Operation of Sine and Cosine Use condition $\displaystyle\sin\theta-\cos\theta=\sqrt{2}$,
Please find the value of $\displaystyle\frac1{\sin^{10}\theta}+\frac1{\cos^{10}\theta}$.
 A: Notice that $\sin \theta - \cos \theta = \sqrt{2} \implies \sin^{2} \theta - \sin (2\theta) + \cos^2\theta = 2 \implies \sin (2 \theta) = -1. $
But this holds iff $2\theta = \frac{4n - 1}{2}\pi \iff \theta = \frac{(4n-1) \pi}{4} $
A: Avoid squaring if possible as it immediately introduces extraneous roots 
We have $$\sin\theta-\cos\theta=\sqrt2\iff \cos\theta-\sin\theta=-\sqrt2$$
$$\iff \cos\theta\cos\frac\pi4-\sin\theta\sin\frac\pi4=-1\left(\text {as }  \sin\frac\pi4=\cos\frac\pi4=\frac1{\sqrt2}\right)$$
$$\iff \cos\left(\theta+\frac\pi4\right)=-1=\cos\pi$$
$$\iff \theta+\frac\pi4=(2n+1)\pi\iff \theta=2n\pi+\pi-\frac\pi4(\text{ where } n \text{ is any integer})$$
$$\implies\sin\theta=\sin\left(2n\pi+\pi-\frac\pi4\right)=\sin\left(\pi-\frac\pi4\right)=\sin\frac\pi4=\frac1{\sqrt2}$$
and 
$$\cos\theta=\cos\left(2n\pi+\pi-\frac\pi4\right)=\cos\left(\pi-\frac\pi4\right)=-\cos\frac\pi4=-\frac1{\sqrt2}$$
A: If we set $a=\sin\theta,b=\cos\theta$
we have $a-b=\sqrt2$ and
as $a^2+b^2=1,(a-b)^2=2\iff -2ab=1$
$\displaystyle (a^2-b^2)^2=(a^2+b^2)^2-4a^2b^2=1^2-(2ab)^2=0$
$\displaystyle\implies a^2-b^2=0$ and we have $a^2+b^2=1$
Can you solve for $a^2,b^2?$
We need to find $\displaystyle\frac1{a^{10}}+\frac1{b^{10}}=\frac1{(a^2)^5}+\frac1{(b^2)^5}$
