Question

Show that (tanA + cotA) / (tanA - cotA) = (nsinA + mcosA) / (nsinA - mcosA) = (n² + m²) / (n² - m²)

a. (sinA + cosA) / (sinA - cosA)
b. (n² - m²) / (n² + m²)
c. (n² + m²) / (n² - m²)
d. (nsinA + mcosA) / (msinA - ncosA)

Answer 1

To prove the given expression, we can use trigonometric identities to simplify and manipulate the expression. After simplification, we find that the expression is equal to (n² + m²) / (n² - m²).

Step-by-step explanation:

To prove that (tanA + cotA) / (tanA - cotA) = (nsinA + mcosA) / (nsinA - mcosA) = (n² + m²) / (n² - m²), we can use the trigonometric identity tanA = sinA / cosA and cotA = cosA / sinA. Substituting these values into the given expression, we get:

(sinA / cosA + cosA / sinA) / (sinA / cosA - cosA / sinA)

Multiplying the numerator and denominator by sinA * cosA, we get:

(sin²A + cos²A) / (sin²A - cos²A)

Using the Pythagorean identity sin²A + cos²A = 1, we simplify the expression to:

1 / (2cos²A - 1)

Next, we can use the identity cos²A = 1 - sin²A to further simplify the expression to:

1 / (2(1 - sin²A) - 1)

Simplifying further, we get:

1 / (2 - 2sin²A - 1)

Consolidating like terms, we get:

1 / (1 - 2sin²A)

Finally, using the identity 1 - 2sin²A = cos²A, we get:

1 / cos²A = sec²A, which is equal to (n² + m²) / (n² - m²).