Prove that: (1+cot^2x)tan^2x = sec^2x

Mathematics

1 answer:

timofeeve [1]3 years ago

8 0

Answer:

Step-by-step explanation:

$1+cot^2x = csc^2x= \frac{1}{sin^2x}$

it becomes

$\frac{1}{sin^2x}*\frac{sin^2x}{cos^2x}=\frac{1}{cos^2x}=sec^2x$

Done

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Sandra rode her bike 9 miles on Monday and 6 miles on Tuesday. How many more miles did

Greeley [361]

Answer:

132 miles

Step-by-step explanation:

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3 years ago

Hailey is scuba diving in the ocean. Her boat is anchored 300 feet away at a 30° angle of depression. If her diving partner is 2

victus00 [196]

Answer:

B)313 ft

Step-by-step explanation:

Hello,I think I can help you with this

to solve this you need to find all the measures of the rigth triangle formed by Hailey, the boat, and the point on the surface of the sea just above the head of Hailey

Step 1

define

triangle

Hypotenuse=300 feet

opposite side=y

adjacent side=x

angle of depression =30°

Step 2

find the distances

$cos\alpha =\frac{adjancent\ side}{hypotenuse}\\\\hypotenuse*cos\alpha =adjancent\ side\\ so\\x=adjancent\ side=hypotenuse*cos\alpha\\\\x=300\ft*cos 30 \\\\x=259.8076$

so, the point on the surface of the sea just above hailey is 259.8076 ft from the boat

$sin\alpha =\frac{opposite\ side}{hypotenuse}\\\\hypotenuse*sin\alpha \\y=opposite\ side\\ so\\y=opposite\ side=hypotenuse*sin\alpha\\\\y=300\ft*sin 30 \\\\y=150$

hailey is at a depth of 150 feet

Step 3

now,her partner is 25 feet directly below her,it is

partner's depth is =hailey is at a depth of 150 feet+25

Y=175

he is at the same distance from de boat ( horizontal),because it is the same point in the sea surface.

X=259.80

Step 4

to find the distance between the partner and the boat we can use

Pythagoras theorem

$adjacentside^{2} +opposite side^{2} =hypotenuse^{2} \\\sqrt[2]{adjacentside^{2} +opposite side^{2}} =Hypotenuse\\Hypotenuse=\sqrt[2]{175^{2} +259.8076^{2}} \\\\Hypotenuse=313.24 ft$