Question

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

Answer 1

The answer is below

Explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

`(√(secA-1) )/(√(secA+1) ) +(√(secA+1) )/(√(secA-1) ) =((√(secA-1))(√(secA-1))+(√(secA+1))(√(secA+1)) )/((√(secA+1))(√(secA-1)) ) \\\\=(secA-1+(secA+1))/(√(sec^2A-secA+secA-1) ) \\\\=(2secA)/(√(sec^2A-1) ) \\\\=(2secA)/(√(tan^2A) ) \\\\=(2secA)/(tanA) \\\\=(2*(1)/(cosA) )/((sinA)/(cosA) )\\\\= 2*(1)/(cosA)*(cosA)/(sinA)\\\\=2*(1)/(sinAA)\\\\=2cosecA`