Question

If 12 sinҨ -- 5 cosҨ = 0 , find the value of 3 cosҨ -- 5 sinҨ​

Answer 1

Step-by-step explanation:

We have,

12 × P (n - 1, 3) = 5 × P (n + 1, 3)

P (n - 1, 3): P(n + 1, 3) = 5 : 12

⇒\dfrac{(n-1)!}{(n-4)!} : [tex]\frac{(n + 1)!}{(n -2)!}

(n−4)!

(n−1)!

:[tex]

(n−2)!

(n+1)!

= 5 : 12

⇒ \dfrac{(n-1)!}{(n-4)!} \times [tex]\frac{(n - 2)!}{(n + 1)!}

(n−4)!

(n−1)!

×[tex]

(n+1)!

(n−2)!

= \frac{5}{12}

12

5

⇒ \frac{(n -2)(n - 3)}{n(n + 1)}

n(n+1)

(n−2)(n−3)

= \frac{5}{12}

12

5

⇒ 7·n^{2}n

2

- 65·n + 72 = 0

⇒ 7·n^{2}n

2

- 56·n - 9·n + 72 = 0

⇒ (n - 8)(7n - 9) =0

∴ n = 8 [∵ n is never fraction)

Hence, n = 8.