If `theta=3alphaa n dsintheta=a/(sqrt(a^2+b^2)),` the value of the expression `acos e calpha-bsecalpha` is `a/(sqrt(a^2+b^2))` (b) `2sqrt(a^2+b^2)` (c) `a+b` (d) none of these

Options

A. `(a)/(sqrt(a^(2)+b^(2))`
B. `2sqrt(a^(2)+b^(2))`
C. a+b
D. none of these.

Answer

B

Explanation

`(sqrt(a^(2)+b^(2)))/(sin alpha cos alpha)[(a)/(sqrt(a^(2)+b^(2)))cosalpha-(b)/sqrt(a^(2)+b^(2))sinalpha]``(sqrt(a^(2)+b^(2)))/(sin alpha cos alpha)[(a)/(sqrt(a^(2)+b^(2)))cosalpha-(b)/sqrt(a^(2)+b^(2))sinalpha]`` (sqrt(a^(2)+b^(2)))/(sin alpha cos alpha)[(a)/(sqrt(a^(2)+b^(2)))cosalpha-(b)/sqrt(a^(2)+b^(2))sinalpha]`
Now, `sin3 alpha=(a)/sqrt(a^(2)+b^(2))` gives
`sqrt(a^(2)+b^(2))` `[(sin3alphacos alpha-cos 3alpha sin alpha)/(sin alpha cos alpha)]=2sqrt(a^(2)+b^(2))`