Question

(SecA+tanA)^2=1+cosecA/1-cosecA

Answer 1

Answer:

dont know

Step-by-step explanation:

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Answer 2

Answer:

Hello !

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) ]

=> \frac{ \frac{cosecA + 1}{cosecA} }{ \frac{cosecA - 1}{cosecA} }

cosecA

cosecA−1

cosecA

cosecA+1

=> cosec A gets cancelled ,on both sides ,

hence ,

LHS becomes ,

cosecA + 1/cosecA - 1