Question

Prove:1/sin²A-1/tan²A=1​

Answer 1

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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Answer 2

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

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