Question

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

• A. 1
• B. -1
• C. 0
• D. 2
• E. -2

Answer 1

Answer: (A)

Given, $\sin \theta -\cos \theta =1$
${\sin }^3\theta -{\cos }^3\theta =(\sin \theta -\cos \theta )<br/><br/>({\sin }^2\theta$
$+{\cos }^2\theta +\sin \theta \cos \theta )$
$=1(1+\sin \theta \cos \theta )$
$=1+\sin \theta \cos \theta$ ...(i)
$[\because \sin \theta -\cos \theta =1]$
On squaring both sides,
$(\sin \theta -\cos \theta {)}^2=(1{)}^2$
$\Rightarrow \left({\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta \right)=1$
$\Rightarrow 1-2\sin \theta \cos \theta -1$
$\Rightarrow \sin \theta \cos \theta =0$ ...(ii)
From Eqs. (ii) and (i), we get
${\sin }^3\theta -{\cos }^3\theta =1$