Question

prove the identity:-
1/ (sec x + tan x) -1/cos x = 1/cos x - 1/(sec x - tan x )

Answer 1

Step-by-step explanation:

Given :

To prove :

$\frac{1}{secX + tanX} - \frac{1}{cosX} = \frac{1}{cosX} - \frac{1}{secX-tanX}$

Proof :

We know that,

$sec^2X - tan^2X = 1$

⇒ $(secX-tanX)(secX+tanX)=1$ ..(i)

_

From (i)

⇒ $secX-tanX = \frac{1}{secX + tanX}$ ..(ii)

_

From (i)

⇒ $secX+tanX = \frac{1}{secX - tanX}$ ..(iii)

__

We know that,

$\frac{1}{cosX} = secX$ ...(iv)

Hence,

$LHS = \frac{1}{secX + tanX}  - \frac{1}{cosX}$

$= secX - tanX - secX = -tanX$ (from (iii) & (iv)

_

$RHS = \frac{1}{cosX} - \frac{1}{secX - tanX}$

$RHS =secX - (secX + tanX)$

$RHS =secX - secX - tanX$

$RHS =- tanX$

_

∴ LHS = RHS,

Hence, proved.