Question

If sec^4theta+sec^2theta=10+tan^4theta+tan^2theta find the value of sin^2 theta​

Answer 1

Answer:

hey here is your answer

pls mark it as brainliest

Step-by-step explanation:

$to \: find = \\ value \: of \: sin \: θ \\ \\ so \: here \\ sec {}^{4} \: θ + sec {}^{2} \: θ = 10 + tan {}^{4} \: θ + tan {}^{2} \: θ \\ = 10 + tan {}^{2} θ(1 + tan {}^{2} \: θ) \\ = 10 + tan {}^{2} \: θ.sec {}^{2} \: θ \\ = 10 + (sec {}^{2} θ - 1).sec {}^{2} \: θ \\sec {}^{4} θ + sec {}^{2} \: θ = 10 + sec {}^{4} \: θ - sec {}^{2} \: θ \\ sec {}^{2} θ + sec {}^{2} θ = 10 \\ 2.sec {}^{2} θ = 10 \\ ie \: \: sec {}^{2} \: θ = 5 \\ \\ hence \: then \\ cos {}^{2} θ = \frac{1}{5} \\ \\ we \: know \: that \\ sin {}^{2} θ + cos {}^{2} θ = 1 \\ sin {}^{2} θ = 1 - cos {}^{2} \: θ \\ \\ = 1 - ( \frac{1}{5} ) {}^{2} \\ \\ = 1 - \frac{1}{25} \\ \\ = \frac{25 - 1}{25} \\ \\ = \frac{24}{25} \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ \\ sin \: θ = ± \: \frac{ \sqrt{24} }{5} = ± \: \frac{2 \sqrt{6} }{5}$