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Prove:1/sin²A-1/tan²A=1​

Sagot :

rehandev390

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

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Corsaquix

Answer:

[tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1[/tex]

Step-by-step explanation:

Prove that:

[tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\tan^2x}=1[/tex]

Recall that by definition:

[tex]\displaystyle \tan x=\frac{\sin x}{\cos x}[/tex]

Therefore,

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

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

[tex]\displaystyle \frac{1}{\sin^2x}-\frac{1}{\frac{\sin^2x}{\cos^2x}}=1[/tex]

Simplify:

[tex]\displaystyle \frac{1}{\sin^2x}-\frac{\cos^2x}{\sin^2x}=1[/tex]

Combine like terms:

[tex]\displaystyle \frac{1-\cos^2x}{\sin^2x}=1[/tex]

Recall the following Pythagorean Identity:

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

Subtract [tex]\cos^2x[/tex] from both sides:

[tex]\sin^2=1-\cos^2x[/tex]

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

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

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