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$$\left| {\cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\}} \right| \leqslant K$$, then the value of K is
A. $\sqrt {1 + {{\cos }^2}\alpha } $
B. $\sqrt {1 + {{\sin }^2}\alpha } $
C. $\sqrt {2 + {{\sin }^2}\alpha } $
D. $\sqrt {2 + {{\cos }^2}\alpha } $

Answer Verified Verified
Hint: We need to find the maximum of the value. For that first we are going to solve$$\cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\} $$and then we can determine using the discriminant by forming a quadratic equation. On solving the value of${b^2} - 4ac \geqslant 0$, we will be able to find the K value in$$\left| {\cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\}} \right| \leqslant K$$.

Complete step-by-step answer:
Given $$\left| {\cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\}} \right| \leqslant K $$
A modulus function is a function which gives the absolute value of a number or variable. The plotting of such graphs is also an easy method where the domain will be all values of input say x (all real numbers) and range will be values of function (y= f(x) = all positive real numbers and 0).
We need to find the maximum of the above given equation.
Let u=$$\cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\} $$
We say that a function f(x) has a relative maximum value at x = a, if f(a) is greater than any value immediately preceding or following. We call it a “relative” maximum because other values of the function may in fact be greater.
$ \Rightarrow u = \cos \theta sin\theta + \cos \theta \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } $
$ \Rightarrow u - \cos \theta sin\theta = \cos \theta \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } $
Squaring on both sides
$ \Rightarrow {\left( {u - \cos \theta sin\theta } \right)^2} = \left( {\cos \theta \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right){ ^2}$
$ \Rightarrow {\left( {u - \cos \theta sin\theta } \right)^2} = {\cos ^2}\theta \left( {{{\sin }^2}\theta + {{\sin }^2}\alpha } \right)$
$ \Rightarrow {u^2} + {\sin ^2}\theta {\cos ^2}\theta - 2u\sin \theta \cos \theta = {\cos ^2}\theta si{n^2}\theta + {\cos ^2}\theta {\sin ^2}\alpha $
$ \Rightarrow {u^2} - 2u\sin \theta \cos \theta = {\cos ^2}\theta {\sin ^2}\alpha $
Divide the entire equation on both sides with${\cos ^2}\theta $
$ \Rightarrow \dfrac{{{u^2}}}{{{{\cos }^2}\theta }} - \dfrac{{2u\sin \theta \cos \theta }}{{{{\cos }^2}\theta }} = \dfrac{{{{\cos }^2}\theta {{\sin }^2}\alpha }}{{{{\cos }^2}\theta }}$
$ \Rightarrow \dfrac{{{u^2}}}{{{{\cos }^2}\theta }} - \dfrac{{2u\sin \theta }}{{\cos \theta }} = si{n^2}\alpha $
$$ \Rightarrow {u^2}{\sec ^2}\theta - 2u\tan \theta = si{n^2}\alpha $$
We know that${\sec ^2}\theta = 1 + {\tan ^2}\theta $
$$ \Rightarrow {u^2}\left( {1 + {{\tan }^2}\theta } \right) - 2u\tan \theta = si{n^2}\alpha $$
$$ \Rightarrow {u^2} + {u^2}{\tan ^2}\theta - 2u\tan \theta - si{n^2}\alpha = 0$$
We can rewrite this equation as
$$ \Rightarrow {u^2}{\tan ^2}\theta - 2u\tan \theta + {u^2} - si{n^2}\alpha = 0$$
Here $$\tan \theta $$ is real
Considering the above one as equation and comparing it with$a{x^2} + bx + c = 0$
Therefore, a= $${u^2}$$
And b= $$ - 2u$$
And c= $${u^2} - si{n^2}\alpha $$
Quadratic formula: x=$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. When the Discriminant ${b^2} - 4ac$is: positive, there are 2 real solutions.
We know that $D \geqslant 0$
${b^2} - 4ac \geqslant 0$
$$ \Rightarrow {\left( { - 2u} \right)^2} - 4 \times {u^2} \times \left( {{u^2} - si{n^2}\alpha } \right) \geqslant 0 $$
$$ \Rightarrow 4{u^2} - 4{u^2}\left( {{u^2} - si{n^2}\alpha } \right) \geqslant 0$$
$$ \Rightarrow 4{u^2}\left( {1 - {u^2} + si{n^2}\alpha } \right) \geqslant 0$$
$$ \Rightarrow 1 - {u^2} + si{n^2}\alpha \geqslant 0$$
$$ \Rightarrow {u^2} \leqslant 1 + si{n^2}\alpha $$

Therefore, $$\left| {\cos \theta \left\{ {\sin \theta + \sqrt {{{\sin }^2}\theta + {{\sin }^2}\alpha } } \right\}} \right| \leqslant \sqrt {1 + si{n^2}\alpha } $$

Note: We can solve the above problem using Cauchy Schwarz Inequality. It is also Cauchy-Bunyakovsky-Schwarz inequality, a useful inequality encountered in many different settings, such as linear algebra, analysis, probability theory, vector algebra and other areas. Therefore, on using this inequality, we get$({\sin ^2}x + {\cos ^2}x).\left[ {{{\cos }^2}x + {{\sin }^2}x + {{\sin }^2}\alpha } \right]$$ \geqslant {\left( {\sin x.\cos x + \cos x.\sqrt {{{\sin }^2}x + {{\sin }^2}\alpha } } \right)^2}$. So we get${y^2} \leqslant \left( {1 + {{\sin }^2}\alpha } \right) \Rightarrow \left| y \right| \leqslant \sqrt {1 + {{\sin }^2}\alpha } $.