Question

if cosec theta - cot theta=a, then find the value of 2cosec^2theta+3cot^2theta

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Answer 1

Answer:

5a²/4 + 5/4a² - 1/2

Step-by-step explanation:

 Using, cosec²θ - cot²θ = 1

         ⇒ (cosecθ + cotθ)(cosecθ - cotθ) = 1

         ⇒ (cosecθ + cotθ)(a) = 1

         ⇒ cosecθ + cotθ = 1/a  

Adding cosecθ - cotθ and cosecθ + cotθ, we get

 2cosecθ = a + 1/a

  cosecθ = 1/2 (a + 1/a)

Substituting this in cosecθ - cotθ = a, we get

 1/2 (a + 1/a) - cotθ = a

 1/2 (1/a - a) = cotθ

Hence,

cosec²θ = [1/2 (a + 1/a)]²  

              = 1/4 (a² + 1/a² + 2)

cot²θ      = [ 1/2 (1/a - a) ]²

              = 1/4 [a² + 1/a² - 2]

∴ 2cosec²θ + 3cot²θ

     ⇒ 2[1/4 (a² + 1/a² + 2)] + 3[1/4 (a² + 1/a² - 2)]

     ⇒ a²/2 + 1/2a² + 1 + 3a²/4 + 3/4a² - 3/2

     ⇒ 5a²/4 + 5/4a² - 1/2

Answer 2

Information provided with us:

• Cosec theta - cot theta = a

What we have to calculate:

• We have to calculate and find out the value of 2cosec²theta + 3cot²theta

Performing Calculations:

____________

Using Formula,

➻ 1 + cot²θ = cosec²θ

By changing the sides we gets,

➻ cosec²θ - cot²θ = 1

And also using Identity,

• (a²-b²) = (a+b)(a-b)

By using the identity we gets,

➻ (cosecθ + cotθ) (cosecθ - cotθ) = 1

According to the question putting the value of (Cosec theta - cot theta) as a,

➻ (cosecθ + cotθ) (a) = 1

Transposing a to R.H.S.,

➻ (cosecθ + cotθ) = 1/a

Here we are adding both (cosecθ + cotθ) and (cosecθ - cotθ) inorder to solve,

➻ (cosecθ - cotθ) + (cosecθ + cotθ) = 1/a

cosecθ - cotθ + cosecθ + cotθ = 1/a

➻ 2 cosecθ = 1/a

Now,

➻ 2cosecθ = (a + 1) / a

Transposing 2 to R.H.S.,

➻ cosecθ = (a + 1 / a) × (1 / 2)

Here we got the value of cosecθ. So we would be putting this value in (cosecθ - cotθ),

➻ (a + 1 / a) × (1 / 2) - cotθ

As it is equal to a,

➻ (a + 1 / a) × (1 / 2) - cotθ = a

Changing the sides,

➻ cotθ = (a + 1 / a) × (1 / 2) - a/1

➻ cotθ = (a + 1 / a - a) × (1/2)

➻ cotθ = (a + 1 / a - a) × (1 / 2)

Again according to the question finding out the value of 2cosec²θ,

• Here we would be doing the squares,

➻ [ (1/2) × (a + 1 / a) ] ²

➻ [ (1×1 / 2×2) × (a² + 1² / a²) ]

➻ [ (1 / 4) × (a² + 1 / a² + 2 ) ]

Now again according to the question finding out the value of cot²θ,

• Here doing the squares,

➻ [ (1/2) × (a + 1 / 2a - a) ] ²

➻ [ (1×1 / 2×2) × (a + 1 / a - a) ] ²

➻ [ (1 / 4) × (a² + 1 / a² - 2) ]

At last finding out the value,

➻ 2cosec²θ + 3cot²θ

Here we have,

• cosec²θ = [ (1 / 4) × (a² + 1 / a² + 2 ) ]
• cot²θ = [ (1 / 4) × (a² + 1 / a² - 2) ]

Substituting the values,

➻ 2 [ (1 / 4) × (a² + 1 / a² - 2) ] + 3 × [ (1 / 4) × (a² + 1 / a² - 2) ]

➻ 2 × [ (1 / 4) × (a² + 1 / a² - 2) ] + 3 × [ (1 / 4) × (a² + 1 / a² - 2) ]

➻ a²/2 + (1/2×a ×a) + (2/2) + (3×a² / 4) + (3×1 / 4×a²) + (-3/2)

➻ (a²/2) + (1/2×a²) + (1) + (3×a²/4) + (3×1/4×a²) + (-3/2)

➻ (a²/2) + (1/2a²) + (1) + (3×a²/4) + (3×1/4×a²) + (-3/2)

➻ (a²/2) + (1/2a²) + (1) + (3a²/4) + (3×1/4×a²) + (-3/2)

➻ (a²/2) + (1/2a²) + (1) + (3a²/4) + (3/4a²) + (-3/2)

On solving them we gets,

➻ (5 / 4a²) - 1/2 + (5a² / 4)

____________

• Hence, the value of cosec²θ and cot²θ is (5 / 4a²) - 1/2 + (5a² / 4)