Question

$(\sin \theta -\sec \theta {)}^2+(\cos \theta -\text{cosec}\theta {)}^2=(1-\sec \theta \cdot \text{cosec}\theta {)}^2$.

Answer 1

Let us begin by simplifying L.H.S and proving it to be equal to R.H.S
L.H.S= $(\sin \theta -\sec \theta {)}^2+(\cos \theta -\text{cosec}\theta {)}^2$
Since $\text{cosec}\theta =\frac{1}{\sin \theta }$ and $\sec \theta =\frac{1}{\cos \theta }$ ................. (1), the above equation can be written as
= ${(\sin \theta -\frac{1}{\cos \theta })}^2+{(\cos \theta -\frac{1}{\sin \theta })}^2$= ${\left(\frac{\sin \theta \cos \theta -1}{\cos \theta }\right)}^2+{\left(\frac{\cos \theta \sin \theta -1}{\sin \theta }\right)}^2$
Taking $(\sin \theta \cos \theta -1{)}^2$ common, we get
$(\sin \theta \cos \theta -1{)}^2$ $(\frac{1}{{\sin }^2\theta }+\frac{1}{co{s}^2\theta })$

= $(sin\theta cos\theta -1{)}^2$ $\left(\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta {\sin }^2\theta }\right)$
since ${\sin }^2\theta +{\cos }^2\theta =1$, the above equation reduces to
${\left(\frac{\sin \theta \cos \theta -1}{\cos \theta \sin \theta }\right)}^2$
=${(1-\frac{1}{\cos \theta \sin \theta })}^2$
Usign equation (1) the equation further reduces to,
= $(1-\sec \theta \cdot \text{cosec}\theta {)}^2$ = R.H.S (Hence proved)