Question

Prove the equation given below,
$\cot \theta +\cot \left( 60+\theta \right)-\cot \left( 60-\theta \right)=3\cot 3\theta$

Answer 1

Hint: First convert ‘cot’ in the form of ‘sin’ and ‘cos’ and then use the formulae of sin (A + B), sin (A – B), cos (A + B), cos (A – B), then put the values $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\cos 60{}^\circ =\dfrac{1}{2}$ and then simplify the equation. Then use the formula $\cot 3\theta =\dfrac{{{\cot }^{3}}\theta -3\cot \theta }{3{{\cot }^{2}}\theta -1}$ to get the final answer.

Complete step-by-step answer:

To solve the given question we will write down the given equation first, therefore,
$\cot \theta +\cot \left( 60+\theta \right)-\cot \left( 60-\theta \right)=3\cot 3\theta$
Now consider the left hand side of the equation, therefore we will get,
Left Hand Side (LHS) = $\cot \theta +\cot \left( 60+\theta \right)-\cot \left( 60-\theta \right)$ …………………………………………… (1)
As we know $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ therefore we can replace $\cot \left( 60+\theta \right)$ by $\dfrac{\cos \left( 60+\theta \right)}{\sin \left( 60+\theta \right)}$ and $\cot \left( 60-\theta \right)$ by $\dfrac{\cos \left( 60-\theta \right)}{\sin \left( 60-\theta \right)}$ in the above equation therefore we will get,
Therefore, Left Hand Side (LHS) = $\cot \theta +\dfrac{\cos \left( 60+\theta \right)}{\sin \left( 60+\theta \right)}-\dfrac{\cos \left( 60-\theta \right)}{\sin \left( 60-\theta \right)}$
Now to proceed further in the solution we should know the formulae given below,
Formulae:

