Question

Simplify ($\sec \theta +\tan \theta \left)\right(1-\sin \theta )$

Answer 1

Now,

$(\sec \theta +\tan \theta )(1-\sin \theta )$

$=(\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta })(1-\sin \theta )$

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

$=\frac{1-{\sin }^2\theta }{\cos \theta }$

$=\frac{{\cos }^2\theta }{\cos \theta }$ [ Since $1-{\sin }^2\theta ={\cos }^2\theta$]

$=\cos \theta$.