Question

If sin theta=1/3 then find the value of(9 cot2 theta+9) is

Answer 1

Answer:

$\frac{567 \sqrt{2} + 576 }{64}$

Step-by-step explanation:

Given :

sinθ = 1/3

To Find :

9cot(2θ) + 9

=> 9(cot(2θ) + 1)

Formulae :

csc²θ = cot²θ + 1 [From sin²θ + cos²θ = 1]

cscθ = 1/sinθ

tanθ = 1/cotθ

cot(2θ) = 1/tan(2θ)

Tan(2θ) =

$\frac{2tanθ}{1 - ta {n}^{2}θ }$

Hence cot(2θ) =

$\frac{1 - ta {n}^{2}θ } {2tanθ}$

We can get tanθ by 1/cotθ.

Procedure :

sinθ = 1/3

sin²θ = 1/9

csc²θ = 9/1 = 9

csc²θ = cot²θ + 1

=> cot²θ = csc²θ - 1

=> cot²θ = 9 - 1

=> cot²θ = 8

Hence tan²θ = 1/8 and tanθ = 1/√8

Hence cot(2θ) =

$\frac{1 - ( \frac{1}{8})^{2} }{2( \frac{1}{ \sqrt{8} } )}$

√8 = 2√2

=>

$\frac{1 - ( \frac{1}{8})^{2} }{( \frac{1}{ \sqrt{2} } )}$

cot(2θ) => √2 × (63/64)

cot(2θ) + 1 =

$\frac{63 \sqrt{2} + 64 }{64}$

9(cot(2θ) + 1)

=>

$\frac{567 \sqrt{2} + 576 }{64}$

Thanks !