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

​$$\sec^2A-\tan^2A=1$$  

​$$\sin^2A + \cos^2 A = 1$$​​

Solution:  

$$b²m² - a²y² + \frac{a²y²}{b²m²} + cos²A$$  

Putting the values of m and y 

​$$= b²(a \sec A)² - a²(b \tan A)² + \frac{a²(b \tan A)²}{b²(a \sec A)²} + cos²A \\ \ \\= b²a^2 \sec ^2A - a²b^2 \tan^2 A + \frac{a²(b^2 \tan^2 A)}{b²(a^2 \sec^2 A)} + cos²A \\ \ \\ = b²a^2( \sec ^2A - \tan^2 A) + \frac{\cancel{a²b^2} \tan^2 A}{\cancel {b²a^2} \sec^2 A} + cos²A \\ \ \\ = b²a^2+ \sin^2A + \cos^2 A \\ = = b²a^2 + 1$$​​