If $4 \sin ^2(2 x-10)^{\circ}=3,0 \leq(2 x-10) \leq 90$, then find the value of $\frac{\sin ^4(x-5)^{\circ}+\cos ^4(x-5)^{\circ}}{1-2 \sin ^2(3 x-15)^{\circ} \cos ^2(3 x-15)^{\circ}}$.
Answer & explanation
Correct answer: option 1
We are given that ,
4 sin² ( 2x - 10 )º = 3
sin² ( 2x - 10 )º = \(\frac{3}{4}\)
sin ( 2x - 10 )º = \(\frac{√3}{2}\)
{ we know, sin60º = \(\frac{√3}{2}\) }
So, ( 2x - 10 )º = 60º
x = 35º
Now,
\(\frac{sin4 (35-5)º + cos4 ( x - 5 )º }{1 - 2sin²(3x-15)º . cos²(3x-15)º }\)
= \(\frac{sin4 (35-5)º + cos4 ( 35 - 5 )º }{1 - 2sin²(105-15)º . cos²(105-15)º }\)
= \(\frac{1/16 + 9/16 }{1 }\)
= \(\frac{10 }{16 }\)
= \(\frac{5}{8}\)