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GuDViN [60]
2 years ago
11

Solve the simultaneous equation:2x+3y=13X+2y=9​

Mathematics
1 answer:
Neko [114]2 years ago
8 0

Answer:

2x + 3y = 13

4x - y = -2

In the second equation, subtract 4x from both sides.

-y = -2 - 4x

Divide both sides by -1.

y = 2 + 4x

Put this into the first equation in place of y.

2x + 3(2 + 4x) = 13

Multiply everything in the parenthesis by 3.

2x + 6 + 12x = 13

Combine like terms.

14x + 6 = 13

Subtract 6 from both sides.

14x = 7

Divide 14 on both sides.

x = 7 / 14

x = 0.5

Put this into the second equation in place of x.

4(0.5) - y = -2

2 - y = -2

Subtract 2 from both sides.

-y = -4

Divide both sides by -1.

y = 4

So x = 0.5 and y = 4.

Step-by-step explanation:

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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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