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Svetlanka [38]
3 years ago
14

Prove:1/sin²A-1/tan²A=1​

Mathematics
2 answers:
Anit [1.1K]3 years ago
7 0

Step-by-step explanation:

1/sin^2A -cos^2A/sin^2 A. ~tan = sin/cos

(1-cos^2)/sin^2A. ~ take lcm

sin^2A/sin^ A. ~ 1-cos^2A = sin^2A

1

for more free ans check bio

yaroslaw [1]3 years ago
4 0

Answer:

\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1

Step-by-step explanation:

Prove that:

\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1

Recall that by definition:

\displaystyle \tan x=\frac{\sin x}{\cos x}

Therefore,

\displaystyle \tan^2x=\left (\frac{\sin^2x}{\cos^2x}\right)^2=\frac{\sin^2x}{\cos^2x}

Substitute \displaystyle \tan^2x=\frac{\sin^2x}{\cos^2x} into \displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1:

\displaystyle \frac{1}{\sin^2x}-\frac{1}{\frac{\sin^2x}{\cos^2x}}=1

Simplify:

\displaystyle \frac{1}{\sin^2x}-\frac{\cos^2x}{\sin^2x}=1

Combine like terms:

\displaystyle \frac{1-\cos^2x}{\sin^2x}=1

Recall the following Pythagorean Identity:

\sin^2x+\cos^2x=1 (derived from the Pythagorean Theorem)

Subtract \cos^2x from both sides:

\sin^2=1-\cos^2x

Finish by substituting \sin^2=1-\cos^2x into \displaystyle \frac{1-\cos^2x}{\sin^2x}=1:

\displaystyle \frac{\sin^2x}{\sin^2x}=1,\\\\1=1\:\boxed{\checkmark\text{ True}}

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Sliva [168]

Hello and Good Morning/Afternoon

<u>Let's take this problem step by step.</u>

        5=\frac{1}{3}(5x-15)\\ 5=\frac{5}{3}x-5\\ 5+5= \frac{5}{3}x-5 +5\\10=\frac{5}{3}x\\ 10 * 3 = 3* \frac{5}{3}x\\30=5x\\\frac{30}{5} =\frac{5}{5}x\\ 6=x\\x=6

<u>What did we do in this problem</u>

  • we ensured that both sides remained equally balanced in value
  • took everything step-by-step to make sure that there were no mistakes
  • to check it, we can plug it back into the original equation to see if it works

                  5=\frac{1}{3}(5(6)-15)\\ 5=\frac{1}{3} (30-15)\\5=\frac{1}{3}*15\\ 5=5 \checkmark\\

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