What point located in quadrant IV has an x-value that is 3 units from the origin and a y-value that is 3 units from the origin?

Mathematics

1 answer:

Bess [88]3 years ago

8 0

Answer: X=27y+3r jkjkljlkjlkjljljkljkjkljlkjlkjlkjlkjlkjlkjkljlkjlkj

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Prove that: (Sec A- cosec A)(1+ tan A+cot A) = tan A× sec A - cot A × cosec A

mash [69]

Answer:

Step-by-step explanation:

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA$

we take the LHS so here goes,

$(sec A-cosecA)(1+tanA+cotA)=tanAsecA-cotAcosecA\\(\frac{1}{cosA} -\frac{1}{sinA})(1+\frac{sinA}{cosA}+\frac{cosA}{sinA})\\\\(\frac{sinA-cosA}{sinAcosA})(\frac{sinAcosA+sin^2A+cos^2A}{sinAcosA})\\$

since , $sin^2A+cos^2A=1$

the identity becomes,

$(\frac{sinA-cosA}{sinAcosA})(\frac{1+sinAcosA}{sinAcosA})\\\\(\frac{sinA+sin^2AcosA-cosA-cos^2AsinA}{sin^2Acos^2A})\\\\$

now, we know,

$sin^2A=1-cos^2A$ and $cos^2A=1-sin^2A$

the identity becomes,

$(\frac{sinA+(1-cos^2A)cosA-cosA-(1-sin^2A)sinA}{sin^2Acos^2A} )\\\\$

$(\frac{sinA+cosA-cos^3A-cosA-sinA+sin^3A}{sin^2Acos^2A})$

sin A and cos A cancel out it becomes zero

$\frac{sin^3A-cos^3A}{sin^2Acos^2A} \\\\$

Splitting the denominator the identity becomes

$\frac{sin^3A}{sin^2Acos^2A}-\frac{cos^3A}{sin^2Acos^2A} \\\\\frac{sinA}{cos^2A} - \frac{cosA}{sin^2A} \\\\\frac{sinA}{cosA}(\frac{1}{cosA})-\frac{cosA}{sinA}(\frac{1}{sinA})\\\\$