Question

prove that cos0+sin0-1/cos0-sin0+1 = cosec0-cot0​

Answer 1

Question:-

$\rm \implies \: \dfrac{ \cos\theta + \sin \theta - 1}{ \cos\theta - \sin\theta+ 1} = \csc\theta - \cot \theta$

Solution:-

$\rm \implies \: \dfrac{ \cos\theta + \sin \theta - 1}{ \cos\theta - \sin\theta+ 1} = \csc\theta - \cot \theta$

$\rm \implies \: \dfrac{ \cos\theta + \sin \theta - 1}{ \cos\theta - \sin\theta+ 1} = \dfrac{1}{ \sin( \theta) } - \dfrac{ \cos( \theta) }{ \sin( \theta) }$

$\rm \implies \: \dfrac{ \cos\theta + \sin \theta - 1}{ \cos\theta - \sin\theta+ 1} = \dfrac{1 - \cos\theta }{ \sin\theta }$

Using cross multiplication method

$\rm \implies \: \: \sin \theta( \cos \theta + \sin\theta - 1) = (1 - \cos \theta)(\cos \theta - \sin\theta + 1)$

$\rm \implies \: \sin\theta\cos \theta + \sin ^{2} \theta - \sin \theta = \cos \theta - \sin\theta + 1 - \cos {}^{2} \theta + \sin \theta \cos\theta - \cos\theta$

$\rm \implies \: \cancel{\sin\theta\cos \theta } + \sin ^{2} \theta - \sin \theta = \cos \theta - \sin\theta + 1 - \cos {}^{2} \theta + \cancel{ \sin \theta \cos\theta} - \cos\theta$

$\rm \implies \: \sin ^{2} \theta - \sin \theta = \cos \theta - \sin\theta + 1 - \cos {}^{2} \theta - \cos\theta$

$\rm \implies \: \sin ^{2} \theta - \sin \theta = 0 - \sin\theta + 1 - \cos {}^{2} \theta$

$\rm \implies \: \sin ^{2} \theta - \sin \theta + \sin\theta = 1 - \cos {}^{2} \theta$

$\rm \implies \: \sin ^{2} \theta = 1 - \cos {}^{2} \theta$

$\rm \implies \: \sin ^{2} \theta +\cos {}^{2} \theta = 1$

$\implies1 = 1$

hence proved