Question

(a) Find the value of $\dfrac{\sec \left( \pi -x \right)\tan \left( \dfrac{\pi }{2}-x \right)\sin \left( -x \right)}{\cot \left( 2\pi -x \right)\sin \left( \pi +x \right)\csc \left( \dfrac{\pi }{2}-x \right)}$
(b) Prove that $\cot x.\cot 2x-\cot 2x.\cot 3x-\cot 3x.\cot x=1$.

Answer 1

In this question there are two sub-questions and we will follow different methods for both problems.
In the first problem, we can observe that there are different trigonometric ratios and all the ratios are not in the direct angles. Most of the angles lie in different quadrants so we will use the All Silver Tea Cups method to reduce each ratio. After reducing each ratio, we will substitute in the given equation to get the result.
In the second problem we will consider the L.H.S and take the term $\cot 3x$ common from the terms $-\cot 2x.\cot 3x-\cot 3x.\cos x$ and then we will simplify the obtained equation by using the known formula $\cot \left( A+B \right)=\dfrac{\cot A.\cot B-1}{\cot A+\cot B}$ and cancel the term which is possible and obtain a result.

Complete step-by-step solution:
Given that, $\dfrac{\sec \left( \pi -x \right)\tan \left( \dfrac{\pi }{2}-x \right)\sin \left( -x \right)}{\cot \left( 2\pi -x \right)\sin \left( \pi +x \right)\csc \left( \dfrac{\pi }{2}-x \right)}$
Consider the term $\sec \left( \pi -x \right)$, we know that $\sec \left( \pi -\theta \right)=-\sec \theta$ $\Rightarrow \sec \left( \pi -x \right)=-\sec x$
Consider the term $\tan \left( \dfrac{\pi }{2}-x \right)$, we know that $\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta$ $\Rightarrow \tan \left( \dfrac{\pi }{2}-x \right)=\cot x$
Consider the term $\sin \left( -x \right)$, we know that $\sin \left( -\theta \right)=-\sin \theta$ $\Rightarrow \sin \left( -x \right)=-\sin x$
Consider the term $\cot \left( 2\pi -x \right)$, we know that $\cot \left( 2\pi -\theta \right)=-\cot \theta$ $\Rightarrow \cot \left( 2\pi -x \right)=-\cot x$
Consider the term $\sin \left( \pi +x \right)$, we know that $\sin \left( \pi +\theta \right)=-\sin \theta$ $\Rightarrow \sin \left( \pi +x \right)=-\sin x$
Consider the term $\csc \left( \dfrac{\pi }{2}-x \right)$, we know that $\csc \left( \dfrac{\pi }{2}-\theta \right)=\sec \theta$ $\Rightarrow \csc \left( \dfrac{\pi }{2}-x \right)=\sec x$
Now substituting all the values, we have in the given equation
$\begin{aligned} & \dfrac{\sec \left( \pi -x \right)\tan \left( \dfrac{\pi }{2}-x \right)\sin \left( -x \right)}{\cot \left( 2\pi -x \right)\sin \left( \pi +x \right)\csc \left( \dfrac{\pi }{2}-x \right)}=\dfrac{\left( -\sec x \right).\left( \cot x \right).\left( -\sin x \right)}{\left( -\cot x \right).\left( -\sin x \right).\left( \sec x \right)} \\\ & \therefore \dfrac{\sec \left( \pi -x \right)\tan \left( \dfrac{\pi }{2}-x \right)\sin \left( -x \right)}{\cot \left( 2\pi -x \right)\sin \left( \pi +x \right)\csc \left( \dfrac{\pi }{2}-x \right)}=1 \\\ \end{aligned}$

(b) Given that, $\cot x.\cot 2x-\cot 2x.\cot 3x-\cot 3x.\cot x=1$
$L.H.S=\cot x.\cot 2x-\cot 2x.\cot 3x-\cot 3x.\cot x$
Taking $\cot 3x$ common from the terms $-\cot 2x.\cot 3x-\cot 3x.\cos x$, then we will get
$\Rightarrow L.H.S=\cot x.\cot 2x-\cot 3x\left( \cot 2x+\cot x \right)$
Writing the term $\cot 3x$ as $\cot \left( 2x+x \right)$ , then we will get
$\begin{aligned} & \Rightarrow L.H.S=\cot x.\cot 2x-\cot 3x\left( \cot 2x+\cot x \right) \\\ & \Rightarrow L.H.S=\cot x.\cot 2x-\cot \left( 2x+x \right).\left( \cot 2x+\cot x \right) \\\ \end{aligned}$
Using the formula $\cot \left( A+B \right)=\dfrac{\cot A.\cot B-1}{\cot A+\cot B}$ in the above equation, then we will get
$\Rightarrow L.H.S=\cot x.\cot 2x-\left( \dfrac{\cot 2x.\cot x-1}{\cot 2x+\cot x} \right)\left( \cot 2x+\cot x \right)$
Cancelling the term $\cot 2x+\cot x$ in the above equation, then we will get
$\Rightarrow L.H.S=\cot x.\cot 2x-\left( \cot 2x.\cot x-1 \right)$
Using the multiplication distribution law $a\left( b+c \right)=ab+ac$ in the above equation, then we will get
$\begin{aligned} & \Rightarrow L.H.S=\cot x.\cot 2x-\cot 2x.\cot x+1 \\\ & \Rightarrow L.H.S=1 \\\ & \therefore L.H.S=R.H.S \\\ \end{aligned}$
Hence proved.

Note: The complete solution for the first part is based on the All Silver Tea Cups Method.

From the above picture, we can find the values of trigonometric ratios for all angles, and this method is simply denoted by All Silver Tea Cups.
For solving the problems like second part we need to know some basic trigonometric formulas which are mentioned below
$\begin{aligned} & \sin \left( A+B \right)=\sin A.\cos B+\cos A.\sin B \\\ & \sin \left( A-B \right)=\sin A.\cos B-\cos A.\sin B \\\ & \cos \left( A+B \right)=\cos A.\cos B-\sin A.\sin B \\\ & \cos \left( A-B \right)=\cos A.\cos B+\sin A.\sin B \\\ \end{aligned}$
$\begin{aligned} & \tan \left( A+B \right)=\dfrac{\sin \left( A+B \right)}{\cos \left( A+B \right)} \\\ & \tan \left( A-B \right)=\dfrac{\sin \left( A-B \right)}{\cos \left( A-B \right)} \\\ \end{aligned}$
$\begin{aligned} & \cot \left( A+B \right)=\dfrac{\cos \left( A+B \right)}{\sin \left( A+B \right)} \\\ & \cot \left( A-B \right)=\dfrac{\cos \left( A-B \right)}{\sin \left( A-B \right)} \\\ \end{aligned}$