Question

If $\sin A + \cos {\text{ec}}A = 3$ then find the value of $\dfrac{{{{\sin }^4}A + 1}}{{{{\sin }^2}A}}$.

Answer 1

Hint:
As we know the condition that $\sin A + \cos {\text{ec}}A = 3$ and we also know that $\sin A.\cos {\text{ec}}A = 1$
Hence $\cos {\text{ec}}A = \dfrac{1}{{\sin A}}$ and by putting this value we can get the value of $\sin A$ and put it in $\dfrac{{{{\sin }^4}A + 1}}{{{{\sin }^2}A}}$ and we will get the answer.

Complete step by step solution:
We are given that $\sin A + \cos {\text{ec}}A = 3$$ - - - - (1)$
Now as we need to find the value of $\dfrac{{{{\sin }^4}A + 1}}{{{{\sin }^2}A}}$ so we need to find the value of $\sin A$ which can be found the following condition which says that $\sin A + \cos {\text{ec}}A = 3$
And we know that$\sin A.\cos {\text{ec}}A = 1$. So we can say that $\cos {\text{ec}}A = \dfrac{1}{{\sin A}}$ and therefore now by replacing the value of $\cos {\text{ec}}A$in the equation (1), we will get
$\sin A + \dfrac{1}{{\sin A}} = 3$
Upon taking LCM we get that
$\dfrac{{{{\sin }^2}A + 1}}{{\sin A}} = 3$
So we get that ${\sin ^2}x + 1 = 3\sin A$
We can write it as
${\sin ^2}x - 3\sin A + 1 = 0$
Hence we get the quadratic equation of the form $a{x^2} + bx + c = 0$ which takes two roots which are given by the formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here we are given that
${\sin ^2}x - 3\sin A + 1 = 0$
So $\sin A = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4(1)(1)} }}{{2(1)}}$
$\Rightarrow \sin A = \dfrac{{ - ( - 3) \pm \sqrt {{{( - 3)}^2} - 4(1)(1)} }}{{2(1)}}$
$\Rightarrow \sin A = \dfrac{{3 \pm \sqrt {9 - 4} }}{2}$
$\Rightarrow \sin A = \dfrac{{3 \pm \sqrt 5 }}{2}$
So we get two values of $\sin A$
$\Rightarrow \sin A = \dfrac{{3 + \sqrt 5 }}{2},\sin A = \dfrac{{3 - \sqrt 5 }}{2}$
But we know that $0 \leqslant \sin A \leqslant 1$
But we know that $\dfrac{{3 + \sqrt 5 }}{2} > 1$
Hence we get that $\sin A = \dfrac{{3 - \sqrt 5 }}{2}$
So we need to find the value of $\dfrac{{{{\sin }^4}A + 1}}{{{{\sin }^2}A}}$
$\Rightarrow {\sin ^2}A + \dfrac{1}{{{{\sin }^2}A}}$
$\Rightarrow \sin A = \dfrac{{3 - \sqrt 5 }}{2}$
$\Rightarrow {\sin ^2}A = {\left( {\dfrac{{3 - \sqrt 5 }}{2}} \right)^2}$
Therefore upon squaring we get that
$\Rightarrow {\sin ^2}A = \left( {\dfrac{{{3^2} + {{(\sqrt 5 )}^2} - 2.3.\sqrt 5 }}{{{2^2}}}} \right)$
$\Rightarrow {\sin ^2}A = \left( {\dfrac{{9 + 5 - 6\sqrt 5 }}{4}} \right)$
$\Rightarrow {\sin ^2}A = \left( {\dfrac{{14 - 6\sqrt 5 }}{4}} \right) = \left( {\dfrac{{7 - 3\sqrt 5 }}{2}} \right)$
Now for $\dfrac{1}{{{{\sin }^2}A}}$$ = \dfrac{2}{{7 - 3\sqrt 5 }} \times \dfrac{{7 + 3\sqrt 5 }}{{7 + 3\sqrt 5 }}$
On rationalising we get that
$\dfrac{1}{{{{\sin }^2}A}}$$ = \dfrac{{2(7 + 3\sqrt 5 )}}{{49 - 45}} = \dfrac{{2(7 + 3\sqrt 5 )}}{4} = \dfrac{{(7 + 3\sqrt 5 )}}{2}$
Now we need to find the value of ${\sin ^2}A + \dfrac{1}{{{{\sin }^2}A}}$
So as we have got ${\sin ^2}A + \dfrac{1}{{{{\sin }^2}A}}$$ = \left( {\dfrac{{7 - 3\sqrt 5 }}{2}} \right) + \left( {\dfrac{{7 + 3\sqrt 5 }}{2}} \right)$
$ = \left( {\dfrac{{7 - 3\sqrt 5 + 7 + 3\sqrt 5 }}{2}} \right)$

$ = \left( {\dfrac{{14}}{2}} \right) = 7$

Note:
We can also solve it in this way that as we know
$\dfrac{{{{\sin }^4}A + 1}}{{{{\sin }^2}A}}$$ = {\sin ^2}A + \dfrac{1}{{{{\sin }^2}A}} = {\sin ^2}A + {\text{cose}}{{\text{c}}^2}A$
And we are given that
$\sin A + \cos {\text{ec}}A = 3$
Upon squaring both sides we get that
$
\Rightarrow {\sin ^2}A + {\text{cose}}{{\text{c}}^2}A + 2\sin A.\cos {\text{ec}}A = 9 \\
\Rightarrow {\sin ^2}A + {\text{cose}}{{\text{c}}^2}A + 2 = 9 \\
\Rightarrow {\sin ^2}A + {\text{cose}}{{\text{c}}^2}A = 7 \\
 $