Question

can u answer this question ?

Answer 1

Answer:

$126 * sin^{5}2\alpha$

Step-by-step explanation:

*PLEASE IGNORE THESE A cap(s) BELOW*

Let the term independent of x be

$T_{r+1}  = 10C_{r}  * (x sin\alpha)^{10-r}  * (\frac{cos\alpha }{x})^{r}\\=10C_{r} * (x)^{10-r} * (sin\alpha)^{10-r} * (cos\alpha)^{r} * (x)^{-r}$

For a term to be independent of x, power of terms containing x should be 0 i.e,

$10 - r - r = 0\\=> r = \frac{10}{2} \\=> r = 5$

Therefore term independent of x,

$T_{5+1}  = 10C_{5}  * (x sin\alpha)^{10-5}  * (\frac{cos\alpha }{x})^{5}\\=10C_{5} * (x)^{10-5} * (sin\alpha)^{10-5} * (cos\alpha)^{5} * (x)^{-5}\\=10C_{5} * (x)^{5} * (sin\alpha)^{5} * (cos\alpha)^{5} * (x)^{-5}\\=252 * (sin\alpha * cos\alpha)^{5}\\=252 * \frac{1}{2} (2sin\alpha * cos\alpha)^{5}\\=126 * (sin2\alpha)^{5}\\=126 * sin^{5}2\alpha$

Answer 2

Answer:

$126 * sin^{5}2\alpha$

Step-by-step explanation:

By using Binomial Theorem. Equate the powers of x to zero, you will get r, and then substitute it.

Formula to be used is:

$T_{r+1} = 10C_{r}  * (x sin\alpha)^{10-r}  * (\frac{cos\alpha }{x})^{r}$

P.S: ignore the A cap