Question

if sintita-costita=0 then the value of (sin^4tita+cos^4tita) is​

Answer 1

$\huge{\mathfrak{\underline{\green{AnSwEr}}}}$

$\large\sf\to\ sinθ - cosθ \: = \: 0 \\ \\ \large\sf\to\ sin θ\: = \: 0 + cosθ \\ \\ \large\sf\to\ sin \: = \: cos$

$\large\to\sf\frac{sinθ}{cosθ} \: = \: 1 \\ \\ \large\sf\to\ Tanθ \: = \: 1$

from this we can conclude that ∅ ɪs 45°

Now; we have to find $\sf\ sin^4θ + cos^4θ$

And from trigonometric ratios,the value of $\sf\ sin45° \: = \: \frac{1}{\sqrt{2} }$ and $\sf\ cos45° \: = \: \frac{1}{\sqrt{2} }$

$\large\sf\to\ (\frac{1}{\sqrt{2} })^4 + (\frac{1}{\sqrt{2} })^4 \: = \: \frac{1}{4} + \frac{1}{4} \: = \: \frac{1}{2}$