50^\circ)$$

Steps: Step 1: Write the given trigonometric equation: We are given the equation involving secant and cosecant functions: $$\sec 4A = \csc(3A - 50^\circ)$$. Step 2: Apply complementary angle relation: We know that $$\sec 	heta = \csc(90^\circ - 	heta)$$. Applying this to the left-hand side, we get $$\sec 4A = \csc(90^\circ - 4A)$$. Step 3: Equate the angles and solve for $A$: Since the cosecant values are equal, their angles must be equal for acute angles: $$90^\circ - 4A = 3A - 50^\circ$$. Rearranging terms gives $$7A = 140^\circ$$, which means $$A = 20^\circ$$.. Step 4: Evaluate the required expression: Substitute $A = 20^\circ$ into $$\csc(A + 25^\circ)$$: $$\csc(20^\circ + 25^\circ) = \csc(45^\circ) = \sqrt{2}$$

Answer: D is correct because evaluating $\csc(A + 25^\circ)$ for $A = 20^\circ$ yields $\csc(45^\circ) = \sqrt{2}$.

Exam Tip: Complementary angle formulas apply similarly to $\sin	heta = \cos(90^\circ - 	heta)$ and $	an	heta = \cot(90^\circ - 	heta)$.

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Discussion (0)

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)