Question

(seca+tana)2=coseca+1/coseca-1(seca+tana)2=coseca+1/coseca-1​

Answer 1

Step-by-step explanation:

LHS :

(secA + tanA)²

( 1/cosA + sinA/cosA)²

(1 + sinA)² / cos²A

we know that ,

cos²A = 1 - sin²A

(1+sinA) (1 + sinA) / {1 - sin²A}

Also ,

1 - sin²A = (1+sinA) (1 - sinA)            [ ∵ a² - b² = (a+b) (a - b) ]

(1+sinA) (1 + sinA) / (1+sinA) (1 - sinA)

(1 + sinA ) gets cancelled on numerator and denominator ,

=> 1 + sinA /  1 - sinA

sinA = 1/cosecA

=>  1 +  (  1/cosecA) / [ 1  - (1/cosecA)  ]

=>

=> cosec A gets cancelled ,on both sides ,

hence ,

LHS becomes ,

cosecA + 1/cosecA - 1