If cosec theta+cot theta=p show that p^2+1/p^-1=sec theta

1.	If cosec theta+cot theta=p show that p^2+1/p^-1=sec theta
Answer» $\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{<klux>COSEC</klux>\theta + cot\theta = p} \end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\begin{gathered}\bf \: To \: prove - \begin{cases} &\sf{\dfrac{ {p}^{2} + 1}{ {p}^{2} - 1} = sec\theta}\end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$ $\boxed{ \sf \: {cosec}^{2} \theta - {cot}^{2} \theta = 1}$ $\boxed{ \sf \: cot\theta = \bigg( \dfrac{<klux>COS</klux>\theta}{sin\theta} \bigg)}$ $\boxed{ \sf \: cosec\theta = \bigg( \dfrac{1}{sin\theta} \bigg)}$ $\large\underline{\sf{Solution-}}$ Consider, $\rm :\longmapsto\:\dfrac{ {p}^{2} + 1}{ {p}^{2} - 1 }$ On substituting the VALUE of p, we get $\sf \: = \: \dfrac{ {(cosec\theta + cot\theta)}^{2} + 1 }{ {(cosec\theta + cot\theta)}^{2} - 1}$ $\sf \: = \: \dfrac{ {cosec}^{2}\theta + {cot}^{2}\theta + 2cot\theta \: cosec\theta + ( {cosec}^{2}\theta - {cot}^{2}\theta)}{{cosec}^{2}\theta + {cot}^{2}\theta + 2cot\theta \: cosec\theta - ( {cosec}^{2}\theta - {cot}^{2}\theta)}$ $\sf \: = \: \dfrac{{cosec}^{2}\theta + \cancel{{cot}^{2}\theta} + 2cot\theta \: cosec\theta + {cosec}^{2}\theta - \cancel{{cot}^{2}\theta}}{ \cancel{{cosec}^{2}\theta} + {cot}^{2}\theta + 2cot\theta \: cosec\theta - \cancel{{cosec}^{2}\theta} + {cot}^{2}\theta}$ $\sf \: = \: \dfrac{2 {cosec}^{2}\theta + 2cosec\theta \: cot\theta }{2 {cosec}^{2}\theta - 2cot\theta \: cosec\theta }$ $\sf \: = \: \dfrac{ \cancel2 \: cosec\theta \: \cancel{(cosec\theta + cot\theta)}}{ \cancel2cot\theta \: \cancel{ (cosec\theta + cot\theta)}}$ $\sf \: = \: \dfrac{cosec\theta}{cot\theta}$ $\sf \: = \: \dfrac{1}{ \cancel{sin\theta} } \times \dfrac{ \cancel{sin\theta}}{cos\theta}$ $\sf \: = \: \dfrac{1}{cos\theta}$ $\sf \: = \: sec\theta$ ${\boxed{\boxed{\bf{Hence, Proved}}}}$ ─━─━─━─━─━─━─━─━─━─━─━─━─ 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

Answer»

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{<klux>COSEC</klux>\theta + cot\theta = p} \end{cases}\end{gathered}\end{gathered}$

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

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

$\boxed{ \sf \: {cosec}^{2} \theta - {cot}^{2} \theta = 1}$

$\boxed{ \sf \: cot\theta = \bigg( \dfrac{<klux>COS</klux>\theta}{sin\theta} \bigg)}$

$\boxed{ \sf \: cosec\theta = \bigg( \dfrac{1}{sin\theta} \bigg)}$

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

Consider,

$\rm :\longmapsto\:\dfrac{ {p}^{2} + 1}{ {p}^{2} - 1 }$

On substituting the VALUE of p, we get

$\sf \: = \: \dfrac{ {(cosec\theta + cot\theta)}^{2} + 1 }{ {(cosec\theta + cot\theta)}^{2} - 1}$

$\sf \: = \: \dfrac{ {cosec}^{2}\theta + {cot}^{2}\theta + 2cot\theta \: cosec\theta + ( {cosec}^{2}\theta - {cot}^{2}\theta)}{{cosec}^{2}\theta + {cot}^{2}\theta + 2cot\theta \: cosec\theta - ( {cosec}^{2}\theta - {cot}^{2}\theta)}$

$\sf \: = \: \dfrac{{cosec}^{2}\theta + \cancel{{cot}^{2}\theta} + 2cot\theta \: cosec\theta + {cosec}^{2}\theta - \cancel{{cot}^{2}\theta}}{ \cancel{{cosec}^{2}\theta} + {cot}^{2}\theta + 2cot\theta \: cosec\theta - \cancel{{cosec}^{2}\theta} + {cot}^{2}\theta}$