Simplify : sin teetha.sec (90- tetha).tan teetha / cosec (90-teetha).cos teetha.cot(90- tetha) - tan (90-teetha)/cot teetha

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Solution

$\frac{\sin θ . \sec (90 - θ) . \tan θ}{cosec (90 - θ) . \cos θ . \cot (90 - θ)} - \frac{\tan (90 - θ)}{\cot θ} = \frac{\sin θ . cosec θ . \tan θ}{\sec θ . \cos θ . \tan θ} - \frac{\cot θ}{\cot θ} = 1 - 1 = 0$

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