Question

If cosecx-cotx=1/x,then prove that secx=(x^2+1)/(x^2-1)
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Answer 1

$\begin{gathered}\begin{gathered}\bf\: Given-\begin{cases} &\sf{cosecx - cotx = \dfrac{1}{x} }\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To \: prove - \begin{cases} &\sf{secx = \dfrac{ {x}^{2} + 1 }{ {x}^{2} - 1} }\end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{{\bf{ {cosec}^{2} - {cot}^{2}x = 1}}}$

$\boxed{{\bf{cosecx = \dfrac{1}{sinx} }}}$

$\boxed{{\bf{secx = \dfrac{1}{cosx} }}}$

$\boxed{{\bf{cotx = \dfrac{cosx}{sinx} }}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:cosecx \: - \: cotx = \dfrac{1}{x} - - - (1)$

We know that

$\rm :\longmapsto\: {cosec}^{2} x - {cot}^{2} x = 1$

$\rm :\longmapsto\:(cosecx + cotx)(cosecx - cotx) = 1$

$\rm :\longmapsto\:(cosecx + cotx) \times \dfrac{1}{x} = 1$

$\rm :\longmapsto\:cosecx + cotx = x - - - (2)$

On adding equation (1) and (2), we get

$\rm :\longmapsto\:2cosecx = x + \dfrac{1}{x}$

$\rm :\longmapsto\:2cosecx = \dfrac{ {x}^{2} + 1}{x}$

$\rm :\longmapsto\:cosecx = \dfrac{ {x}^{2} + 1}{2x} - - - (3)$

Now,

On Subtracting equation (1) from equation (2), we get

$\rm :\longmapsto\:2cotx = x - \dfrac{1}{x}$

$\rm :\longmapsto\:2cotx = \dfrac{ {x}^{2} - 1}{x}$

$\rm :\longmapsto\:cotx = \dfrac{ {x}^{2} - 1}{2x} - - - (4)$

Now, divide equation (3) by equation (4), we get

$\rm :\longmapsto \: \dfrac{cosecx}{cotx} = \dfrac{\dfrac{ {x}^{2} + 1 }{ \cancel{2x}} }{ \: \: \: \: \: \dfrac{ {x}^{2} - 1}{ \cancel{2x}} \: \: \: \: \: }$

$\rm :\longmapsto\:\dfrac{1}{sinx} \times \dfrac{sinx}{cosx} = \dfrac{ {x}^{2} + 1}{ {x}^{2} - 1}$

$\rm :\longmapsto\:\dfrac{1}{cosx} = \dfrac{ {x}^{2} + 1}{ {x}^{2} - 1}$

$\rm :\longmapsto\:secx = \dfrac{ {x}^{2} + 1}{ {x}^{2} - 1}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1