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)

Given, $\sin \theta-\cos \theta=1$
$\sin ^{3} \theta-\cos ^{3} \theta=(\sin \theta-\cos \theta)
\left(\sin ^{2} \theta\right.$
$+\left.\cos ^{2} \theta+\sin \theta \cos \theta\right)$
$=1(1+\sin \theta \cos \theta)$
$=1+\sin \theta \cos \theta$ ...(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
$\sin ^{3} \theta-\cos ^{3} \theta=1$