& \cos \left( A+B \right)=\cos A\cos B-\sin A\sin B \\\ & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B \\\ & \sin \left( A+B \right)=\sin A\cos B+\cos A\sin B \\\ & \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\\ \end{aligned}$$ If we use the above four formulae in ‘LHS’ we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\cos 60{}^\circ \cos \theta -\sin 60{}^\circ \sin \theta }{\sin 60{}^\circ \cos \theta +\cos 60{}^\circ \sin \theta }-\dfrac{\cos 60{}^\circ \cos \theta +\sin 60{}^\circ \sin \theta }{\sin 60{}^\circ \cos \theta -\cos 60{}^\circ \sin \theta }$$ To proceed further in the solution we should know the values of trigonometric functions given below, Formulae: $$\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$$ $$\cos 60{}^\circ =\dfrac{1}{2}$$ If we put the above values in ‘LHS’ we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\dfrac{1}{2}\cos \theta -\dfrac{\sqrt{3}}{2}\sin \theta }{\dfrac{\sqrt{3}}{2}\cos \theta +\dfrac{1}{2}\sin \theta }-\dfrac{\dfrac{1}{2}\cos \theta +\dfrac{\sqrt{3}}{2}\sin \theta }{\dfrac{\sqrt{3}}{2}\cos \theta -\dfrac{1}{2}\sin \theta }$$ As the denominator is same in all terms therefore above equation can also be simplified as, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\dfrac{\cos \theta -\sqrt{3}\sin \theta }{2}}{\dfrac{\sqrt{3}\cos \theta +\sin \theta }{2}}-\dfrac{\dfrac{\cos \theta +\sqrt{3}\sin \theta }{2}}{\dfrac{\sqrt{3}\cos \theta -\sin \theta }{2}}$$ Above equation can be rearranged as follows, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\cos \theta -\sqrt{3}\sin \theta }{2}\times \dfrac{2}{\sqrt{3}\cos \theta +\sin \theta }-\dfrac{\cos \theta +\sqrt{3}\sin \theta }{2}\times \dfrac{2}{\sqrt{3}\cos \theta -\sin \theta }$$ Further simplification in the above equation will give, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\cos \theta -\sqrt{3}\sin \theta }{\sqrt{3}\cos \theta +\sin \theta }-\dfrac{\cos \theta +\sqrt{3}\sin \theta }{\sqrt{3}\cos \theta -\sin \theta }$$ If we take the LCM in the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\left( \sqrt{3}\cos \theta -\sin \theta \right)\left( \cos \theta -\sqrt{3}\sin \theta \right)-\left( \cos \theta +\sqrt{3}\sin \theta \right)\left( \sqrt{3}\cos \theta +\sin \theta \right)}{\left( \sqrt{3}\cos \theta +\sin \theta \right)\left( \sqrt{3}\cos \theta -\sin \theta \right)}$$ As we know that $$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$$ therefore if we use the formula in the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\left( \sqrt{3}\cos \theta -\sin \theta \right)\left( \cos \theta -\sqrt{3}\sin \theta \right)-\left( \cos \theta +\sqrt{3}\sin \theta \right)\left( \sqrt{3}\cos \theta +\sin \theta \right)}{{{\left( \sqrt{3}\cos \theta \right)}^{2}}-{{\left( \sin \theta \right)}^{2}}}$$ By simplifying the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\left( \sqrt{3}\cos \theta -\sin \theta \right)\left( \cos \theta -\sqrt{3}\sin \theta \right)-\left( \cos \theta +\sqrt{3}\sin \theta \right)\left( \sqrt{3}\cos \theta +\sin \theta \right)}{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ If we multiply the brackets in the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\sqrt{3}\cos \theta \left( \cos \theta -\sqrt{3}\sin \theta \right)-\sin \theta \left( \cos \theta -\sqrt{3}\sin \theta \right)-\left[ \cos \theta \left( \sqrt{3}\cos \theta +\sin \theta \right)+\sqrt{3}\sin \theta \left( \sqrt{3}\cos \theta +\sin \theta \right) \right]}{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ Further simplification in the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\sqrt{3}\cos \theta \cos \theta -\sqrt{3}\cos \theta \sqrt{3}\sin \theta -\left( \sin \theta \cos \theta -\sin \theta \sqrt{3}\sin \theta \right)-\left[ \cos \theta \sqrt{3}\cos \theta +\cos \theta \sin \theta +\sqrt{3}\sin \theta \sqrt{3}\cos \theta +\sqrt{3}\sin \theta \sin \theta \right]}{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ If we give negative sign inside the bracket we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\sqrt{3}\cos \theta \cos \theta -\sqrt{3}\cos \theta \sqrt{3}\sin \theta -\sin \theta \cos \theta +\sin \theta \sqrt{3}\sin \theta -\cos \theta \sqrt{3}\cos \theta -\cos \theta \sin \theta -\sqrt{3}\sin \theta \sqrt{3}\cos \theta -\sqrt{3}\sin \theta \sin \theta }{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ Further simplification in the above equation will give, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\sqrt{3}{{\cos }^{2}}\theta -3\cos \theta \sin \theta -\sin \theta \cos \theta +\sqrt{3}{{\sin }^{2}}\theta -\sqrt{3}{{\cos }^{2}}\theta -\cos \theta \sin \theta -3\sin \theta \cos \theta -\sqrt{3}{{\sin }^{2}}\theta }{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ If we observe the above equation carefully then we will see that there as some terms which are same in value by different in sign ad these terms can easily be cancelled, therefore we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{-3\cos \theta \sin \theta -\sin \theta \cos \theta -\cos \theta \sin \theta -3\sin \theta \cos \theta }{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ Further simplification in the above equation will give, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{-8\cos \theta \sin \theta }{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }$$ As we know that our right hand side is in ‘cot’ and therefore we have to convert our ‘LHS’ into ‘cot’ and for that we will divide both the numerator and denominator of the second term of the above equation by $${{\sin }^{2}}\theta $$, therefore we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\dfrac{-8\cos \theta \sin \theta }{{{\sin }^{2}}\theta }}{\dfrac{3{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }}$$ By separating the denominator in the denominator of second term and further simplifying the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{\dfrac{-8\cos \theta }{\sin \theta }}{\dfrac{3{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }}$$ As we know that $$\dfrac{\cos \theta }{\sin \theta }=\cot \theta $$ and if we put that value in the above equation we will get, Therefore, Left Hand Side (LHS) = $$\cot \theta +\dfrac{-8\cot \theta }{3{{\cot }^{2}}\theta -1}$$ Above equation can also be written as, Therefore, Left Hand Side (LHS) = $$\cot \theta -\dfrac{8\cot \theta }{3{{\cot }^{2}}\theta -1}$$ by cross multiplying in the above equation we will get, Therefore, Left Hand Side (LHS) = $$\dfrac{\cot \theta \left( 3{{\cot }^{2}}\theta -1 \right)-8\cot \theta }{3{{\cot }^{2}}\theta -1}$$ Therefore, Left Hand Side (LHS) = $$\dfrac{3{{\cot }^{2}}\theta \cot \theta -\cot \theta -8\cot \theta }{3{{\cot }^{2}}\theta -1}$$ Therefore, Left Hand Side (LHS) = $$\dfrac{3{{\cot }^{3}}\theta -9\cot \theta }{3{{\cot }^{2}}\theta -1}$$ By taking 3 common from the above equation we will get, Therefore, Left Hand Side (LHS) = $$\dfrac{3\left( {{\cot }^{3}}\theta -3\cot \theta \right)}{3{{\cot }^{2}}\theta -1}$$ Therefore, Left Hand Side (LHS) = $$3\left( \dfrac{{{\cot }^{3}}\theta -3\cot \theta }{3{{\cot }^{2}}\theta -1} \right)$$ Now to proceed further in the solution we should know the formula given below, Formula: $$\cot 3\theta =\dfrac{{{\cot }^{3}}\theta -3\cot \theta }{3{{\cot }^{2}}\theta -1}$$ If we use this formula in ‘LHS’ we will get, Therefore, Left Hand Side (LHS) = $$3\cot 3\theta $$ If we compare above equation with equation (1) we will get, $$\cot \theta +\cot \left( 60+\theta \right)-\cot \left( 60-\theta \right)=3\cot 3\theta $$ Hence proved. Note: While proving the given equation you have to be very careful about the sigh as it makes very much confusion. Also you can solve this question by using the formulae $$\cot \left( A+B \right)=\dfrac{\cot B\cot A-1}{\cot B+\cot A}$$ and $$\cot \left( A-B \right)=\dfrac{\cot B\cot A+1}{\cot B-\cot A}$$.