Question

Prove that
cosec A+1/cosecA-1 =(secA+tanA)^2
​

Answer 1

Answer:

We need to prove that ,

[math](cosec A + 1)\(cosec A - 1) = (sec A + tan A )^2 [/math]

L.H.S

cosec A = 1 / sin A

[math]cosec A + 1 => 1/sin A + 1 [/math]

=> [math](1+sin A ) / sin A --- (i)[/math]

[math]cosec A - 1 => 1/sin A - 1[/math]

=> [math](1-sin A) / sin A ---(ii)[/math]

Divide (i) and (ii)

[math][ (1+sin A)/sin A ] / [ (1- sin A)/ sin A ] [/math]

[math](1+sin A) / (1 - sin A) [/math]

Multiplying by ( 1+sin A ) in numerator and denominator

(1+sinA)2]/(1−sin2A)

1 - sin^2 A = cos ^ 2 A = ( cos A ) ^ 2

(1+sinA)2/(cosA)2

a^m ÷ b^m = ( a ÷ b ) ^m

(1+sinA/cosA)2

1 / cos A = sec A

sin A / cos A = tan A

(secA+tanA)2

L.H.S = R.H.S

Edits are welcome