Question

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

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{cosec\theta + cot\theta = p} \end{cases}\end{gathered}\end{gathered}$

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

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

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

$\boxed{ \sf \: cot\theta = \bigg( \dfrac{cos\theta}{sin\theta} \bigg)}$

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

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

Consider,

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

On substituting the value of p, we get

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

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

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

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

$\sf \: = \: \dfrac{ \cancel2 \: cosec\theta \: \cancel{(cosec\theta + cot\theta)}}{ \cancel2cot\theta \: \cancel{ (cosec\theta + cot\theta)}}$

$\sf \: = \: \dfrac{cosec\theta}{cot\theta}$

$\sf \: = \: \dfrac{1}{ \cancel{sin\theta} } \times \dfrac{ \cancel{sin\theta}}{cos\theta}$

$\sf \: = \: \dfrac{1}{cos\theta}$

$\sf \: = \: sec\theta$

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

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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