provar (1-cos(x))/(1+cos(x))=(cot(x)-csc(x))^2

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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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Steps Examples

Generated by AI

inverse (

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)

Solution

(

−2512	1112	−56	13
56	−16	13	−13
3112	−1712	116	−13
−1912	1112	−43	13

)

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Solution steps

One step at a time

(

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

)⁻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 4x4 identity matrix

[

1	2	3	4	∣	1	0	0	0
5	6	7	8	∣	0	1	0	0
3	2	3	2	∣	0	0	1	0
3	1	7	8	∣	0	0	0	1

]

Reduce the matrix on the left to the identity matrix

[

1	0	0	0	∣	−2512	1112	−56	13
0	1	0	0	∣	56	−16	13	−13
0	0	1	0	∣	3112	−1712	116	−13
0	0	0	1	∣	−1912	1112	−43	13

]

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

−2512	1112	−56	13
56	−16	13	−13
3112	−1712	116	−13
−1912	1112	−43	13

)

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Hi, ask a question about inverse (

1	2	3	4
5	6	7	8
3	2	3	2
3	1	7	8

) or choose below

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

provar (1-cos(x))/(1+cos(x))=(cot(x)-csc(x))^2

Solution

Solution steps

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