Question

If sec theta+tan+=p that find the value of case theta

Answer 1

Given,

$sec \: θ + tan \: θ = p$

$\frac{1}{cos \: \: θ} + \frac{sinθ \: }{cos θ } = p$

$1 + \sinθ = p.cos \: θ$

Since,

$cosθ = \sqrt{1 - {sin}^{2} \: θ}$

$= > 1 - sin \: θ = p. \sqrt{1 - {sin \: \: ^{2} }$

Squaring both sides; We get

$( {1 + sin})^{2} = {p}^{2} .( {1 - sin}^{2} )$

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

$= > (1 + {p}^{2} ) {sin}^{2} + 2sin + (1 - {p}^{2} ) = 0$

From , Quadratic formula.We have

$D = {b}^{2} - 4ac$

$= > D = 4 - 4(1 + {p}^{2} )(1 - {p}^{2} )$

$= > D = 4 {p}^{4}$

$= > sin = \frac{ - 2 + \sqrt{4 {p}^{4} } }{2(1 + {p}^{2} } ) \: \: or \frac{ - 2 - \sqrt{4 {p}^{4} } }{2(1 + {p}^{2}) }$

$= > sin = \frac{ {p}^{2} + 1}{ {p}^{2} - 1 } \: \: or \: \: \: - 1$