Question

For the equation $1 - 2x - {x^2} = ta{n^2}(x + y) + co{t^2}(x + y)$
A.Exactly one of $x$ exists
B.Exactly two value of $x$ exists
C.$y = - 1 + n\pi \pm \dfrac{\pi }{4}$
D.$y = 1 + n\pi \pm \dfrac{\pi }{4}$

Answer 1

Hint : In order to determine the value of equation, we need to expand the equation $1 - 2x - {x^2}$ and ${\tan ^2}(x + y) + {\cot ^2}(x + y)$. From the equation we get $x$ and $y$ value by substitution method and trigonometry rules. We use reciprocal trigonometric function: $\tan x = \dfrac{1}{{\cot x}}$ and $\cot x = \dfrac{1}{{\tan x}}$.
We need to know the function, $\tan \dfrac{\pi }{4} = 1$ and We must know about the trigonometry definitions, identities and formulas which are involving the trigonometry ratios, while solving the trigonometry based questions and also we have to compare the equation with the algebraic formulas are, ${a^2} + {b^2} = {(a - b)^2} + 2ab$, ${(a + b)^2} = {a^2} + 2ab + {b^2}$. Finally, from this formula, ${\tan ^2}\theta = 1 \Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}$ is used for finding the required solution.

Complete step-by-step answer :
In mathematics, a series of simultaneous equations, also known as a system of equations or an equation system, is a finite set of equations for which common solutions are found by using the substitution method.
We are given the trigonometric function,
$\Rightarrow 1 - 2x - {x^2} = {\tan ^2}(x + y) + {\cot ^2}(x + y)$
On comparing the algebraic formula ${a^2} + {b^2} = {(a - b)^2} + 2ab$ with RHS of above equation, then
$\Rightarrow 1 - 2x - {x^2} = {(\tan (x + y) - \cot (x + y))^2} + 2\tan (x + y)\cot (x + y)$
Now, We use reciprocal trigonometric function: $\tan x = \dfrac{1}{{\cot x}}$ and $\cot x = \dfrac{1}{{\tan x}}$.
Since $x = x + y$.
$\Rightarrow 1 - 2x - {x^2} = {(\tan (x + y) - \cot (x + y))^2} + 2\tan (x + y)\dfrac{1}{{\tan (x + y)}}$
$\Rightarrow 1 - 2x - {x^2} = {(\tan (x + y) - \cot (x + y))^2} + 2$
By expanding $2$ from RHS to LHS, the positive sign can be changed as negative, we get
$\Rightarrow 1 - 2x - {x^2} - 2 = {(\tan (x + y) - \cot (x + y))^2}$
On further simplification, we get
$\Rightarrow - 1 - 2x - {x^2} = {(\tan (x + y) - \cot (x + y))^2}$
Taking out the negative sign from the algebraic expression, then
$\Rightarrow - ({x^2} + 2x + 1) = {(\tan (x + y) - \cot (x + y))^2}$
Here, we have to compare the formula, ${(a + b)^2} = {a^2} + 2ab + {b^2}$ with $({x^2} + 2x + 1)$.
Since, $a = x,b = 1$
So, we can write the LHS as follows
$\Rightarrow - {(x + 1)^2} = {(\tan (x + y) - \cot (x + y))^2}$
On cancelling the like terms square on both sides , then we get
$\Rightarrow - (x + 1) = \tan (x + y) - \cot (x + y)$
By solving the LHS and RHS of the above equation, we can get the value of $x$, we get
LHS:

$$x + 1 = 0 \\\ x = - 1 \;$$

RHS:
$\tan (x + y) - \cot (x + y) = 0$
Expanding the term $\cot (x + y)$ from LHS to RHS, then
$\tan (x + y) = \cot (x + y)$
By using a Reciprocal trigonometric function, $\cot x = \dfrac{1}{{\tan x}}$ with above function
$\tan (x + y) = \dfrac{1}{{\tan (x + y)}}$
${\tan ^2}(x + y) = 1$
Substitute the value of $x = - 1$ and $\tan \dfrac{\pi }{4} = 1$, then
${\tan ^2}( - 1 + y) = \tan \dfrac{\pi }{4}$
We know that, ${\tan ^2}\theta = 1 \Rightarrow \theta = n\pi \pm \dfrac{\pi }{4}$
$\Rightarrow - 1 + y = n\pi \pm \dfrac{\pi }{4}$, Since, $x = - 1$
Therefore, $y = 1 + n\pi \pm \dfrac{\pi }{4}$
Hence, the option (A) Exactly one of $x$ exists and Option (D) $y = 1 + n\pi \pm \dfrac{\pi }{4}$
is the correct answer.
So, the correct answer is “OptionA and D”.

Note : Trigonometric ratios: Some ratios of the sides of a right-angle triangle with respect to its acute angle called trigonometric ratios of the angle.
The ratios defined are abbreviated as sin A, cos A, tan A, csc A or cosec A, sec A and cot A
These functions are defined as the reciprocal of the standard trigonometric functions: sine, cosine, and tangent, and hence they are called the reciprocal trigonometric functions.