Question

13. Find the value of (sec 0 + tan 0) (sec 0 – tan 0).
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Answer 1

$\sf{\underline{\boxed{\purple{\large{\bold{ Solution }}}}}}$

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$\sf :\implies\:{\bold{( Sec\theta + Tan\theta)( Sec\theta - Tan\theta) }}$

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$\sf{\red{\boxed{\bold{(a + b ) ( a - b ) = a^2 - b^2}}}}$

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$\sf :\implies\:{\bold{Sec^2\theta - Tan^2\theta }}$

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$\sf{\red{\boxed{\bold{Sec^2A - Tan^2A = 1 }}}}$

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$\sf :\implies\:{\bold{ 1 }}$

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Some important formulae :-

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$\sf{\underline{\underline{\large{\bold{ Trigonometrical \: Ratio}}}}}$

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$\sf :\implies\:{\bold{Sin = \dfrac{ Perpendicular }{ Hypotenuse } }}$

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$\sf :\implies\:{\bold{Cos = \dfrac{ Base }{ Hypotenuse } }}$

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$\sf :\implies\:{\bold{Tan = \dfrac{ Perpendicular }{ Base} }}$

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$\sf :\implies\:{\bold{Cosec = \dfrac{ Hypotenuse }{ Perpendicular } }}$

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$\sf :\implies\:{\bold{Sec = \dfrac{ Hypotenuse } {Base}}}$

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$\sf :\implies\:{\bold{Cot = \dfrac{Base} { Perpendicular } }}$

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$\sf{\underline{\underline{\large{\bold{Reciprocal\: identities }}}}}$

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$\sf :\implies\:{\bold{Sin = \dfrac{ 1 }{ Cosec } }}$

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$\sf :\implies\:{\bold{Cos = \dfrac{ 1 }{ Sec} }}$

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$\sf :\implies\:{\bold{Tan = \dfrac{ 1 }{ Cot } }}$

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Answer 2

$\huge\mathfrak{\color{green}{\underline{Solution:}}}$

$\sf { \blue{\implies\:{\bold{( Sec\theta + Tan\theta)( Sec\theta - Tan\theta) }}}}$

$\sf{\red{\boxed{\bold{(a + b ) ( a - b ) = a^2 - b^2}}}}$

$\sf { \blue{\implies\:{\bold{Sec^2\theta - Tan^2\theta }}}}$

$\sf{\red{\boxed{\bold{Sec^2A - Tan^2A = 1 }}}}$

$\sf { \blue{\implies\:{\bold{ 1 }}}}$

$\Large \mathfrak{ \orange{ \underline{Some important formulae :-}}}$

$\sf{ \purple{{\underline{\underline{\large{\bold{ Trigonometrical \: \: Ratio}}}}}}}$

$\begin{gathered}\small\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \sf {{\implies{Sin = \dfrac{ Perpendicular }{ Hypotenuse }}}} \\ \\ </p><p>\sf{{\implies{Cos = \dfrac{ Base }{ Hypotenuse } }}} \\ \\ \sf{{\implies{Tan = \dfrac{ Perpendicular }{ Base} }}} \\ \\ \sf \implies \: {Cosec = \dfrac{ Hypotenuse }{ Perpendicular } } \\ \\ \sf \implies{Sec = \dfrac{ Hypotenuse } {Base}} \\ \\ \sf \implies{Cot = \dfrac{Base} { Perpendicular } } \\ \\ \end{array}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$

$\sf{ \purple{{\underline{\underline{\large{\bold{Reciprocal\: \: identities }}}}}}}$

$\begin{gathered}\small\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c} \sf \implies{Sin = \dfrac{ 1 }{ Cosec } } \\ \\ \sf \implies{Cos = \dfrac{ 1 }{ Sec} } \\ \\ \sf \implies{Tan = \dfrac{ 1 }{ Cot } }</p><p>\end{array}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}$