Math, asked by lawmkimahmar328, 1 month ago

if sin Theta=4/5 find the value of 4tan theta-5coa theta sec theta + 4cit theta​

Answers

Answered by tennetiraj86
4

Step-by-step explanation:

Given :-

Sin θ = 4/5

To find :-

Find the value of 4Tan θ-5Cos θ Sec θ+4 Cot θ ?

Solution :-

Given that

Sin θ = 4/5

On squaring both sides then

=> (Sin θ )²= (4/5)²

=> Sin² θ = 16/25

On Subtracting above equation from 1 then

=>1- Sin² θ = 1-(16/25)

=> 1- Sin² θ = (25-16)/25

=> 1- Sin² θ = 9/25

We know that

Sin² A + Cos² A = 1

=> Cos² θ = 9/25

=> Cos θ = √(9/25)

Since θ is acute angle then

=> Cos θ = 3/5

We know that

Tan θ = Sin θ / Cos θ

=> Tan θ = (4/5)/(3/5)

=> Tan θ = (4×5)/(3×5)

=> Tan θ = 4/3

and

1/ Tan θ = 1/(4/3)

=> Cot θ = 3/4

Now,

4Tan θ-5Cos θ Sec θ+4 Cot θ

=> 4Tan θ-5Cos θ (1/Cos θ)+4 Cot θ

=> 4Tan θ-5(Cos θ/Cos θ)+4 Cot θ

=> 4Tan θ-5(1)+4 Cot θ

=> 4Tan θ-5+4 Cot θ

=> 4(4/3) -5 +4(3/4)

=> (16/3)-5+(12/4)

=> (16/3)-5+3

=> (16/3) -2

=> (16-6)/3

=> 10/3

Answer:-

The value of 4Tan θ-5Cos θ Sec θ+4 Cot θ is 10/3

Used formulae:-

→ Sin² A + Cos² A = 1

→ Tan θ = Sin θ / Cos θ

→ Sec θ = 1/ Cos θ

→ Cot θ = 1/ Tan θ

Answered by gyaneshwarsingh882
0

Answer:

Given :-

Sin θ = 4/5

To find :-

Find the value of 4Tan θ-5Cos θ Sec θ+4 Cot θ ?

Solution :-

Given that

Sin θ = 4/5

On squaring both sides then

=> (Sin θ )²= (4/5)²

=> Sin² θ = 16/25

On Subtracting above equation from 1 then

=>1- Sin² θ = 1-(16/25)

=> 1- Sin² θ = (25-16)/25

=> 1- Sin² θ = 9/25

We know that

Sin² A + Cos² A = 1

=> Cos² θ = 9/25

=> Cos θ = √(9/25)

Since θ is acute angle then

=> Cos θ = 3/5

We know that

Tan θ = Sin θ / Cos θ

=> Tan θ = (4/5)/(3/5)

=> Tan θ = (4×5)/(3×5)

=> Tan θ = 4/3

and

1/ Tan θ = 1/(4/3)

=> Cot θ = 3/4

Now,

4Tan θ-5Cos θ Sec θ+4 Cot θ

=> 4Tan θ-5Cos θ (1/Cos θ)+4 Cot θ

=> 4Tan θ-5(Cos θ/Cos θ)+4 Cot θ

=> 4Tan θ-5(1)+4 Cot θ

=> 4Tan θ-5+4 Cot θ

=> 4(4/3) -5 +4(3/4)

=> (16/3)-5+(12/4)

=> (16/3)-5+3

=> (16/3) -2

=> (16-6)/3

=> 10/3

Answer:-

The value of 4Tan θ-5Cos θ Sec θ+4 Cot θ is 10/3

Used formulae:-

→ Sin² A + Cos² A = 1

→ Tan θ = Sin θ / Cos θ

→ Sec θ = 1/ Cos θ

→ Cot θ = 1/ Tan θ

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