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# If $a\cos x+\cot x+1=cosecx$ then possible values of x can be(a) $x=n\pi ,a\in R,n\in I$,(b) $x=\left( 4n+1 \right)\dfrac{\pi }{2},a\in R,n\in I$,(c) $x=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4\left( a+1 \right)}{{{a}^{2}}} \right),a\in \left( -\infty ,2-2\sqrt{2} \right)\cup \left[ 2+2\sqrt{2,}\infty \right]-\left\{ -2 \right\}$.(d) None of these

Last updated date: 20th Jun 2024
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Hint: When an equation is given in terms of Sine, Cosine, tangent, we must use any of the trigonometric identities to make the equation solvable. There are many inter-relations between Sine, Cosine, tan, cosecant and secant functions. These are inter-relations called as identities. Whenever you see conditions such that $\theta \in R$ , that means inequality is true for all angles. So, directly think of identity which will make your work easy. Use: - $\cot x=\dfrac{\cos x}{\sin x}$ , $cosecx=\dfrac{1}{\sin x}$.

An equality trigonometric function in them is called trigonometric equality. These are solved by some inter-relations known before-hand.
All the inter-relations which relate Sine, Cosine, tangent, Cotangent, Secant, Cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.
The equation in the question is in trigonometric terms: $a\cos x+\cot x+1=cosecx$.
By general knowledge of trigonometry, we know these identities: $\cot x=\dfrac{\cos x}{\sin x}$ , $cosecx=\dfrac{1}{\sin x}$.
By substituting these into our original equation, we turn it into:
$\Rightarrow a\cos x+\dfrac{\cos x}{\sin x}+1=\dfrac{1}{\sin x}$.
By multiplying $\sin x$ on both the sides of equation, we get it as:
$\Rightarrow a\sin x\cos x+\cos x+\sin x=1$.
By subtracting $a\sin x\cos x$ on both sides and squaring on both sides, we get: $\Rightarrow {{\left( \sin x+\cos x \right)}^{2}}={{\left( 1-a\sin x\cos x \right)}^{2}}$.
$\Rightarrow {{\sin }^{2}}x+{{\cos }^{2}}x+2\sin x\cos x=1+{{a}^{2}}{{\sin }^{2}}x{{\cos }^{2}}x-2a\sin x\cos x$.
By using identities ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ and simplifying, we get equation
$\Rightarrow {{a}^{2}}{{\sin }^{2}}x{{\cos }^{2}}x-2\left( a+1 \right)\sin x\cos x=0$.
We know $2\sin x\cos x=\sin 2x$ , by substituting this, we get it as:
$\Rightarrow \dfrac{{{a}^{2}}}{4}{{\sin }^{2}}2x-\left( a+1 \right)\sin 2x=0$.
By taking $\sin 2x$ common from left hand side of equation, we get:
$\Rightarrow \sin 2x\left( \dfrac{{{a}^{2}}}{4}\left( \sin 2x \right)-\left( a+1 \right) \right)=0$.
From above we can say roots of the equation are:
$\sin 2x=0$ ……………………………………….(i)
$\dfrac{{{a}^{2}}}{4}\left( \sin 2x \right)-\left( a+1 \right)=0$.
By adding $\left( a+1 \right)$ and multiply $\dfrac{4}{{{a}^{2}}}$ on both sides, we get
$\sin 2x=\dfrac{4\left( a+1 \right)}{{{a}^{2}}}$…………………………………….(ii)
From equation (i), we say roots $x=\dfrac{n\pi }{2}\text{,n}\in \text{I}$.
From equation (ii), we say roots $x=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4\left( a+1 \right)}{{{a}^{2}}} \right)$, we get them by applying ${{\sin }^{-1}}$ on both sides for (i) and (ii) as $\dfrac{4\left( a+1 \right)}{{{a}^{2}}}=\sin \left( something \right)$ it must be $-1\le \dfrac{4\left( a+1 \right)}{{{a}^{2}}}\le 1$.
Now, let us consider $-1\le \dfrac{4\left( a+1 \right)}{{{a}^{2}}}$.
$\Rightarrow -{{a}^{2}}\le 4a+4$.
$\Rightarrow {{a}^{2}}+4a+4\ge 0$.
$\Rightarrow \left( a+2 \right)\left( a+2 \right)\ge 0$.
We know that if $\left( x-a \right)\left( x-b \right)\ge 0$ and $a < b$, then $x\in \left( -\infty ,a \right]\cup \left[ b,\infty \right)$.
$\Rightarrow a\in \left( -\infty ,-2 \right]\cup \left[ -2,\infty \right)$.
Since we have both values as equal, we get $a\in \left( -\infty ,\infty \right)$ ---(iii).
Now, let us consider $\dfrac{4\left( a+1 \right)}{{{a}^{2}}}\le 1$.
$\Rightarrow {{a}^{2}}\ge 4a+4$.
$\Rightarrow {{a}^{2}}-4a-4\ge 0$.
$\Rightarrow {{a}^{2}}-\left( 2+2\sqrt{2} \right)a-\left( 2-2\sqrt{2} \right)a-4\ge 0$.
$\Rightarrow a\left( a-\left( 2+2\sqrt{2} \right) \right)-\left( 2-2\sqrt{2} \right)\left( a-\left( 2+2\sqrt{2} \right) \right)\ge 0$.
$\Rightarrow \left( a-\left( 2-2\sqrt{2} \right) \right)\left( a-\left( 2+2\sqrt{2} \right) \right)\ge 0$.
We know that if $\left( x-a \right)\left( x-b \right)\ge 0$ and $a < b$, then $x\in \left( -\infty ,a \right]\cup \left[ b,\infty \right)$.
$\Rightarrow a\in \left( -\infty ,2-2\sqrt{2} \right]\cup \left[ 2+2\sqrt{2},\infty \right)$ ---(iv).
From equation (iii) and (iv), we get the common interval as $a\in \left( -\infty ,2-2\sqrt{2} \right)\cup \left( 2+2\sqrt{2},\infty \right)$….(v).
We know that $\cot x$ and $\operatorname{cosec}x$ are not defined at $x=0$ and $x=n\pi$.
So, at $x=0$, we have $\dfrac{4\left( a+1 \right)}{{{a}^{2}}}=0$.
$\Rightarrow a+1=0$.
$\therefore a=-1$ ---(vi).
From equation (v) and (vi), we get $a\in \left( -\infty ,2-2\sqrt{2} \right)\cup \left( 2+2\sqrt{2},\infty \right)-\left\{ -1 \right\}$.
Therefore, option (c) is correct.

Note:
We should not just choose option (b), as the solution $x=\dfrac{n\pi }{2}$ is not satisfied by the option (b). We should make calculation mistakes while solving these problems. We should keep in mind that the range of $\sin x$ lies between –1 and +1 (including both). We should not say both the infinities we get in$\cot 0=\infty$ and $cosec0=\infty$ are equal as this is the wrong explanation.