Question

( cosec theta - sin theta) ( sec theta - cos theta) ( tan theta + cot theta) = 1​

Answer 1

Step-by-step explanation:

(1+cotθ−cosecθ)(1+tanθ+secθ)=(1+cosθsinθ−1sinθ)(1+sinθcosθ+1cosθ)

=(sinθ+cosθ−1sinθ)(sinθ+cosθ+1cosθ)

=(sinθ+cosθ)2−12sinθcosθ

=sin2θ+cos2θ+2sinθcosθ−1sinθcosθ

=1+2sinθcosθ−1sinθcosθ

=2sinθcosθsinθcosθ

=2

Answer 2

Question:-

$\: \bf \: ( \csc \theta \: - \: \sin\theta )( \sec\theta - \cos\theta)( \tan\theta + \cot \theta = 1$

Solution:-

$\bf \: ( \csc \theta \: - \: \sin\theta )( \sec\theta - \cos\theta)( \tan\theta + \cot \theta = 1$

Use some trigonometry identities

$\to \: \bf \: \csc( \theta) = \frac{1}{ \sin( \theta) }$

$\to \: \bf \: \sec( \theta) = \frac{1}{ \cos( \theta) }$

$\to \bf \: \tan( \theta) = \frac{ \sin( \theta) }{ \cos( \theta) }$

$\to \bf \: \cot( \theta) \: = \frac{ \cos( \theta) }{ \sin( \theta) }$

$\to \bf \: \sin {}^{2} ( \theta) + \cos {}^{2} ( \theta) = 1$

$\to \bf \: \sin {}^{2} ( \theta) = 1 - \cos {}^{2} ( \theta)$

$\bf \to \: \cos {}^{2} ( \theta) = 1 - \sin {}^{2} ( \theta)$

By using this identities we get

$\bf \: ( \frac{1}{ \sin\theta } - \sin \theta) ( \frac{1}{ \cos\theta } - \cos\theta)( \frac{ \sin \theta }{ \cos \theta } + \frac{ \cos \theta }{ \sin\theta } )$

By taking LCM we get

$\bf \: ( \frac{1 - \sin {}^{2} \theta }{ \sin \theta} )( \frac{1 - \cos {}^{2} \theta }{ \cos \theta } )( \frac{ \sin {}^{2} \theta+ \cos {}^{2} \theta }{ \sin \theta \times \cos \theta } )$

$\bf \frac{ \cos {}^{2} \theta }{ \sin \theta } \times \frac{ \sin {}^{2} \theta}{ \cos\theta} \times \frac{1}{ \sin\theta \cos\theta}$

$\bf \frac{ 1}{ \sin \theta } \times \frac{ \sin {}^{2} \theta}{ 1} \times \frac{1}{ \sin\theta }$

$\bf \large \: 1$

Hence proved