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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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simplify (2ab³)⁻¹20· (2⁻¹· ba⁵)² · (4ab⁻¹5b )⁻²

Solution

532a¹³b

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Solution steps

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tan²(x)−1sec²(x) =tan(x)−cot(x)tan(x)+cot(x)

Manipulating left side

tan²(x)−1sec²(x)

Express with sin, cos

=−cos²(x)+sin²(x)

Manipulating right side

tan(x)−cot(x)tan(x)+cot(x)

Express with sin, cos

=−cos²(x)+sin²(x)cos²(x)+sin²(x)

Rewrite using trig identities

=−cos²(x)+sin²(x)

We showed that the two sides could take the same form

⇒ True

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Notebook

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

1	2	1
6	−1	0
−1	−2	−1

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