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Kisachek [45]
3 years ago
8

Factor 6 z^3/3 + 1 z^1/3

Mathematics
1 answer:
balandron [24]3 years ago
7 0

Use Rule of One: x^1 = x

6 × x^3 + 1 × z/3

Simplify 6 × z^3/3 to 2z^3

2z^3 + 1 × z/3

Simplify 1 × z/3 to z/3

<u>= 2z^3 + z/3</u>

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Prove that: [1 + 1/tan²theta] [1 + 1/cot² thata] = 1/(sin²theta - sin⁴theta]
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Step-by-step explanation:

<h3><u>Given :-</u></h3>

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

<h3><u>Required To Prove :-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Proof :-</u></h3>

On taking LHS

[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]

We know that

Tan θ = 1/ Cot θ

and

Cot θ = 1/Tan θ

=> (1+Cot²θ)(1+Tan²θ)

=> (Cosec² θ) (Sec²θ)

Since Cosec²θ - Cot²θ = 1 and

Sec²θ - Tan²θ = 1

=> (1/Sin² θ)(1/Cos² θ)

Since , Cosec θ = 1/Sinθ

and Sec θ = 1/Cosθ

=> 1/(Sin²θ Cos²θ)

We know that Sin²θ+Cos²θ = 1

=> 1/[(Sin²θ)(1-Sin²θ)]

=> 1/(Sin²θ-Sin²θ Sin²θ)

=> 1/(Sin²θ - Sin⁴θ)

=> RHS

=> LHS = RHS

<u>Hence, Proved.</u>

<h3><u>Answer:-</u></h3>

[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)

<h3><u>Used formulae:-</u></h3>

→ Tan θ = 1/ Cot θ

→ Cot θ = 1/Tan θ

→ Cosec θ = 1/Sinθ

→ Sec θ = 1/Cosθ

<h3><u>Used Identities :-</u></h3>

→ Cosec²θ - Cot²θ = 1

→ Sec²θ - Tan²θ = 1

→ Sin²θ+Cos²θ = 1

Hope this helps!!

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