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Question

Question: How do you simplify \({{\tan }^{2}}x-{{\cot }^{2}}x\)?...

How do you simplify tan2xcot2x{{\tan }^{2}}x-{{\cot }^{2}}x?

Explanation

Solution

In this problem we need to simplify the given trigonometric equation. For this we will convert the whole equation in terms of sinx\sin x, cosx\cos x by using the basic definitions of trigonometric ratios i.e., tanx=sinxcosx\tan x=\dfrac{\sin x}{\cos x}, cotx=cosxsinx\cot x=\dfrac{\cos x}{\sin x}. Now we will substitute those values in the given equation and simplify them by taking the LCM and calculating the difference. Now we will apply the algebraic formula a2b2=(a+b)(ab){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) and use the trigonometric identities and formulas to simplify the obtained equation. After simplifying the equation, we will get our required solution.

Complete step by step answer:
Given that, tan2xcot2x{{\tan }^{2}}x-{{\cot }^{2}}x.
From the basic definitions of trigonometric ratios substituting the values tanx=sinxcosx\tan x=\dfrac{\sin x}{\cos x}, cotx=cosxsinx\cot x=\dfrac{\cos x}{\sin x} in the above equation, then we will get
tan2xcot2x=(sinxcosx)2(cosxsinx)2\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x={{\left( \dfrac{\sin x}{\cos x} \right)}^{2}}-{{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}
Simplifying the above equation, then we will get
tan2xcot2x=sin2xcos2xcos2xsin2x\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}-\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}
Taking the LCM and doing the subtraction in the above equation, then we will get
tan2xcot2x=sin2x×sin2xcos2x×cos2xsin2xcos2x tan2xcot2x=(sin2x)2(cos2x)2(sinxcosx)2 \begin{aligned} & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{{{\sin }^{2}}x\times {{\sin }^{2}}x-{{\cos }^{2}}x\times {{\cos }^{2}}x}{{{\sin }^{2}}x{{\cos }^{2}}x} \\\ & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{{{\left( {{\sin }^{2}}x \right)}^{2}}-{{\left( {{\cos }^{2}}x \right)}^{2}}}{{{\left( \sin x\cos x \right)}^{2}}} \\\ \end{aligned}
Using the algebraic formula a2b2=(a+b)(ab){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) in the above equation, then we will get
tan2xcot2x=(sin2x+cos2x)(sin2xcos2x)(sinxcosx)2\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)}{{{\left( \sin x\cos x \right)}^{2}}}
We have the trigonometric identity sin2x+cos2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1, trigonometric formulas sin2xcos2x=cos2x{{\sin }^{2}}x-{{\cos }^{2}}x=-\cos 2x, sinxcosx=sin2x2\sin x\cos x=\dfrac{\sin 2x}{2}. Substituting these values in the above equation, then we will get
tan2xcot2x=(1)(cos2x)(sin2x2)2\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\dfrac{\left( 1 \right)\left( -\cos 2x \right)}{{{\left( \dfrac{\sin 2x}{2} \right)}^{2}}}
Simplifying the above equation, then we will get
tan2xcot2x=cos2xsin22x4 tan2xcot2x=4cos2xsin22x \begin{aligned} & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-\dfrac{\cos 2x}{\dfrac{{{\sin }^{2}}2x}{4}} \\\ & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-4\dfrac{\cos 2x}{{{\sin }^{2}}2x} \\\ \end{aligned}
Writing the denominator sin22x=sin2x.sin2x{{\sin }^{2}}2x=\sin 2x.\sin 2x, then we will get
tan2xcot2x=4×cos2xsin2x×1sin2x\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-4\times \dfrac{\cos 2x}{\sin 2x}\times \dfrac{1}{\sin 2x}
We have the trigonometric formulas cosxsinx=cotx\dfrac{\cos x}{\sin x}=\cot x, 1sinx=cscx\dfrac{1}{\sin x}=\csc x. Applying those formulas in the above equation, then we will get
tan2xcot2x=4cot2xcsc2x\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=-4\cot 2x\csc 2x

Hence the simplified form of the given equation tan2xcot2x{{\tan }^{2}}x-{{\cot }^{2}}x is 4cot2xcsc2x-4\cot 2x\csc 2x.

Note: In some cases, students may use the trigonometric identities sec2xtan2x=1{{\sec }^{2}}x-{{\tan }^{2}}x=1, csc2xcot2x=1{{\csc }^{2}}x-{{\cot }^{2}}x=1 to simplify the equation. From the identity sec2xtan2x=1{{\sec }^{2}}x-{{\tan }^{2}}x=1, the value of tan2x{{\tan }^{2}}x is sec2x1{{\sec }^{2}}x-1 and from the identity csc2xcot2x=1{{\csc }^{2}}x-{{\cot }^{2}}x=1, the value of cot2x{{\cot }^{2}}x is csc2x1{{\csc }^{2}}x-1. Substituting these values in the above equation, then we will get
tan2xcot2x=(sec2x1)(csc2x1)\Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x=\left( {{\sec }^{2}}x-1 \right)-\left( {{\csc }^{2}}x-1 \right)
Simplifying the above equation, then we will get
tan2xcot2x=sec2x1csc2x+1 tan2xcot2x=sec2xcsc2x \begin{aligned} & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x={{\sec }^{2}}x-1-{{\csc }^{2}}x+1 \\\ & \Rightarrow {{\tan }^{2}}x-{{\cot }^{2}}x={{\sec }^{2}}x-{{\csc }^{2}}x \\\ \end{aligned}
I think the above form is not the simplified form of the given equation. So, I suggest students don’t follow the procedure in this note. Please use the procedure which we discussed in the solution part.