Question

If $3x=\cos ec\theta $ and $\dfrac{3}{x}=\cot \theta $, find the value of $3\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)$.

Answer 1

Hint: Assume the equations as equation (i) and equation (ii). Divide both the sides of each equation by 3. Now, square both the equations and take the difference: $\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)$. Use the trigonometric identity, $\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$ to simplify the expression. Finally, multiply both sides of the simplified form by 3 to get the answer.

Complete step-by-step answer:
In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined.

We have been provided with two equations:

$3x=\cos ec\theta .....................(i)$

And, $\dfrac{3}{x}=\cot \theta ..................(ii)$

Dividing both sides of each equation by 3, we get,

$x=\dfrac{\cos ec\theta }{3}$

And, $\dfrac{1}{x}=\dfrac{\cot \theta }{3}$

On squaring both the sides we get,

${{x}^{2}}=\dfrac{\cos e{{c}^{2}}\theta }{9}$

And, $\dfrac{1}{{{x}^{2}}}=\dfrac{{{\cot }^{2}}\theta }{9}$

Taking the difference: $\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)$, we get,

$\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)=\dfrac{\cos e{{c}^{2}}\theta }{9}-\dfrac{{{\cot }^{2}}\theta }{9}$

Taking L.C.M on both the sides, we get,

$\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)=\dfrac{\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta }{9}$

Now, using the trigonometric identity: $\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$, we get,
$\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)=\dfrac{1}{9}$

Now, multiplying both the sides of the equation by 3, we get,

$\begin{align}

  & 3\times \left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)=3\times \dfrac{1}{9} \\

 & \Rightarrow 3\left( {{x}^{2}}-\dfrac{1}{{{x}^{2}}} \right)=\dfrac{1}{3} \\

\end{align}$



Note: We can also solve this question by an alternate method. We have to substitute: $\cos ec\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. Now, divide both sides of each equation by 3 and take the difference of squares of these equations. Use the formula: $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ in the numerator to get the simplified expression. Finally, multiply both sides of the simplified form by 3 to get the answer.