1. cot x sec4x = cot x + 2 tan x + tan3x
1 answer:
1. cot(x)sec⁴(x) = cot(x) + 2tan(x) + tan(3x)
cot(x)sec⁴(x) cot(x)sec⁴(x)
0 = cos⁴(x) + 2cos⁴(x)tan²(x) - cos⁴(x)tan⁴(x)
0 = cos⁴(x)[1] + cos⁴(x)[2tan²(x)] + cos⁴(x)[tan⁴(x)]
0 = cos⁴(x)[1 + 2tan²(x) + tan⁴(x)]
0 = cos⁴(x)[1 + tan²(x) + tan²(x) + tan⁴(4)]
0 = cos⁴(x)[1(1) + 1(tan²(x)) + tan²(x)(1) + tan²(x)(tan²(x)]
0 = cos⁴(x)[1(1 + tan²(x)) + tan²(x)(1 + tan²(x))]
0 = cos⁴(x)(1 + tan²(x))(1 + tan²(x))
0 = cos⁴(x)(1 + tan²(x))²
0 = cos⁴(x) or 0 = (1 + tan²(x))²
⁴√0 = ⁴√cos⁴(x) or √0 = (√1 + tan²(x))²
0 = cos(x) or 0 = 1 + tan²(x)
cos⁻¹(0) = cos⁻¹(cos(x)) or -1 = tan²(x)
90 = x or √-1 = √tan²(x)
i = tan(x)
(No Solution)
2. sin(x)[tan(x)cos(x) - cot(x)cos(x)] = 1 - 2cos²(x)
sin(x)[sin(x) - cos(x)cot(x)] = 1 - cos²(x) - cos²(x)
sin(x)[sin(x)] - sin(x)[cos(x)cot(x)] = sin²(x) - cos²(x)
sin²(x) - cos²(x) = sin²(x) - cos²(x)
+ cos²(x) + cos²(x)
sin²(x) = sin²(x)
- sin²(x) - sin²(x)
0 = 0
3. 1 + sec²(x)sin²(x) = sec²(x)
sec²(x) sec²(x)
cos²(x) + sin²(x) = 1
cos²(x) = 1 - sin²(x)
√cos²(x) = √(1 - sin²(x))
cos(x) = √(1 - sin²(x))
cos⁻¹(cos(x)) = cos⁻¹(√1 - sin²(x))
x = 0
4. -tan²(x) + sec²(x) = 1
-1 -1
tan²(x) - sec²(x) = -1
tan²(x) = -1 + sec²
√tan²(x) = √(-1 + sec²(x))
tan(x) = √(-1 + sec²(x))
tan⁻¹(tan(x)) = tan⁻¹(√(-1 + sec²(x))
x = 0
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Answer:
(f+g)(x)=f(x)+g(x)
=(-3x-5)+(4x-2)
= -3x-5+4x-2. collect like term
= -3x+4x-5-2
= x-7. the final answer
F= 9 x 7 = 63
G= 19 x 4 = 76
H= 12 x 4 = 48
J = 32 x 2 = 64
J is the answer
Answer:
27.968 mins
Step-by-step explanation:
1) we need to find 8% of 32:
32*0.08=2.56
so 8% of 32 is 2.56 mins.
we need to subtract 2.56 by 32 since its says the times was cut by 8%
so
32-2.56=29.44
now 5% of 29.44:
29.44*0.05=1.472
1.472 mins
so
29.44-1.472=27.968
Hope this helps!
Radiation from the sun. That depends on the sun.
Let smaller no be x and bigger be y
- x+y=39--(1)
- 3x=y+81
- y=3x-81
- y=3(x-27)--(2)
Put it in eq(1)
- x+3(x-27)=39
- x+3x-81=39
- 4x=81+39
- 4x=120
- x=30
Now
The numbers are 9 and 30
