Question

${\sin }^8\theta -{\cos }^8\theta -({\sin }^2\theta -{\cos }^2\theta )(1-{\sin }^2\theta )$=0

• A. True
• B. False

Answer 1

The correct option is B False

${\sin }^4\theta +{\cos }^4\theta =({\sin }^2\theta +{\cos }^2\theta {)}^2-2{\sin }^2\theta {\cos }^2\theta =1-2{\sin }^2\theta {\cos }^2\theta$

${\sin }^4\theta -{\cos }^4\theta =({\sin }^2\theta -{\cos }^2\theta )({\sin }^2\theta +{\cos }^2\theta )={\sin }^2\theta -{\cos }^2\theta$

${\sin }^8\theta -{\cos }^8\theta -({\sin }^2\theta -{\cos }^2\theta )(1-{\sin }^2\theta )=({\sin }^4\theta +{\cos }^4\theta )({\sin }^4\theta -{\cos }^4\theta )-({\sin }^2\theta -{\cos }^2\theta )(1-{\sin }^2\theta )$

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

$=({\sin }^2\theta -{\cos }^2\theta )\left({\sin }^2\theta \right)(1-2{\cos }^2\theta )$

So the given relation is $\text{False}$