Question

if sec A-tanA=x then prove x^2+1/x^2-l=-cosec A
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Answer 1

Sorry for the drawing on the corner.

Hope this helps you

Answer 2

Step-by-step explanation:

$Given: secA - tanA = x$

LHS:

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

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

$= \frac{sec^2A+tan^2A-2secAtanA+1}{sec^2A+tan^2A-2secAtanA-1}$

$= \frac{sec^2A+sec^2A-2secAtanA}{tan^2A+tan^2A-2secAtanA}$

$= \frac{2sec^2A-2secAtanA}{2tan^2A-2secAtanA}$

$= \frac{2secA(secA-tanA)}{2tanA(tanA-secA)}$

$= \frac{-2secA(tanA-secA)}{2tanA(tanA-secA)}$

$= \frac{-secA}{tanA}$

$= \frac{-1}{cosA} *\frac{cosA}{sinA}$

$= -\frac{1}{sinA}$

$= \textbf {-cosecA}$

Hope it helps!