JEE Trigonometry

Let $y=k \sin^2 x$ :

$y+\dfrac{1}{y}=2$

$y^2-2y+1=0 \Rightarrow (y-1)^2=0 \Rightarrow y=1$

Hence, $k \sin^2 x=1$ . Now, obtain $\sin x$ and $\cos x$ :

$\sin x = \dfrac{1}{\sqrt{k}}$

$\cos x = \sqrt{1-\sin^2 x}=\dfrac{\sqrt{k-1}}{\sqrt{k}}$

And substitute them in the given expression:

$\left(\dfrac{\sqrt{k-1}}{\sqrt{k}}\right)^2+5\left(\dfrac{1}{\sqrt{k}}\right)\left(\dfrac{\sqrt{k-1}}{\sqrt{k}}\right)+6\left(\dfrac{1}{\sqrt{k}}\right)^2$

$\dfrac{k+5+5\sqrt{k-1}}{k}$

So the answer is None of these .

Whenever $y+\frac{1}{y}$ =1 remember that $y=1$ is the only case. We get $\sin x=\frac{1}{k}$ Now since there is $\sin x•\cos x$ term in the required expression, it can be deduced that the final expression will contain a term of the form $\sqrt {f(k)}$ This form is absent in options. So none of these should be the answer.

good question as many like me would have solved the question but would have kept scratching their head , thinking -"how can it be NONE OF THESE"

from given condition arrive at sin^2x=1/k and substitute in the question.