Question

if sec^4θ+sec^2θ=10+tan^4θ+tan^2θ then the value of sin^2θ is

Answer 1

Answer:

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Step-by-step explanation:

$so \: here \: given \: trigonometric \: equation \: is \\ sec {}^{4} + sec^2θ = 10 + tan {}^{2} θ + tan {}^{4} θ$

$thus \: ie \: \: sec {}^{4} θ + sec {}^{2} θ = 10 + tan {}^{2} θ(1 + \tan {}^{2} θ) \\ ie \: sec {}^{4} θ + sec {}^{2} θ = 10 + (sec {}^{2} θ - 1).sec {}^{2} θ \\ since \: 1 + tan {}^{2} θ = sec {}^{2} θ$

$sec {}^{4} θ + sec {}^{2} θ = 10 + sec {}^{4} θ - sec {}^{2} θ \\ ie \: sec {}^{2} θ = 10 - sec ^{2} θ \\ 2sec {}^{2} θ = 10 \\ so \: thus \: sec {}^{2} θ = 5 \\ so \: cos {}^{2} θ = 1 \div 5 \\ since \: secθ = 1 \div cos \: θ$

$now \: using \\ sin {}^{2} θ + cos {}^{2} θ = 1 \\ ie \: sin {}^{2} θ = 1 - cos {}^{2} θ \\ = 1 - 1 \div 5 \\ = (5 - 1) \div 5 \\ = 4 \div 5 \\ ie \: sin {}^{2} θ = 4 \div 5$