Prove that:
cosec A +1 /cosec A-1 =(sec A + tan A)^2
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Step-by-step explanation:
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
[math][ (1+sin A)^2 ] / ( 1-sin^2 A)[/math]
1 - sin^2 A = cos ^ 2 A = ( cos A ) ^ 2
[math]( 1+sin A ) ^ 2 / ( cos A )^ 2 [/math]
a^m ÷ b^m = ( a ÷ b ) ^m
[math](1+sin A/cos A) ^ 2 [/math]
1 / cos A = sec A
sin A / cos A = tan A
[math](sec A + tan A ) ^ 2[/math]
L.H.S = R.H.S
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