Question

if cot theta=3 evaluate cosec theta(cos theta+sec theta) by sec theta sin theta.

Answer 1

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Answer 2

$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Question}}}}}{\bigstar}$

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$\purple{If \:cot \theta=3 \:evaluate \displaystyle{\frac{cosec \theta(cos \theta+sec \theta)}{sec \theta sin \theta}}}$

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$\huge{\bigstar}{ \green{ \mathfrak{ \underline{ \underline{Answer}}}}}{\bigstar}$

$\boxed{\red{\bold{Simplifying:}}}$

$\purple{\implies \displaystyle{\frac{cosec \theta(cos \theta+sec \theta)}{sec \theta sin \theta}}}$

$\implies \displaystyle{\frac{1 × (cos \theta+sec \theta)}{sin \theta ×sec \theta × sin \theta}}$

$\blue{\left( \because cosec \theta = \displaystyle{\frac{1}{sin \theta}} \right)}$

$\implies\displaystyle{\frac{cos \theta(cos \theta+sec \theta)}{sin^2 \theta}}$

$\blue{\left( \because sec \theta =\displaystyle{\frac{1}{cos \theta}} \right)}$

$\boxed{\red{\bold{By\:distributive\: property: }}}$

$\implies \displaystyle{\frac{cos \theta × cos \theta+ cos \theta × sec \theta}{sin^2 \theta}}$

$\implies \displaystyle{\frac{cos^2 \theta + cos \theta × \frac{1}{cos \theta}}{sin^2 \theta}}$

$\purple{\implies \displaystyle{\frac{cos^2 \theta + 1}{sin^2 \theta}}}$

$\purple{\implies \displaystyle{\frac{cos^2 \theta}{sin^2 \theta}} + \frac{1}{sin^2 \theta}}$

$\implies cot^2 \theta + cosec^2 \theta$

$\blue{ \left( \because \displaystyle{\frac{cos^2 \theta}{sin^2 \theta}} = cot^2 \theta \right)}$

$\implies cot^2 \theta + 1 + cot^2 \theta$

$\blue {\left( \because 1 + cot^2 \theta = cosec^2 \theta; \: by \: identity \right)}$

$\implies 2cot^2 \theta + 1$

$\boxed{\red{\bold{Putting \: value \: of \: cot\theta:}}}$

$\green{\implies 2(3)^2 + 1}$

$\green{\implies 2×9+1}$

$\green{\implies 18 +1}$

$\green{\bold{\implies The \: answer \: is \:{\boxed{\boxed{\bold{19}}}}}}$

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