Question

The general solution of $\tan 5\theta =\cot 2\theta$ is $$$$
A.\theta =\dfrac{n\pi }{7}+\dfrac{\pi }{14}$$$$$ B. \theta =\dfrac{n\pi }{7}+\dfrac{\pi }{5} C. $\theta =\dfrac{n\pi }{7}+\dfrac{\pi }{2}
D. $\theta =\dfrac{n\pi }{7}+\dfrac{\pi }{3}$$$$$

Answer 1

We use the complimentary angle relation $\tan \theta =\cot \left( \dfrac{\pi }{2}-\theta \right)$ and convert cotangent in the given equation $\tan 5\theta =\cot 2\theta$ to tangent and the find the solution for $\theta$ using the information that the solution of the equation $\tan x=\tan \alpha$ is given by $x=n\pi +\alpha$ for some arbitrary integer $n$.$$$$

Complete step by step answer:

We know that in right angled triangle the side opposite to right angled triangle is called hypotenuse denoted as $h$, the vertical side is called perpendicular denoted as $p$ and the horizontal side is called the base denoted as $b$.$$$$
Here in the above diagram of right angled triangle ABC we have,
$AC=h,AB=p,BC=b$
We know from the trigonometric ratios in a right angled triangle the tangent of the angle is the ratio of opposite side to the adjacent side (excluding hypotenuse, also called leg adjacent) . So we have tangent of the angle of angle \theta $$$$$ $$\tan \theta =\dfrac{AB}{AC}=\dfrac{p}{b}$$ The ratio reciprocal to tangent of the angle is called co-tangent which means of ratio of leg adjacent to the opposite side is denoted by \cot \theta and is given by $$\cot \theta =\dfrac{AC}{AB}=\dfrac{b}{p}$$ We know that the sum of the angles in a triangle is{{180}^{\circ }}$. So we have

& A+B+C={{180}^{\circ }} \\\ & \Rightarrow A+C={{90}^{\circ }} \\\ & \Rightarrow A={{90}^{\circ }}-C={{90}^{\circ }}-\theta \\\ \end{aligned}$$ We use definition of tangent and cotangent to for angles $A,C$ to have, $$\begin{aligned} & \tan C=\tan \theta =\dfrac{p}{b},\cot C=\cot \theta =\dfrac{b}{p} \\\ & \cot A=\cot \left( 90-\theta \right)=\dfrac{p}{b},\cot A=\cot \left( 90-\theta \right)=\dfrac{b}{p} \\\ \end{aligned}$$ We have from above, $$\tan \left( {{90}^{\circ }}-\theta \right)=\cot \theta ,\cot \left( {{90}^{\circ }}-\theta \right)=\sin \theta $$ The above equation is called complementary angle relation of tangent and cotangent or reduction formula of tangent-cotangent.$$$$ We know the solutions of the equation $\tan x=\tan \alpha $ where $x$ is the unknown and $\alpha $ is the known angle is given with some arbitrary integer $n$ as $$x=n\pi +\alpha $$ We are given in the question the trigonometric equation with unknown angle $\theta $ as, $$\tan 5\theta =\cot 2\theta $$ We use the complimentary angle relation for $\cot \left( 2\theta \right)$ and convert it into tangent of the angle $2\theta $ to have, $$\begin{aligned} & \Rightarrow \tan 5\theta =\tan \left( {{90}^{\circ }}-2\theta \right) \\\ & \Rightarrow \tan 5\theta =\tan \left( \dfrac{\pi }{2}-2\theta \right) \\\ \end{aligned}$$ We find the solutions of above equation in tangent function taking unknown $x=5\theta $ and the known $\alpha =\dfrac{\pi }{2}-2\theta $ as $$\begin{aligned} & \Rightarrow 5\theta =n\pi +\dfrac{\pi }{2}-2\theta \\\ & \Rightarrow 7\theta =n\pi +\dfrac{\pi }{2} \\\ \end{aligned}$$ We divide the above equation by 7 to have the required result as, $$\Rightarrow \theta =\dfrac{n\pi }{7}+\dfrac{\pi }{14},n\in Z$$ **So, the correct answer is “Option A”.** **Note:** We note that the tangent function is not defined for $\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$ and cotangent function is not defined for $\theta =n\pi $ and hence the obtained result well defined. We must not confuse between complementary and supplementary angle relation which is given by a shift $\pi $ radian as $\tan \left( \pi +\theta \right)=-\tan \theta ,\cot \left( \pi +\theta \right)=-\cot \theta $.