Question

Within what respective limits must $\dfrac{A}{2}$ lies when
1). $2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A}$
2). $2\sin \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A}$
3). $2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A}$ and
4). $2\cos \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A}$

Answer 1

To solve this question we have to convert $\sin A$ into the half-angle formula and put the values in the options and simplify all those. If the right-hand side is equal to the left-hand side then that option is the correct answer. Use the trigonometry property in the place of $1$ then try to make it a perfect square and take them outside of the root.

Complete step-by-step solution:
Given,
Few options are given
$2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A}$
$2\sin \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A}$
$2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A}$ and
$2\cos \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A}$
To find,
which option is correct;
Here, all these options take a look to the right hand side then we observe that only $\sqrt {1 + \sin A}$ and $\sqrt {1 - \sin A}$ are there. All other things are the different mathematical operators are used at different operators.
So, applying formula on only $\sqrt {1 + \sin A}$ and $\sqrt {1 - \sin A}$
First we solve,
$\sqrt {1 + \sin A}$
Let this be $a$
$a = \sqrt {1 + \sin A}$
On putting the identity $1 = {\sin ^2}\dfrac{A}{2} + {\cos ^2}\dfrac{A}{2}$ and half angle of $\sin A$
$a = \sqrt {{{\sin }^2}\dfrac{A}{2} + {{\cos }^2}\dfrac{A}{2} + 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}$ ($\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$)
On making the terms in the form of ${(a + b)^2}$
$a = \sqrt {{{(\sin \dfrac{A}{2} + \cos \dfrac{A}{2})}^2}}$
Taking the terms outside of bracket
$a = \sin \dfrac{A}{2} + \cos \dfrac{A}{2}$
Again put the value of $a$
$\sqrt {1 + \sin A} = \sin \dfrac{A}{2} + \cos \dfrac{A}{2}$ …………………………(i)
Now we solve
$\sqrt {1 - \sin A}$
Let, this be $b$
$b = \sqrt {1 - \sin A}$
On putting the identity $1 = {\sin ^2}\dfrac{A}{2} + {\cos ^2}\dfrac{A}{2}$ and half angle of $\sin A$
$b = \sqrt {{{\sin }^2}\dfrac{A}{2} + {{\cos }^2}\dfrac{A}{2} - 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}$ ($\sin A = 2\sin \dfrac{A}{2}\cos \dfrac{A}{2}$)
On making the terms in the form of ${(a - b)^2}$
$b = \sqrt {{{(\sin \dfrac{A}{2} - \cos \dfrac{A}{2})}^2}}$
Taking the terms outside of bracket
$b = \sin \dfrac{A}{2} - \cos \dfrac{A}{2}$
Again put the value of $a$
$\sqrt {1 - \sin A} = \sin \dfrac{A}{2} - \cos \dfrac{A}{2}$ ……………………(ii)
Option 1.
$2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} + \sqrt {1 - \sin A}$
Putting the values form equation (i) and (ii)
$2\sin \dfrac{A}{2} = \sin \dfrac{A}{2} + \cos \dfrac{A}{2} + \sin \dfrac{A}{2} - \cos \dfrac{A}{2}$
On further solving
$2\sin \dfrac{A}{2} = 2\sin \dfrac{A}{2}$
Option 1 is the correct answer
Option 2.
$2\sin \dfrac{A}{2} = - \sqrt {1 + \sin A} + \sqrt {1 - \sin A}$
Putting the values form equation (i) and (ii)
$2\sin \dfrac{A}{2} = - (\sin \dfrac{A}{2} + \cos \dfrac{A}{2}) + \sin \dfrac{A}{2} - \cos \dfrac{A}{2}$
On further solving
$2\sin \dfrac{A}{2} = - 2\cos \dfrac{A}{2}$
Option 2 is not the correct answer
Option 3.
$2\sin \dfrac{A}{2} = \sqrt {1 + \sin A} - \sqrt {1 - \sin A}$
Putting the values form equation (i) and (ii)
$2\sin \dfrac{A}{2} = \sin \dfrac{A}{2} + \cos \dfrac{A}{2} - (\sin \dfrac{A}{2} - \cos \dfrac{A}{2})$
On further solving
$2\sin \dfrac{A}{2} = 2\cos \dfrac{A}{2}$
Option 3 is not the correct answer but the answer is matched with option 4 so option 4 is also the correct answer
Final answer:
Option 1 and option 4 both are correct.

Note: To solve these types of questions we have to use the property of trigonometry and must know all the identity and formulas of trigonometry and have a good practice to use all the formulas. In this particular question, we use two formulas of trigonometry. Make the term inside the root a perfect square.