beweisen tan^2(x)-sin^2(x)=tan^2(x)sin^2(x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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Generated by AI

row echelon (

4	4	2	3	−2
0	1	−2	−2	2
6	12	11	2	−4
9	20	10	10	−6
15	28	14	5	−3

)

Solution

(

4	4	2	3	−2
0	1	−2	−2	2
0	0	20	192	−13
0	0	0	19516	−458
0	0	0	0	2113

)

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Solution steps

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(

4	4	2	3	−2
0	1	−2	−2	2
6	12	11	2	−4
9	20	10	10	−6
15	28	14	5	−3

)⁻¹

Inverse matrix

A matrix is invertible if and only if there is a sequence of elementary row (or column) operations taking it to the identity matrix.

Augment with a 5x5 identity matrix

[

4	4	2	3	−2	∣	1	0	0	0	0
0	1	−2	−2	2	∣	0	1	0	0	0
6	12	11	2	−4	∣	0	0	1	0	0
9	20	10	10	−6	∣	0	0	0	1	0
15	28	14	5	−3	∣	0	0	0	0	1

]

Reduce the matrix on the left to the identity matrix

[

1	0	0	0	0	∣	1235	−435	−235	−17105	221
0	1	0	0	0	∣	0	715	215	215	−215
0	0	1	0	0	∣	−25	−45	−15	−215	415
0	0	0	1	0	∣	−1335	−92105	−46105	−2105	27
0	0	0	0	1	∣	−2735	−148105	−74105	−23105	1321

]

The inverse matrix is on the right side of the augmented matrix

1235	−435	−235	−17105	221
0	715	215	215	−215
−25	−45	−15	−215	415
−1335	−92105	−46105	−2105	27
−2735	−148105	−74105	−23105	1321

)

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Hi, ask a question about eigenvalues (

4	4	2	3	−2
0	1	−2	−2	2
6	12	11	2	−4
9	20	10	10	−6
15	28	14	5	−3

) or choose below

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

beweisen tan^2(x)-sin^2(x)=tan^2(x)sin^2(x)

Solution

Solution steps

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