Question

Prove:1/sin²A-1/tan²A=1​

Answer 1

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

`\displaystyle (1)/(\sin^2x)-(1)/(\tan^2x)=1`

Explanation:

Prove that:

`\displaystyle (1)/(\sin^2x)-(1)/(\tan^2x)=1`

Recall that by definition:

`\displaystyle \tan x=(\sin x)/(\cos x)`

Therefore,

`\displaystyle \tan^2x=\left ((\sin^2x)/(\cos^2x)\right)^2=(\sin^2x)/(\cos^2x)`

Substitute
`\displaystyle \tan^2x=(\sin^2x)/(\cos^2x)` into
`\displaystyle (1)/(\sin^2x)-(1)/(\tan^2x)=1`:

`\displaystyle (1)/(\sin^2x)-(1)/((\sin^2x)/(\cos^2x))=1`

Simplify:

`\displaystyle (1)/(\sin^2x)-(\cos^2x)/(\sin^2x)=1`

Combine like terms:

`\displaystyle (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 (1-\cos^2x)/(\sin^2x)=1`:

`\displaystyle (\sin^2x)/(\sin^2x)=1,\\\\1=1\:\boxed{\checkmark\text{ True}}`