Question

If (sec A-tan A) = x then prove that 1+x^2/1-x^2=cosec A​

Answer 1

Given :

(secA - tanA) = x

To prove :

(1 + x²)/(1 - x²) = cosecA

Proof :

We have ;

secA - tanA = x

=> x = secA - tanA

Now ,

Squaring both the sides , we have ;

=> x² = (secA - tanA)²

=> x² = sec²A + tan²A + 2•secA•tanA

Now ,

Applying componendo and dividendo in above equation , we get ;

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 } = \frac{ {sec}^{2}A + {tan}^{2} A + 2. secA.tanA + 1 }{{sec}^{2}A + {tan}^{2}A + 2. secA.tanA - 1}$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 } = \frac{ {sec}^{2}A + ( {tan}^{2} A + 1) + 2. secA.tanA }{({sec}^{2}A - 1) + {tan}^{2} A + 2. secA.tanA }$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= \frac{ {sec}^{2}A + {sec}^{2} A+ 2. secA.tanA }{{tan}^{2}A + {tan}^{2} A + 2. secA.tanA }$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= \frac{ 2.{sec}^{2} A+ 2. secA.tanA }{2.{tan}^{2}A + 2. secA.tanA }$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= \frac{2.secA(secA + tanA)}{2.tanA(tanA + secA)}$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= \frac{secA}{tanA}$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= \frac{1}{cosA} \times \frac{cosA}{sinA}$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= \frac{1}{sinA}$

$= > \frac{ {x}^{2} + 1}{ {x}^{2} - 1 }= cosecA$

Hence proved .