Question

If $\sin \theta -\cos \theta =1$, then the value of ${\sin }^3\theta -{\cos }^3\theta$ is equal to

• A. $1$
• B. $-1$
• C. $0$
• D. $2$
• E. $-2$

Answer 1

Answer: (A)

$1$

Given, $\sin \theta -\cos \theta =1$

${\sin }^3\theta -{\cos }^3\theta =(\sin \theta -\cos \theta )({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta )$

$=1(1+\sin \theta \cos \theta )$

$=1+\sin \theta \cos \theta$.........(i)

$[\because \sin \theta -\cos \theta =1]$

On squaring both sides,

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

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

$\Rightarrow 1-2\sin \theta \cos \theta =1$

$\sin \theta \cos \theta =0$.........(ii)

From Equations (ii) and (i), we get ${\sin }^3\theta -{\cos }^3=1$