Search any question & find its solution
Question:
Answered & Verified by Expert
If $x=\sin \theta+\cos \theta$ and $y=\sin \theta \cdot \cos \theta$, then what is the value
of $x^{4}-4 x^{2} y-2 x^{2}+4 y^{2}+4 y+1 ?$
Options:
of $x^{4}-4 x^{2} y-2 x^{2}+4 y^{2}+4 y+1 ?$
Solution:
1792 Upvotes
Verified Answer
The correct answer is:
0
Let $x=\sin \theta+\cos \theta$ and $y=\sin \theta \cdot \cos \theta$
Now, $x^{4}-4 x^{2} y-2 x^{2}+4 y^{2}+4 y+1$
$=(\sin \theta+\cos \theta)^{4}-4(\sin \theta+\cos \theta)^{2} y-$
$\quad 2(\sin \theta+\cos \theta)^{2}+4 y^{2}+4 y+1$
$=\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)^{2}-$
$\quad 4\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right) y$
$\quad-2\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)+4 y^{2}+4 y+1$
$=(1+2 y)^{2}-4(1+2 y) y-2(1+2 y)+4 y^{2}+4 y+1$
$=1+4 y^{2}+4 y-4 y-8 y^{2}-2-4 y+4 y^{2}+4 y+1=0$
Now, $x^{4}-4 x^{2} y-2 x^{2}+4 y^{2}+4 y+1$
$=(\sin \theta+\cos \theta)^{4}-4(\sin \theta+\cos \theta)^{2} y-$
$\quad 2(\sin \theta+\cos \theta)^{2}+4 y^{2}+4 y+1$
$=\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)^{2}-$
$\quad 4\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right) y$
$\quad-2\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)+4 y^{2}+4 y+1$
$=(1+2 y)^{2}-4(1+2 y) y-2(1+2 y)+4 y^{2}+4 y+1$
$=1+4 y^{2}+4 y-4 y-8 y^{2}-2-4 y+4 y^{2}+4 y+1=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.