Question

$\sf If \: sec θ + tan θ = p \: then \: prove \: that : - \\ \\ \bigstar \boxed{\sf sin \theta = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 } } \bigstar\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Thank you :) ​

Answer 1

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

Given that,

$\rm \: p = sec\theta + tan\theta$

Now, Consider

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

On substituting the value of p, we get

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

We know,

$\boxed{\sf{ \:(x + y)^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{ {sec}^{2} \theta + {tan}^{2}\theta + 2sec\theta \: tan\theta \: - \: 1}{ {sec}^{2} \theta + {tan}^{2} \theta + 2sec\theta \: tan\theta \: + \: 1}$

can be re-arranged as

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

We know that,

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

So, using this, we get

$\rm \: = \: \dfrac{ {tan}^{2}\theta + {tan}^{2}\theta + 2sec\theta \: tan\theta }{ {sec}^{2} \theta + {sec}^{2} \theta + 2sec\theta \: tan\theta}$

$\rm \: = \: \dfrac{ 2{tan}^{2}\theta + 2sec\theta \: tan\theta }{2{sec}^{2} \theta + 2sec\theta \: tan\theta}$

$\rm \: = \: \dfrac{ 2tan\theta (tan\theta + sec\theta) }{2sec\theta (sec\theta + tan\theta)}$

$\rm \: = \: \dfrac{tan\theta }{sec\theta }$

$\rm \: = \: \dfrac{sin\theta }{cos\theta } \div \dfrac{1}{cos\theta }$

$\rm \: = \: sin\theta$

Hence,

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

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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

Answer 2

Step-by-step explanation:

$\color{blue} \[ \begin{array}{l} \text {Sol. Given,} \sec \theta+\tan \theta=p \\ \\ \tt Now, \text { H.S. }=\dfrac{p^{2}-1}{p^{2}+1} \\ \\ \tt =\dfrac{(\sec \theta+\tan \theta)^{2}-1}{(\sec \theta+\tan \theta)^{2}+1} \\ \\ \tt =\dfrac{\sec ^{2} \theta+\tan ^{2} \theta+2 \sec \theta \tan \theta-1}{\sec ^{2} \theta+\tan ^{2} \theta+2 \sec \theta \tan \theta+1} \\ \\ \qquad \qquad \qquad\tt[ \because \: (a + b) {}^{2} = {a}^{2} + 2ab + {b}^{2} ]\\ \\ \tt =\dfrac{\left(\sec ^{2} \theta-1\right)+\tan ^{2} \theta+2 \sec \theta \tan \theta}{\sec ^{2} \theta+\left(1+\tan ^{2} \theta\right)+2 \sec \theta \tan \theta} \end{array} \]$

$\tt \color{red}\begin{aligned}=& \frac{\tan ^{2} \theta+\tan ^{2} \theta+2 \sec \theta \tan \theta}{\sec ^{2} \theta+\sec ^{2} \theta+2 \sec \theta \tan \theta} \\ \\ & \qquad \qquad \qquad\left[\begin{array}{l}\because \sec ^{2} \theta-1=\tan ^{2} \theta \\ \sec ^{2} \theta=1+\tan ^{2} \theta\end{array}\right] \end{aligned}$

$\color{purple}\[ \begin{array}{l} =\dfrac{2 \tan ^{2} \theta+2 \sec \theta \tan \theta}{2 \sec ^{2} \theta+2 \sec \theta \tan \theta} \\ \\ =\dfrac{2 \tan \theta(\tan \theta+\sec \theta)}{2 \sec \theta(\sec \theta+\tan \theta)} \\ \\ =\dfrac{\tan \theta}{\sec \theta} \\ \\ =\dfrac{\dfrac{\sin \theta}{\cos \theta}}{\cfrac{1}{\cos \theta}} \\ \\ =\sin \theta=\text { L.H.S. } \\ \\ \text{Hence Proved. }\end{array} \]$