Verified answer

Answer:

[tex] \boxed{\bf \: (1). \: \: \dfrac{ {x}^{2} - 1}{ {x}^{2} + 1} \: } \\ [/tex]

Step-by-step explanation:

Given that,

[tex]\sf \: sec\phi + tan\phi = x - - - (1) \\ [/tex]

We know,

[tex]\sf \: {sec}^{2}\phi - {tan}^{2}\phi = 1 \\ [/tex]

[tex]\sf \: (sec\phi + tan\phi)(sec\phi - tan\phi) = 1 \\ [/tex]

[tex]\sf \: x(sec\phi - tan\phi) = 1 \: \: \: \left[ \because \:of \: (1) \right]\\ [/tex]

[tex]\implies\sf \: sec\phi - tan\phi = \dfrac{1}{x} - - - (2) \\ [/tex]

On adding equation (1) and (2), we get

[tex]\sf \: 2sec\phi = x + \dfrac{1}{x} \\ [/tex]

[tex]\sf \: 2sec\phi = \dfrac{ {x}^{2} + 1}{x} \\ [/tex]

[tex]\implies\sf \: sec\phi = \dfrac{ {x}^{2} + 1}{2x} - - - (3)\\ [/tex]

On Subtracting equation (2) from (1), we get

[tex]\sf \: 2tan\phi = x - \dfrac{1}{x} \\ [/tex]

[tex]\sf \: 2tan\phi = \dfrac{ {x}^{2} - 1}{x} \\ [/tex]

[tex]\implies\sf \: tan\phi = \dfrac{ {x}^{2} - 1}{2x} - - - (4) \\ [/tex]

On dividing equation (4) by (3), we get

[tex]\sf \: tan\phi \div sec\phi= \dfrac{ {x}^{2} - 1}{2x} \div \dfrac{ {x}^{2} + 1 }{x} \\ [/tex]

[tex]\sf \: \dfrac{sin\phi}{cos\phi} \div \dfrac{1}{cos\phi} = \dfrac{ {x}^{2} - 1}{ {x}^{2} + 1} \\ [/tex]

[tex]\implies\sf \: \boxed{\bf \: sin\phi = \dfrac{ {x}^{2} - 1}{ {x}^{2} + 1} \: } \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sinx = \dfrac{1}{cosecx} }\\ \\ \bigstar \: \bf{cosx = \dfrac{1}{secx} }\\ \\ \bigstar \: \bf{tanx = \dfrac{sinx}{cosx} = \dfrac{1}{cotx} }\\ \\ \bigstar \: \bf{cot x= \dfrac{cosx}{sinx} = \dfrac{1}{tanx} }\\ \\ \bigstar \: \bf{cosec x = \dfrac{1}{sinx} }\\ \\ \bigstar \: \bf{secx = \dfrac{1}{cosx} }\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 } \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\\: \end{array} }}\end{gathered}\end{gathered}\end{gathered} \\ \\[/tex]

secθ+tanθ=x

cosθ1+cosθsinθ=x

1+sinθ=xcosθ

Squaring both sides,

1+sin2θ+2sinθ=x2cos2θ

1+sin2θ+2sinθ=x2(1−sin2θ)

(1+x2)sin2θ+2sinθ+(1−x2)=0hus, sinθ=2(1+x2)−2±4−4(1−x2)(1+x2)

sinθ=2(1+x2)−2(4−4+4x4)

sinθ=2(1+x2)−2±2x2

sinθ=−1,

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