If m = a secA and y = b tanA, then find the value of b²m² - a²y² + (a²y² / b²m²) + cos²A.

Question

Accepted Answer

Let us substitute the given values:

m = a secA ⇒ m² = a² sec²A
y = b tanA ⇒ y² = b² tan²A

Now, compute each term:

1. b²m² = b² × (a² sec²A) = a²b² sec²A
2. a²y² = a² × (b² tan²A) = a²b² tan²A
3. (a²y²) / (b²m²) = [a²b² tan²A] / [a²b² sec²A] = tan²A / sec²A
   But tan²A / sec²A = (sin²A / cos²A) / (1 / cos²A) = sin²A
4. cos²A (as is)

Now, sum up:

b²m² - a²y² + (a²y² / b²m²) + cos²A
= a²b² sec²A - a²b² tan²A + sin²A + cos²A

Recall: sec²A - tan²A = 1
So, a²b² sec²A - a²b² tan²A = a²b² (sec²A - tan²A) = a²b² × 1 = a²b²

So, the expression becomes:
a²b² + sin²A + cos²A
But sin²A + cos²A = 1

Final answer: a²b² + 1

Answer

Answer

Answer