Question

If $\sin x + \cos x + \tan x + \cot x + \sec x + \cos ecx = 7$ and $\sin 2x = a - b\sqrt 7$ then $a - b + 14$ is divisible by
A. 5
B. 3
C. 0
D. 7

Answer 1

Hint: Simplify the first equation by grouping two terms in order using trigonometric relations. See if we can reduce the first equation in the form of the second equation to then compare the results to find values of $a$ and $b$.

Complete step-by-step answer:
We will try to simplify the first equation and try to bring in a form similar to the second equation. Let us consider first equation from the question:
$\sin x + \cos x + \tan x + \cot x + \sec x + \ co sec x = 7$
We write all the terms in sine and cosine expressions. So we will move ahead as follows:
$\Rightarrow (\sin x + \cos x) + \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{\sin x}} + \dfrac{1}{{\cos x}} + \dfrac{1}{{\sin x}} = 7$
We will club together two terms in order and see if we get terms in common.
$\Rightarrow (\sin x + \cos x) + \dfrac{{({{\sin }^2}x + {{\cos }^2}x)}}{{\sin x\cos x}} + \dfrac{{(\sin x + \cos x)}}{{\sin x\cos x}} = 7$
But we know that, ${\cos ^2}\theta + {\sin ^2}\theta = 1$, we get,
$\Rightarrow (\sin x + \cos x) + \dfrac{2}{{2\sin x\cos x}} + \dfrac{{2(\sin x + \cos x)}}{{2\sin x\cos x}} = 7 \\\ \Rightarrow (\sin x + \cos x)\left( {1 + \dfrac{2}{{\sin 2x}}} \right) = 7 - \dfrac{2}{{\sin 2x}} \\\$
In the last step we used the identity, $\sin 2\theta = 2\sin \theta \cos \theta$. Now squaring on both sides, we get,
$\Rightarrow ({\sin ^2}x + {\cos ^2}x + 2\sin x\cos x){\left( {1 + \dfrac{2}{{\sin 2x}}} \right)^2} = 49 + \dfrac{4}{{{{\sin }^2}2x}} - \dfrac{{28}}{{\sin 2x}} \\\ \Rightarrow (1 + \sin 2x){\left( {1 + \dfrac{2}{{\sin 2x}}} \right)^2} = \dfrac{{49{{\sin }^2}2x + 4 - 28\sin 2x}}{{{{\sin }^2}2x}} \\\$
Consider $\sin 2x = t$, we will get,
$\Rightarrow \left( {1 + t} \right){(t + 2)^2} = 49{t^2} + 4 - 28t \\\ \Rightarrow (1 + t)({t^2} + 4t + 4) = 49{t^2} + 4 - 28t \\\ \Rightarrow {t^2} + 4t + 4 + {t^3} + 4{t^2} + 4t = 49{t^2} + 4 - 28t \\\ \Rightarrow {t^3} - 44{t^2} + 36t = 0 \\\ \Rightarrow {t^2} - 44t + 36 = 0 \\\$
This is a quadratic equation whose roots are:
$t = \sin 2x = \dfrac{{ - ( - 44) \pm \sqrt {{{44}^2} - 4 \times 36} }}{2} \\\ \sin 2x = 22 \pm \sqrt {\dfrac{{(44 - 12)(44 + 12)}}{4}} \\\ \sin 2x = 22 \pm \sqrt {32 \times 14} = 22 \pm 8\sqrt 7 \\\$
After comparing it with the second equation of the problem, $\sin 2x = a - b\sqrt 7$, we get,
$a = 22;b = 8$.
$a - b + 14 = 22 - 8 + 14 = 28$ which is a multiple of 7. Thus option D is correct.

Note: Be careful with the algebra while performing simplification of the trigonometric equations. Often there are chances of making mistakes with multiples and sign changes. Pairing of terms in the first question was because most trigonometric identities have two terms involved.