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# Let x and y be 2 real numbers which satisfy the equations $\left( {{\tan }^{2}}x-{{\sec }^{2}}y \right)=\dfrac{5a}{6}-3$ and $\left( -{{\sec }^{2}}x+{{\tan }^{2}}y \right)={{a}^{2}}$, then the value of a can be equal to:(a) $\dfrac{2}{3}$(b) $\dfrac{-2}{3}$(c) $\dfrac{3}{2}$(d) $\dfrac{-3}{2}$

Last updated date: 05th Aug 2024
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Hint: We start solving the problem by adding the given two trigonometric equations and then using the trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$. We then make the necessary arrangements to get a quadratic equation in ‘a’. We then factorize the obtained quadratic equation and then equate the factors to zero to find the possible values of ‘a’.

According to the problem, we are given that x and y be 2 real numbers which satisfy the equations $\left( {{\tan }^{2}}x-{{\sec }^{2}}y \right)=\dfrac{5a}{6}-3$ and $\left( -{{\sec }^{2}}x+{{\tan }^{2}}y \right)={{a}^{2}}$. We need to find the value of a.
Let us add the given equations $\left( {{\tan }^{2}}x-{{\sec }^{2}}y \right)=\dfrac{5a}{6}-3$ and $\left( -{{\sec }^{2}}x+{{\tan }^{2}}y \right)={{a}^{2}}$.
So, we have $\left( {{\tan }^{2}}x-{{\sec }^{2}}y \right)+\left( -{{\sec }^{2}}x+{{\tan }^{2}}y \right)=\dfrac{5a}{6}-3+{{a}^{2}}$.
$\Rightarrow \left( {{\tan }^{2}}x-{{\sec }^{2}}x \right)+\left( {{\tan }^{2}}y-{{\sec }^{2}}y \right)=\dfrac{5a}{6}-3+{{a}^{2}}$.
$\Rightarrow -\left( {{\sec }^{2}}x-{{\tan }^{2}}x \right)-\left( {{\sec }^{2}}y-{{\tan }^{2}}y \right)=\dfrac{5a}{6}-3+{{a}^{2}}$ ---(1)
From the trigonometric identities we have ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$. Let us substitute this identity in equation (1).
$\Rightarrow -1-1=\dfrac{5a}{6}-3+{{a}^{2}}$.
$\Rightarrow -2=\dfrac{5a}{6}-3+{{a}^{2}}$.
$\Rightarrow {{a}^{2}}+\dfrac{5a}{6}-1=0$.
$\Rightarrow \dfrac{6{{a}^{2}}+5a-6}{6}=0$.
$\Rightarrow 6{{a}^{2}}+5a-6=0$.
Now, let us factorize this quadratic equation to find the value(s) of ‘a’.
$\Rightarrow 6{{a}^{2}}+9a-4a-6=0$.
$\Rightarrow \left( 3a-2 \right)\left( 2a+3 \right)=0$.
$\Rightarrow 3a-2=0$ or $2a+3=0$.
$\Rightarrow 3a=2$ or $2a=-3$.
$\Rightarrow a=\dfrac{2}{3}$ or $a=\dfrac{-3}{2}$.
So, the possible values of ‘a’ are $\dfrac{2}{3}$ or $\dfrac{-3}{2}$.

So, the correct answer is “Option a and d”.

Note: Whenever we get this type of problems, we try to make use of the trigonometric identities which reduces our calculation time and avoids confusion. We can also find the roots of the quadratic equation $6{{a}^{2}}+5a-6=0$ by using the fact that the roots of the quadratic equation $p{{x}^{2}}+qx+r=0$ is $\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$. We should not confuse signs with trigonometric identities while solving this type of problem.