pareekvinit21
Answered

Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Get quick and reliable answers to your questions from a dedicated community of professionals on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.

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

for more free ans check bio

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]

Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.