Question

x = a tan^2 A + b sec^2 A then x+a/x-b =

Answer 1

Answer:-

Given:

x = a tan² A + b sec² A

→ x + a = a tan² A + b sec² A + a

→ x + a = a (1 + tan² A) + b sec² A

We know that,

sec² A - tan² A = 1

→ sec² A = (1 + tan² A)

→ tan² A = sec² A - 1

Hence,

→ x + a = a sec² A + b sec² A

→ x + a = sec² A(a + b)

→ x - b = a tan² A + b sec² A - b

→ x - b = a tan² A + b(Sec² A - 1)

Putting the value of "sec² A - 1 " as tan² A we get,

→ x - b = a tan² A + b tan² A

→ x - b = tan² A(a + b)

Hence,

(x + a)/(x - b) = [sec² A(a + b)]/[tan² A(a + b)]

→ (x + a)/(x - b) = Sec² A/tan² A

Putting the values of sec² A = 1/Cos² A and tan² A = Sin² A/Cos² A we get,

→ (x + a)/(x - b) = (1/Cos² A)/(Sin² A /Cos² A)

→ (x + a)/(x - b) = 1/Cos² A × Cos² A/Sin² A

→ (x + a)/(x - b) = 1/Sin² A

Putting 1/Sin² A = Cosec² A in RHS we get,

→ (x + a)/(x - b) = Cosec² A