Question

Solve:-
$\frac{{\sin }^3({180}^{\circ }+A){\cos }^3(-A)\tan ({360}^{\circ }-A){\sec }^2({180}^{\circ }-A)}{{{\cos }^2({90}^{\circ }+A){\sin }^3({180}^{\circ }-A)cosec}^{2}A\sin ({360}^{\circ }+A)}$

Answer 1

$\frac{{\sin }^3(180+A){\cos }^3(-A)\tan ({360}^o-A){\sec }^2(180-A)}{{\cos }^2(90+A){\sin }^3(180-A)cose{c}^{2}A\sin (360+A)}$

we know that

$\sin (180+A)=-\sin A$

$\cos (-A)=\cos A$

$\tan (360-A)=-\tan A$

$\sec (180-A)=-\sec A$

$\cos (90+A)=-\sin A$

$\sin (180-A)=\sin A$

$\sin (360+A)=sinA$

by putting values

$=\frac{(-\sin A{)}^3{\cos }^{3}A(-\tan A\left)\right(-\sec A{)}^2}{(-\sin A{)}^2.{\sin }^{3}A.cose{c}^{2}A.\sin A}$

$=\frac{-si{n}^{3}A.{\cos }^{3}A.(-\tan A).{\sec }^{2}A}{{\sin }^{2}A.{\sin }^{3}A.cose{c}^{2}A.sinA}$

$=\frac{{\sin }^{3}A.{\cos }^{3}A.\frac{\sin A}{\cos A}.\frac{1}{{\cos }^{2}A}}{{\sin }^{2}A.{\sin }^{3}A.\frac{1}{{\sin }^{2}A}\sin A}$

$=\frac{{\sin }^{3}A.\sin A}{{\sin }^{3}A.\sin A}$

$=1$