Find the acute angles A and B satisfying secAcotB-secA-2cotB+2=0
Ans : secAcotB-secA-2cotB+2=0
2cotB-secAcotB+secA=2
secAcotB-secA-2cotB+2=0
secA(cotB-1) + 2(cotB-1) = 0
(secA+2)(cotB-1) = 0
Now it should be clear what the angles are.
Not got the answer
To solve the equation sec(A)cot(B) - sec(A) - 2cot(B) + 2 = 0, we can simplify it further using some trigonometric identities.
Step 1: Factor out common terms.
sec(A)(cot(B) - 1) - 2(cot(B) - 1) = 0
Step 2: Notice that we have a common factor of (cot(B) - 1). Factor it out.
(cot(B) - 1)(sec(A) - 2) = 0
Step 3: Set each factor equal to zero and solve for the acute angles.
cot(B) - 1 = 0
cot(B) = 1
Using the definition of cotangent as adjacent over opposite, we have:
cot(B) = 1 = adjacent/opposite
Since we are looking for acute angles, we can choose a right-angled triangle where the adjacent side is equal to the opposite side. So, let's say the adjacent and opposite sides are both 1.
Using the Pythagorean theorem, we can find the hypotenuse:
hypotenuse^2 = adjacent^2 + opposite^2
hypotenuse^2 = 1^2 + 1^2
hypotenuse^2 = 2
hypotenuse = √2
Now, we can use the definitions of trigonometric functions to find the acute angle B:
cot(B) = adjacent/opposite = 1/1 = 1
Since cot(B) = 1, we can say that B = 45 degrees or B = π/4 radians.
sec(A) - 2 = 0
sec(A) = 2
Using the definition of secant as hypotenuse over adjacent, we can say:
sec(A) = hypotenuse/adjacent = √2 / 1 = √2
Now, we can use the definition of trigonometric functions to find the acute angle A:
sec(A) = √2
By taking the inverse secant (or arcsec) of both sides, we get:
A = sec^(-1)(√2)
So, the acute angles A and B satisfying the equation sec(A)cot(B) - sec(A) - 2cot(B) + 2 = 0 are:
A = sec^(-1)(√2)
B = 45 degrees (or π/4 radians).