Question

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

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

$\sin \theta -\cos \theta =1$ ....(1)

Squaring both sides

${\left(\sin \theta -\cos \theta \right)}^2=1^2\Rightarrow {\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta =1\Rightarrow \sin \theta \cos \theta =0$

Now taking cube on both sides of (1)

${\left(\sin \theta -\cos \theta \right)}^3=1^3\Rightarrow {\sin }^3\theta -{\cos }^3\theta -3\sin \theta \cos \theta \left(\sin \theta -\cos \theta \right)=1$

............. Using $(a-b{)}^3=a^3-b^3-3ab(a-b)$

$\Rightarrow {\sin }^3\theta -{\cos }^3\theta =1$