Question

5. If sec theta+tan theta=p, then find the value of sin theta interms of p

Answer 1

Answer:

$\bold{sin \alpha = \dfrac{ {p}^{2} \: - \: 1 }{ {p}^{2} \: + \: 1} }$

Step-by-step explanation:

$\bold{Given = >} \\ sec \alpha \: + \: tan \alpha \: = \: p$

$\bold{To \: find = > } \\ value \: of \: sin \alpha$

$\bold{Concept \: used = > } \\ 1) \: sec ^{2} \alpha \: - tan ^{2} \alpha = 1 \\ 2)tan \alpha = \dfrac{sin \alpha }{cos \alpha } \\ 3)sec \alpha = \dfrac{1}{cos \alpha }$

$\bold{Solution = >} \\ sec \alpha + tan \alpha = p \: ...........(1)$

$multiplying \: both \: sides \: by \: (sec \alpha - tan \alpha ) \\ = > (sec \alpha + tan \alpha ) \: (sec \alpha - tan \alpha ) = p \: (sec \alpha - tan \alpha )$

$= > ( {sec}^{2} \alpha - {tan}^{2} \alpha ) = p \:(sec \alpha - tan \alpha )$

$= > 1 \: = \: p \: (sec \alpha \: - \: tan \alpha \: )$

$= > sec \alpha \: - tan \alpha \: = \dfrac{1}{p} .......(2)$

$adding \: equation \: (1) \: and \: (2) \: we \: get$

$sec \alpha + tan \alpha + sec \alpha - tan \alpha = p + \dfrac{1}{p}$

$= > 2 \: sec \alpha = \dfrac{1}{p} + p$

$= > sec \alpha = \dfrac{1 + {p}^{2} }{2p}$

$now \:subtracting \: equation \: (1) \: and \: (2)$

$= > sec \alpha + tan \alpha - (sec \alpha - tan \alpha ) = p - \dfrac{1}{p}$

$= > sec \alpha + tan \alpha - sec \alpha + tan \alpha = p - \dfrac{1}{p}$

$= > 2tan \alpha = \dfrac{ {p}^{2} - 1}{p}$

$= > tan \alpha = \dfrac{ {p}^{2} - 1 }{2p}$

$now$

$sin \alpha = \dfrac{sin \alpha }{cos \alpha } \: cos \alpha$

$sin \alpha = \dfrac{ \dfrac{sin \alpha }{cos \alpha } }{ \dfrac{1}{cos \alpha } }$

$= > sin \alpha = \dfrac{tan \alpha }{sec \alpha }$

$= > sin \alpha = \dfrac{ \dfrac{ {p}^{2} - 1 }{2p} }{ \dfrac{ {p}^{2} + 1 }{2p} }$

$= > sin \alpha = \dfrac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$