Question

hey guys...
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tan theta /1- cot theta + cot theta /1- tan theta = 1 + sec theta cosec theta .....

Answer 1

hope this helps hiiiiii

Answer 2

HEYA!!
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$\frac{tan \alpha }{1 - cot \: \alpha } + \frac{cot \: \alpha }{1 - tan \: \alpha } = 1 + sec \: \alpha \: \: cosec \: \alpha \\ \\ < < taking \: \: lhs \: \: > > \\ \\ \\ \frac{tan \: \alpha }{1 - \frac{1}{tan \: \alpha } } + \frac{ \frac{1}{tan \: \alpha } }{1 - tan \alpha } \\ \\ \\ \frac{tan {}^{2} \alpha }{tan \: \alpha - 1} + \frac{1}{tan \alpha (1 - tan \: \alpha )} \\ \\ \\ \frac{ - tan {}^{2} \alpha }{1 - tan \alpha } + \frac{1}{tan \: \alpha (1 - tan \: \alpha )} \\ \\ \frac{ - tan {}^{3} \alpha + 1}{(1 - tan \: \alpha )(tan \: \alpha )} \\ \\ \\ < using \: a {}^{3} - b {}^{3} > \\ \\ \\ \\ \frac{(1 - tan \alpha )(tan {}^{2} \alpha + 1 + tan \alpha ) }{(1 - tan \alpha )(tan \alpha )} \\ \\ \frac{sec {}^{2} \alpha + tan \alpha }{tan \alpha } \\ \\ \frac{sec {}^{2} \alpha }{tan \alpha } + \frac{tan \alpha }{tan \alpha } \\ \\ \frac{1}{cos {}^{2} \alpha } \times \frac{cos \alpha }{sin \alpha } + 1 \\ \\ \frac{1}{cos \alpha } \frac{1}{sin \: \alpha } + 1 \\ \\ sec \alpha \: \: cosec \alpha + 1 = rhs$
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