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What would be the solution to this Pythagorean/Trigonometric equation? (1+ctg²B) sen²B = 1.

a. B = 0
b. B = 45°
c. B = 90°
d. Not Mentioned

User Ari
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1 Answer

5 votes

Final answer:

The given equation is a trigonometric identity that is true for any angle B where sin(B) is defined. The options given do not encompass all possible solutions, so the correct answer is 'Not Mentioned'.

Step-by-step explanation:

To solve the given Pythagorean/trigonometric equation (1 + ctg²B) sen²B = 1, we can make use of trigonometric identities to simplify the equation. The cotangent squared (ctg²B) can be written as (cos²B)/(sin²B). Substituting this into the initial equation, we obtain (1 + (cos²B)/(sin²B)) sin²B = 1. Simplifying, we notice that the equation becomes sin²B + cos²B = 1, which is a fundamental trigonometric identity that holds true for all angles B.

Therefore, the equation is an identity and is true for all angles in a right triangle. However, we must consider the original equation which includes the term 'sen²B', implying that sin(B) must be defined. Given the options available, we can rule out B = 90° because sin(90°) is 1, and 1 + ctg²90° is not defined as ctg(90°) is undefined. Also, we rule out B = 0° because sen²0° is 0, and thus the equation would not hold. This leaves us with B = 45° as a possibility, where sin(45°) = cos(45°), both equal to √2/2, making the equation hold true.

However, the equation is true for any angle B where sin(B) is defined, so the correct answer is 'Not Mentioned'. Therefore, none of the options a, b, or c provided are strictly correct as they do not encompass all possible solutions.

User Neomega
by
7.6k points

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