How do I prove that 1/(sec A+1)+1/(sec A-1)1secA+1+1secA1=2 csc A cot A2cscAcotA ?

1 Answer
Oct 4, 2015

1 / (sec A + 1) + 1 / (Sec A - 1)1secA+1+1secA1

Taking the Lowest common Multiple,

(Sec A - 1 + Sec A + 1) / (Sec A +1) * (Sec A - 1)secA1+secA+1secA+1(secA1)

As you may be aware, a^2 - b^2 = (a + b) * (a - b)a2b2=(a+b)(ab)

Simplifying, (2 Sec A) / (Sec^2 A - 1)2secAsec2A1

Now Sec^2 A - 1 = tan^2 A = Sin^2A / Cos^2Asec2A1=tan2A=sin2Acos2A
and Sec A = 1 / Cos AsecA=1cosA

Substituting,

2 / Cos A * Cos^2A / Sin^2A = 2 * Cos A / Sin^2A2cosAcos2Asin2A=2cosAsin2A

which can be written as 2 * Cos A / Sin A * ( 1 / Sin A)2cosAsinA(1sinA)

Now Cos A / Sin A = Cot A and 1 / Sin A = Cosec AcosAsinA=cotAand1sinA=cosecA
Substituting, we get 2 Cot A * Cosec A2cotAcosecA