For `for A=133^(@),2cos""A/2`is equal to

Options

A. `-sqrt(1+sinA)-sqrt(1-sinA)`
B. `-sqrt(1+sinA)+sqrt(1-sinA)`
C. `sqrt(1+sinA)-sqrt(1-sinA)`
D. `sqrt(1+sinA)+sqrt(1-sinA)`

Answer

C

Explanation

We have,
`sqrt(1+sinA)=sqrt((cos""(A)/(2)+sin""(A)/(2))^(2))`
`impliessqrt(1+sinA)=|cos""(1)/(2)+sin""(A)/(2)|`
`impliessqrt(1+sinA)=cos""(A)/(2)+sin""(A)/(2)" "[because(A)/(2)"is an acute angle"]`
`and, sqrt(1-sinA)=sqrt((cos""(A)/(2)-sin(A)/(2))^(2))`
`impliessqrt(1-sinA)=|cos ""(A)/(2)-sin""(1)/(2)|`
`impliessqrt(1-sinA)=(cos""(A)/(2)-sin""(A)/(2))" "[{:(,because=66(1)/(2)""^(@)),(,thereforecos""(A)/(2)ltsin""(A)/(2)):}]`
`impliessqrt(1-sinA)=sin""(A)/(2)cos""(A)/(2)`
`therefore2 cos""(A)/(2)=sqrt(1+sinA)-sqrt(1-sinA)`