Question

Sec theta+tan theta=k then find sin theta

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$to \: find \: = \\ value \: of \: sin \: θ \: (in \: terms \: of \: k) \\ \\ so \: here \: \\ sec \: θ + tan \: θ = k \: \: \: \: \: \: (1)\\ so \: we \: know \: that \\ sec {}^{2} θ - tan {}^{2} θ = 1 \\ ie \: (sec \: θ + tan \: θ)(sec \: θ - tan \: θ) = 1 \\ k.(sec \: θ - tan \: θ) = 1 \\ \\ sec \: θ - tan \: θ = \frac{1}{k} \: \: \: \: \: \: \: (2)$

$adding \: (1) \: and \: (2) \\ we \: get \: \\ sec \: θ + sec \: θ = 1 + \frac{1}{k} \\ \\ 2.sec \: θ = \frac{k {}^{2} + 1}{k} \\ \\ ie \: sec \: θ = \frac{k {}^{2} + 1}{2k} \\ \\ thus \: then \\ cos \: θ = \frac{2k}{k {}^{2} + 1}$

$we \: know \: that \\ sin {}^{2} θ + cos {}^{2} θ = 1 \\ ie \: \: sin {}^{2} θ = 1 - cos {}^{2} θ \\ \\ = 1 - ( \frac{2k}{k {}^{2} + 1 } ) {}^{2} \\ \\ = 1 - \frac{4k {}^{2} }{k {}^{4} + 1 + 2k {}^{2} } \\ \\ = \frac{k {}^{4} + 1 + 2k {}^{2} - 4k {}^{2} }{k {}^{4} + 1 + 2k {}^{2} } \\ \\ = \frac{k {}^{4} + 1 - 2k {}^{2} }{k {}^{4} + 1 + 2k {}^{2} } \\ \\ = \frac{(k - 1) {}^{2} }{(k + 1) {}^{2} } \\ \\ = ( \frac{k - 1}{k + 1} \: ) {}^{2} \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ \\ sin \: θ = ± \: \frac{k - 1}{k + 1}$