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# If $x=\sec\phi-\tan\phi$ and $y=cosec\phi+\cot \phi$ then show that $xy+x-y+1=0$

Can you answer this question?

Given :
$x=\sec\phi-\tan\phi$
$y=cosec\phi+\cot \phi$
$xy+x-y+1=0$
$\Rightarrow x-y=-(xy+1)$
L.H.S
$x-y$
$\sec\phi-\tan\phi-cosec \phi-\cot \phi$
$\sec\phi-cosec\phi-(\tan\phi+\cot\phi)$
$\sec\phi-cosec\phi-(\large\frac{\sin^2\phi+\cos^2\phi}{\sin \phi.\cos \phi})$
$\sec\phi-cosec\phi-\large\frac{1}{\sin \phi.\cos\phi}$
$\sec\phi-cosec\phi-cosec\phi.\sec\phi$
R.H.S
-(xy+1)
$-[(\sec\phi-\tan\phi)(cosec\phi+\cot\phi)+1]$
$-[\sec\phi.cosec\phi+\sec\phi.\cot\phi-\tan\phi cosec\phi-\tan\phi\cot\phi+1]$
$-[sec\phi.cosec\phi+cosec\phi-sec\phi]$
$\sec\phi-cosec\phi-cosec\phi.\sec\phi$
$\therefore$L.H.S=R.H.S
Hence proved.
answered Apr 20, 2014