Question

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

Answer 1

Given:$x=\sec \varphi -\tan \varphi =\frac{1-\sin \varphi }{\cos \varphi }$ …(i)

$y=cosec\varphi +\cot \varphi =\frac{1+\cos \varphi }{\sin \varphi }$ …(ii)

Now, let’s prove $xy+x-y+1=0$

$LHS=xy+x-y+1$

$LHS=\left(\frac{1-\sin \varphi }{\cos \varphi }\right)\left(\frac{1+\cos \varphi }{\sin \varphi }\right)+\left(\frac{1-\sin \varphi }{\cos \varphi }\right)-\left(\frac{1+\cos \varphi }{\sin \varphi }\right)+1$

$LHS=\frac{\left(\right(1-\sin \varphi \left)\right(1+\cos \varphi )+(1-\sin \varphi )\sin \varphi -(1+\cos \varphi \left)\cos \varphi \right)}{\cos \varphi \sin \varphi }+1$

$LHS=\frac{(1-\sin \varphi +\cos \varphi -\cos \varphi \sin \varphi +\sin \varphi -{\sin }^2\varphi -\cos \varphi -{\cos }^2\varphi )}{\cos \varphi \sin \varphi }+1$

$LHS=\frac{1-\cos \varphi \sin \varphi -({\sin }^2\varphi +co{s}^2\varphi )}{\cos \varphi \sin \varphi }+1$

$LHS=\frac{1-\cos \varphi \sin \varphi -1}{\cos \varphi \sin \varphi }+1$

$[\because {\sin }^2\theta +{\cos }^2\theta =1]$

$LHS=-1+1$

$LHS=0=RHS$

Hence proved.