Question

Pls answer, god will bless you with ₹999999999

Answer 1

Given :

secA + tanA = p

To prove :

sinA = (p² - 1)/(p² + 1)

Proof :

We have ;

secA + tanA = p

Now ,

Squaring both the sides , we get ;

=> (secA + tanA)² = p²

=> sec²A + tan²A + 2secA.tanA = p²

Now ,

Applying componendo and dividendo rule , we get ;

$= > \frac{ {sec}^{2}A + {tan}^{2}A + 2secA.tanA - 1}{{sec}^{2}A + {tan}^{2}A + 2secA.tanA + 1} = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

$= > \frac{ ({sec}^{2}A - 1)+ {tan}^{2}A + 2secA.tanA }{{sec}^{2}A + ({tan}^{2}A + 1)+ 2secA.tanA } = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

$= > \frac{ {tan}^{2}A+ {tan}^{2}A+ 2secA.tanA }{{sec}^{2}A + {sec}^{2} A + 2secA.tanA } = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

$= > \frac{ 2{tan}^{2} A + 2secA.tanA }{2{sec}^{2}A + 2secA.tanA } = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

$= > \frac{ 2tanA(tanA+ secA) }{2secA(secA + tanA) } = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

$= > \frac{ tanA }{secA } = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1}$

$= > tanA \times \frac{1}{secA} = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$

$= > \frac{sinA}{cosA} \times cosA = \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$

$= > sinA= \frac{ {p}^{2} - 1 }{ {p}^{2} + 1 }$

Hence proved .