Question

If sin0 = a/b then prove that (sec 0 + tan 0) =​

Answer 1

Answer:

√( (b + a)/(b - a) )

Step-by-step explanation:

sinA = a/b , therefore, using,

sin²A + cos²A = 1

=> (a/b)² + cos²A = 1

=> cos²A = 1 - a²/b² = (b² - a²)/b²

=> cosA = √{(b² - a²)/b²} = √(b² - a²)/b

Hence,

secA = b/√(b² - a²) ...(1)

Square on both sides :

=> sec²A = b²/(b² - a²)

=> 1 + tan²A = b²/(b² - a²)

=> tan²A = b²/(b² - a²) - 1

=> tan²A = a²/(b² - a²)

=> tanA = a/√(b² - a²) ...(2)

Thus, adding (1) & (2):

secA + tanA

=> b/√(b² - a²) + a/√(b² - a²)

=> (b + a)/√(b² - a²)

=> √(b + a)² / √(b - a)(b + a)

=> √(b + a) / √(b - a)

=> √( (b + a)/(b - a) )