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iren2701 [21]
3 years ago
12

Please help me! It’s for my homework

Mathematics
1 answer:
vodka [1.7K]3 years ago
5 0
Which question? 21 or 22?
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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})\\\\

Hence,

tanAsecA-cotAcosecA

3 0
3 years ago
What is this answer to<br> +3-2+4
otez555 [7]

Answer: I’m pretty sure it’s 5

Step-by-step explanation:

6 0
3 years ago
What is Pythagorean Therom and how is it used?
stealth61 [152]

Answer:

The Pythagorean Theorem in a^2 + b ^2 = c^2

It is used to solve for one side of a right triangle (any triangle where one angle is 90 degrees). Most of the times it is used to solve for the hypotenuse, which is represented by c. A and B represent the other sides of the triangle.

6 0
3 years ago
Read 2 more answers
​​​​If the unit rate is constant, what are the total sales for 12 pounds of asparagus?
Mandarinka [93]
How many r sold per day
7 0
3 years ago
If the hypotenuse of a right triangle is 10 inches and the side opposite to aq equals 5 inches, what is the measurement of aq ?
Debora [2.8K]
A^2 + b^2 = c^2.....a and b are ur legs and c is ur hypotenuse
5^2 + b^2 = 10^2
25 + b^2 = 100
b^2 = 100 - 25
b^2 = 75...now take the square root of both sides...this gets rid of the ^2 on the left side.
b = sq rt 75
b = 8.66 <===measure of aq

*** the pythagorean theorem (a^2 + b^2 = c^2) can only be used on right triangles
3 0
3 years ago
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