Question

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Answer 1

Question:-

$\dfrac{1 + \cos\theta - \sin {}^{2} \theta }{ \sin\theta(1 + \cos \theta) } = \cot\theta$

Solution:-

$\dfrac{1 + \cos\theta - \sin {}^{2} \theta }{ \sin\theta(1 + \cos \theta) }$

$\dfrac{ \cos\theta + 1- \sin {}^{2} \theta }{ \sin\theta(1 + \cos \theta) }$

$\dfrac{ \cos\theta + \ \cos {}^{2} \theta }{ \sin\theta(1 + \cos \theta) }$

$\rm \dfrac{ \cos \theta(1 + \cos \theta) }{ \sin\theta(1 + \cos\theta) }$

$\rm \dfrac{ \cos \theta \cancel{(1 + \cos \theta)} }{ \sin\theta \cancel{(1 + \cos\theta) } }$

$\dfrac{ \cos \theta }{ \sin\theta }$

$\cot\theta$

Hence proved

Some trigonometry identities

$\to \csc \theta = \dfrac{1}{ \sin\theta}$

$\to \cot \theta = \dfrac{ \cos \theta}{ \sin \theta}$

$\to \tan\theta = \dfrac{ \sin \theta }{ \cos \theta}$

$\to \sec \theta = \dfrac{1}{ \cos\theta }$

$\to \sin {}^{2} \theta + \cos {}^{2} \theta = 1$

Answer 2

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Answer♡

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