Question

√sec² theta + cosec² theta = tan theta + cot theta​

Answer 1

Answer:

$\sf\green{We\: use\: the \: following \: trigonometric\: identities: }$

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$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: \sec^{2} θ \: = \: \tan^{2} θ \: + \: 1}} }\\ \\\end{gathered}\end{gathered}$

⠀⠀⠀⠀⠀ AND

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: \cosec ^{2} θ \: = \: \cot^{2}θ \: + \: 1}} }\\ \\\end{gathered}\end{gathered}$

$\sf\green{On \: adding \: these, \: we \: get }$

$\begin{gathered}\begin{gathered}\\ : \implies \displaystyle \sf \: \sec^{2} θ \: + \: \cosec^{2} θ \: = \: \tan^{2}θ \: + \cot^{2} θ \: + \: 2\\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\\ : \implies \displaystyle \sf \: \sec^{2} θ \: + \: \cosec^{2} θ \: = \: \tan^{2}θ \: + \cot^{2} θ \: + \\ \ \: 2 \tanθ \: \cotθ \: = ( \tanθ \: + \: \cotθ)^{2} \\ \\ \\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}: \implies \underline{ \boxed{\displaystyle \sf \bold{\: \sqrt{ \sec^{2} θ} \: + \: \cosec^{2} θ \: = \: \tan^{2}θ \: + \cot^{2} θ \: }} }\\ \\\end{gathered}\end{gathered}$

$\huge\color{red}{Hence,\:Proved}$

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Answer 2

Answer:

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Step-by-step explanation: