csc^4-cot^4= 1+cos^2/sin^2
i need help! please help me
LS = (csc^2 Ø + cot^2 Ø)(csc^2 Ø - cot^2 Ø
= (1 + cot^2 Ø + cot^2 Ø)(1+cot^2 Ø - cot^2 Ø
= 1 + 2cot^2 Ø
RS = (1 + cos^2 Ø)/sin^2 Ø
= (sin^2 Ø + cos^2 Ø + cos^2 Ø)/sin^2 Ø
= (sin^2 Ø + 2cos^2 Ø)/sin^2 Ø
= 1 + 2 cot^2 Ø
= LS
forgot to mention that you cannot just skip the "angle" after a trig operator.
What you typed would be like having an equation of the type ...
√ = ∫ + ∑
Certainly! I can help you with the given equation: csc^4 - cot^4 = (1 + cos^2) / sin^2.
To solve it, we can start by expressing csc^4 and cot^4 in terms of sin and cos. Remember that csc^2 is the reciprocal of sin^2 and cot^2 is the reciprocal of tan^2:
csc^4 = (1 / sin^2)^2 = 1 / sin^4
cot^4 = (cos^2 / sin^2)^2 = cos^4 / sin^4
Now, substituting these values into the equation, we get:
1 / sin^4 - cos^4 / sin^4 = (1 + cos^2) / sin^2
Next, we need to find a common denominator for the fractions on the left side. The common denominator will be sin^4, so we multiply the first fraction by sin^4 / sin^4:
(1 - cos^4) / sin^4 = (1 + cos^2) / sin^2
Now, we can cross multiply:
sin^2 * (1 - cos^4) = sin^4 * (1 + cos^2)
Simplifying both sides:
sin^2 - sin^2 * cos^4 = sin^4 + sin^4 * cos^2
Note that sin^2 is equal to 1 - cos^2 (from the Pythagorean identity sin^2 + cos^2 = 1). Replacing sin^2 with 1 - cos^2:
(1 - cos^2) - (1 - cos^2) * cos^4 = (1 - cos^2)^2 + (1 - cos^2) * cos^2
Expanding:
1 - cos^2 - (1 - cos^2) * cos^4 = (1 - cos^2)(1 - cos^2 + cos^2)
Canceling out similar terms:
1 - cos^2 - cos^4 + cos^6 = 1 - cos^2 + cos^2 - cos^4
Simplifying further:
1 - cos^2 - cos^4 + cos^6 = 1 - cos^4
Now, we can cancel out the common terms on both sides, leaving us with:
cos^6 = 0
Taking the cube root of both sides:
cos^2 = 0
And finally, taking the square root of both sides:
cos = 0
Therefore, the solution to the original equation is cos = 0.