Question

→ Prove the above question using suitable Trigonometric Identities.

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Answer 1

Question:-

$\rm \: \to \: \dfrac{ \tan \theta }{1 - \cot \theta} + \dfrac{ \cot \theta }{1 - \tan \theta } = \sec \theta. \csc \theta + 1$

Solution:-

Taking LHS

$\rm \: \to \: \dfrac{ \tan \theta }{1 - \cot \theta} + \dfrac{ \cot \theta }{1 - \tan \theta }$

Using trigonometry identities

$\to \dfrac{ \dfrac{ \sin\theta}{ \cos\theta } }{1 - \dfrac{ \cos\theta}{ \sin \theta} } + \dfrac{ \dfrac{ \cos\theta}{ \sin\theta} }{1 - \dfrac{ \sin\theta }{ \cos\theta} }$

$\to \dfrac{ \dfrac{ \sin\theta}{ \cos\theta } }{ \dfrac{ \sin\theta - \cos\theta}{ \sin \theta} } + \dfrac{ \dfrac{ \cos\theta}{ \sin\theta} }{\dfrac{ \cos\theta - \sin\theta }{ \cos\theta} }$

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

$\rm \to\dfrac{ \sin {}^{2} \theta}{ \cos\theta( \sin\theta - \cos \theta)} - \dfrac{ \cos {}^{2} \theta }{ \sin \theta( \sin \theta - \cos\theta )}$

$\rm \to \: \dfrac{ \sin {}^{3} \theta- \cos {}^{3} \theta }{ \sin\theta \cos\theta ( \sin \theta - \cos \theta) }$

Now we can write as

$\rm \: \to \: \dfrac{( \sin \theta - \cos \theta)( \sin {}^{2} \theta + \sin \theta \cos\theta + \cos {}^{2} \theta)}{\sin \theta \cos\theta (\sin \theta - \cos \theta)}$

$\rm \: \to \: \dfrac{ \cancel{( \sin \theta - \cos \theta)}( \sin {}^{2} \theta + \sin \theta \cos\theta + \cos {}^{2} \theta)}{\sin \theta \cos\theta \cancel{ (\sin \theta - \cos \theta)}}$

$\to \: \dfrac{ 1+ \sin \theta \cos \theta}{\sin \theta \cos \theta}$

$\to \dfrac{1}{\sin \theta \cos \theta} + \dfrac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$\rm \: \to \sec\theta \csc \theta + 1$

Hence proved