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taurus [48]
2 years ago
9

SOMEONE PLEASE HELP ME I MEED IT

Mathematics
2 answers:
Minchanka [31]2 years ago
7 0
Answer is d (mark as brainliest)
GarryVolchara [31]2 years ago
6 0

Answer:

answer is D

Step-by-step explanation:

your welcome, have a nice day

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Verify the trigonometric identities
snow_lady [41]
1)

here, we do the left-hand-side

\bf [sin(x)+cos(x)]^2+[sin(x)-cos(x)]^2=2
\\\\\\\
[sin^2(x)+2sin(x)cos(x)+cos^2(x)]\\\\+~ [sin^2(x)-2sin(x)cos(x)+cos^2(x)]
\\\\\\
2sin^2(x)+2cos^2(x)\implies 2[sin^2(x)+cos^2(x)]\implies 2[1]\implies 2

2)

here we also do the left-hand-side

\bf \cfrac{2-cos^2(x)}{sin(x)}=csc(x)+sin(x)
\\\\\\
\cfrac{2-[1-sin^2(x)]}{sin(x)}\implies \cfrac{2-1+sin^2(x)}{sin(x)}\implies \cfrac{1+sin^2(x)}{sin(x)}
\\\\\\
\cfrac{1}{sin(x)}+\cfrac{sin^2(x)}{sin(x)}\implies csc(x)+sin(x)

3)

here, we do the right-hand-side

\bf \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}=\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}
\\\\\\
\cfrac{csc(x)-tan(x)}{sec(x)+cot(x)}\implies \cfrac{\frac{1}{sin(x)}-\frac{sin(x)}{cos(x)}}{\frac{1}{cos(x)}-\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{cos(x)-sin^2(x)}{sin(x)cos(x)}}{\frac{sin(x)+cos^2(x)}{sin(x)cos(x)}}
\\\\\\
\cfrac{cos(x)-sin^2(x)}{\underline{sin(x)cos(x)}}\cdot \cfrac{\underline{sin(x)cos(x)}}{sin(x)+cos^2(x)}\implies \cfrac{cos(x)-sin^2(x)}{sin(x)+cos^2(x)}
8 0
3 years ago
I got 16/63 but i am not sure if its correct or not .
QveST [7]

Answer:

i think i got

sin x =16/65 not 16/63

3 0
2 years ago
Select the equivalent expression.
Fittoniya [83]

Answer:

B

Step-by-step explanation:

8 0
3 years ago
Someone help me!! i’ll mark brain list!!
algol [13]
Hello this is super easy i can help you
5 0
2 years ago
Find the equation of the line that is parallel to the line x+3y=-15 that passes through the origin
Monica [59]
You would subtract x from each side. It would then be 3y=-x-15. Then you would divide all of the equation by 3. Your final answer would be y=-3x-5.
5 0
3 years ago
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