Question

If $cosec\theta -sin\theta =l$ and $sec\theta -cos\theta$ = m, prove that $l^2{m}^2(l^2+m^2+3)=1$

Answer 1

We have ,

$L.H.S=l^2{m}^2(l^2+m^2+3)$

$L.H.S=(cosec\theta -\sin \theta {)}^2(\sec \theta -\cos \theta {)}^2\left[\right(cosec\theta -\sin \theta {)}^2+(\sec \theta -\cos \theta {)}^2+3]$

$L.H.S={(\frac{1}{\sin \theta }-\sin \theta )}^2{(\frac{1}{\cos \theta }-\cos \theta )}^2[{(\frac{1}{\sin \theta }-\sin \theta )}^2+{(\frac{1}{\cos \theta }-\cos \theta )}^2+3]$

$L.H.S={\left(\frac{1-{\sin }^2\theta }{\sin \theta }\right)}^2{\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)}^2[{\left(\frac{1-{\sin }^2\theta }{\sin \theta }\right)}^2+{\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)}^2+3]$

$L.H.S={\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^2{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^2[{\left(\frac{{\cos }^2\theta }{\sin \theta }\right)}^2+{\left(\frac{{\sin }^2\theta }{\cos \theta }\right)}^2+3]$

$L.H.S=\frac{{\cos }^4\theta }{{\sin }^2\theta }\times \frac{{\sin }^4\theta }{{\cos }^2\theta }[\frac{{\cos }^4\theta }{{\sin }^2\theta }+\frac{{\sin }^4\theta }{{\cos }^2\theta }+3]$

$L.H.S={\cos }^2\theta \times {\sin }^2\theta \left[\frac{{\cos }^6\theta +{\sin }^6\theta +3{\cos }^2\theta {\sin }^2\theta }{{\cos }^2\theta {\sin }^2\theta }\right]$

$L.H.S={\cos }^6\theta +{\sin }^6\theta +3{\cos }^2\theta {\sin }^2\theta$

$L.H.S=\left[\right({\cos }^2\theta {)}^3+\left({\sin }^2\theta {)}^3\right]+3{\cos }^2\theta {\sin }^2\theta$

Using, $(a^3+b^3)=(a+b{)}^3-3ab(a+b)$, we get

$L.H.S=\left[\right({\cos }^2\theta +{\sin }^2\theta {)}^3-3{\cos }^2\theta {\sin }^2\theta ({\cos }^2\theta +{\sin }^2\theta )]+3{\cos }^2\theta {\sin }^2\theta$

$L.H.S=[1-3{\cos }^2\theta {\sin }^2\theta ]+3{\cos }^2\theta {\sin }^2\theta$ -------------[$\because {\cos }^2\theta +{\sin }^2\theta =1$]

$L.H.S=1=R.H.S$