UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use trigonometric identities to simplify the given equation and solve for $\sin 2x$.Explanation:Given: $\sin x + \cos x + \tan x + \cot x + \sec x + \csc x = 7$.Convert all terms to $\sin x$ and $\cos x$:$\tan x = \frac{\sin x}{\cos x},\; \cot x = \frac{\cos x}{\sin x},\; \sec x = \frac{1}{\cos x},\; \csc x = \frac{1}{\sin x}$.Multiply the whole equation by $\sin x \cos x$ and simplify to obtain:$\sin x \cos x (\sin x + \cos x) + \sin x + \cos x + 1 = 7 \sin x \cos x$.Rearrange to isolate $\sin x + \cos x$:$\sin x + \cos x = \frac{7 \sin x \cos x - 1}{1 + \sin x \cos x}$.Since $\sin 2x = 2 \sin x \cos x$, we rewrite:$\sin x + \cos x = \frac{7 \sin 2x - 2}{2 + \sin 2x}$.Square both sides:$(\sin x + \cos x)^2 = 1 + \sin 2x = \left(\frac{7 \sin 2x - 2}{2 + \sin 2x}\right)^2$.Simplify to get a quadratic in $\sin 2x$:$\sin^2 2x - 44 \sin 2x + 36 = 0$.Solve using quadratic formula:$\sin 2x = \frac{44 \pm \sqrt{1936 - 144}}{2} = \frac{44 \pm 16\sqrt{7}}{2} = 22 \pm 8\sqrt{7}$.Since $\sin 2x$ lies in $[-1,1]$, we take $\sin 2x = 22 - 8\sqrt{7}$.Hence $a = 22$, $b = 8$, $c = 7$.Compute $a - b + 2c = 22 - 8 + 2(7) = 28$.Answer:$28$ (Option D).