Consider the following:
1. ${\tan }^2\theta -{\sin }^2\theta ={\tan }^2\theta {\sin }^2\theta$
2.$(1+{\cot }^2\theta )(1-\cos \theta )(1+\cos \theta )=1$
Which of the statements given below is correct ?

The correct option is C Both 1 and 2 are identities
$1.{\tan }^2\theta -{\sin }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta$
$=\frac{{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\cos }^2\theta }$

$=\frac{{\sin }^2\theta (1-{\cos }^2\theta )}{{\cos }^2\theta }$
$=\frac{{\sin }^2\theta }{{\cos }^2\theta }\times {\sin }^2\theta ={\tan }^2\theta {\sin }^2\theta$
$2.(1+{\cot }^2\theta )(1-\cos \theta )(1+cos\theta )$
$=(1+{\cot }^2\theta )(1-{\cos }^2\theta )$
$=(1+\frac{{\cos }^2\theta }{{\sin }^2\theta })(1-{\cos }^2\theta )$
$=\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta }\right)\times {\sin }^2\theta$ .......[Since, $(1-{\cos }^2\theta ={\sin }^2\theta )$]
$=1$ .......[Since, $({\cos }^2\theta +{\sin }^2\theta =1)$]
Therefore, both are Identities.