prove-that-sec-2-cosec-2-sec-2-cosec-2-

Question Number 31679 by pieroo last updated on 12/Mar/18

prove that \sec^{2} θ {cosec}^{2} θ = \sec^{2} θ + {cosec}^{2} θ

Answered by Tinkutara last updated on 12/Mar/18

\sec^{2} θ {cosec}^{2} θ = \frac{1}{\sin^{2} θ \cos^{2} θ}

= \frac{\sin^{2} θ + \cos^{2} θ}{\sin^{2} θ \cos^{2} θ}

= \frac{1}{\cos^{2} θ} + \frac{1}{\sin^{2} θ} = \sec^{2} θ + {cosec}^{2} θ

Commented by pieroo last updated on 12/Mar/18

thanks boss

Answered by Joel578 last updated on 12/Mar/18

\sec^{2} θ {cosec}^{2} θ = (1 + \tan^{2} θ) (1 + \cot^{2} θ)

= 1 + \cot^{2} θ + \tan^{2} θ + \tan^{2} θ \cot^{2} θ

= {cosec}^{2} θ + \sec^{2} θ

Commented by pieroo last updated on 12/Mar/18

i am grateful

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