Math, asked by mvpbvamsirithvik, 1 month ago

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

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Answered by MysticSohamS
0

Answer:

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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}

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