Question

to prove: 1/(sec A-1) + 1/(sec A+1)= 2 cosec A cot A

Answer 1

Solution :

LHS = 1/(secA -1) + 1/(secA+1)

= [secA+1+secA-1]/[(secA-1)(secA+1)]

= (2secA)/[sec²A-1]

=(2secA)/tan²A

= (2/cosA)/(sin²A/cos²A)

= ( 2cosA)/sin²A

= 2(1/sinA)(cosA/sinA)

= 2cosecAcotA

= RHS

••••

Answer 2

$Given : \frac{1}{secA - 1} + \frac{1}{secA + 1}$

$=> \frac{secA - 1 + secA + 1}{(secA - 1)(secA + 1)}$

We know that (a + b)(a - b) = a^2 - b^2

$=> \frac{2secA}{(sec^2A - 1)}$

We know that sec^2θ - tan^2θ = 1 (or) sec^2θ - 1 = tan^2θ

$=> \frac{2secA}{tan^2A}$

$=> \frac{2}{cosA} * \frac{cos^2A}{sin^2A}$

$=> 2 * \frac{1}{sinA} *\frac{cosA}{sinA}$

⇒ 2 cosecA cotA.

Hope it helps!