Question

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

Answer 1

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)  ]

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

=> cosec A gets cancelled ,on both sides ,

hence ,

LHS becomes ,

cosecA + 1/cosecA - 1

Answer 2

I did it with second method

I hope you liked it

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