Question

. If tan e + sec e = p, then prove that sin e = (p2 -1) / (p²+1)
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Answer 1

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

Given that,

$\rm \: tane + sece = p$

Now, Consider

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

On substituting the value of p, we get

$\rm \: =  \:\dfrac{ {(tane + sece)}^{2} - 1}{ {(tane + sece)}^{2} + 1}$

$\rm \: =  \:\dfrac{ {tan}^{2} e + {sec}^{2} e + 2tane \: sece \: - \: 1}{ {tan}^{2} e + {sec}^{2} e + 2tane \: sece \: + \: 1}$

can be re-arranged as

$\rm \: =  \:\dfrac{ {tan}^{2} e + 2tane \: sece \: + ({sec}^{2}e \: - \: 1)}{ {sec}^{2} e + 2tane \: sece \: + \: (1 + {tan}^{2} e)}$

We know that,

$\boxed{\sf{  \:\rm \: {sec}^{2}x - {tan}^{2}x = 1 \: \: }}$

So, using this identity, we get

$\rm \: =  \:\dfrac{ {tan}^{2} e + 2tane \: sece \: + {tan}^{2} e}{ {sec}^{2} e + 2tane \: sece \: + \: {sec}^{2} e}$

$\rm \: =  \:\dfrac{2 {tan}^{2} e + 2tane \: sece \: }{2 {sec}^{2} e + 2tane \: sece \: }$

$\rm \: =  \:\dfrac{2tane(tane + sece)}{2sece(sece + tane)}$

$\rm \: =  \:\dfrac{tane}{sece}$

$\rm \: =  \:\dfrac{sin \: e}{cos \: e} \times cos \: e$

$\rm \: =  \:sin \: e$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \dfrac{ {p}^{2} - 1}{ {p}^{2} + 1} \: = \: sin \: e \: \:}}$

$\rule{190pt}{2pt}$

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 2

Step-by-step explanation:

Solution.

Given,

$\color{red}\tt \sec e+\tan e=p \Rightarrow \dfrac{1}{\cos e}+\dfrac{\sin e}{\cos e}=p$

$\color{brown}\[ \begin{array}{l} \tt \Rightarrow \quad \dfrac{1+\sin e}{\cos e}=p \\ \\ \tt\Rightarrow \dfrac{(1+\sin e)^{2}}{\cos ^{2} e}=p^{2} \\ \\ \tt \Rightarrow \quad \dfrac{(1+\sin e)^{2}}{1-\sin ^{2} e}=p^{2} \\ \\ \tt \Rightarrow \dfrac{1+\sin e}{1-\sin e}=\dfrac{p^{2}}{1} \\ \\ \tt \Rightarrow \quad 1+\sin e=p^{2}-p^{2} \sin e \\ \\ \tt \Rightarrow\left(1+p^{2}\right) \sin e=p^{2}-1 \\ \\ \tt \Rightarrow \quad \sin e=\dfrac{p^{2}-1}{p^{2}+1} . \end{array} \]$