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Question: Prove the following : \[\cot^{2} x-\tan^{2} x=4\cot 2x\csc 2x\]....

Prove the following : cot⁡2x−tan⁡2x=4cot⁡2xcsc⁡2x\cot^{2} x-\tan^{2} x=4\cot 2x\csc 2x.

Explanation

Solution

Hint: In this question it is given that we have to prove cot⁡2x−tan⁡2x=4cot⁡2xcsc⁡2x\cot^{2} x-\tan^{2} x=4\cot 2x\csc 2x. So for this we have to take the LHS part and we have to use appropriate trigonometric formulas in order to get the solution. So before proceeding to the solution part you have to know that whenever cot⁡x or tan⁡x\cot x\ or\ \tan x is given then you have to convert it into cos⁡xsin⁡x or sin⁡xcos⁡x\dfrac{\cos x}{\sin x} \ or\ \dfrac{\sin x}{\cos x}, which helps to solve any trigonometric expression in a simpler way.
Complete step-by-step solution:
Since as we know that cot⁡x=cos⁡xsin⁡x, tan⁡x=sin⁡xcos⁡x\cot x=\dfrac{\cos x}{\sin x} ,\ \tan x=\dfrac{\sin x}{\cos x}
So we can write,
LHS,
cot⁡2x−tan⁡2x\cot^{2} x-\tan^{2} x
=(cos⁡xsin⁡x)2−(sin⁡xcos⁡x)2\left( \dfrac{\cos x}{\sin x} \right)^{2} -\left( \dfrac{\sin x}{\cos x} \right)^{2}
=cos⁡2xsin⁡2x−sin⁡2xcos⁡2x\dfrac{\cos^{2} x}{\sin^{2} x} -\dfrac{\sin^{2} x}{\cos^{2} x}
=cos⁡4x−sin⁡4xsin⁡2xcos⁡2x\dfrac{\cos^{4} x-\sin^{4} x}{\sin^{2} x\cos^{2} x}
=(cos⁡2x)2−(sin⁡2x)2sin⁡2xcos⁡2x\dfrac{\left( \cos^{2} x\right)^{2} -\left( \sin^{2} x\right)^{2} }{\sin^{2} x\cos^{2} x}......(1)
Since as we know the identity, a2−b2=(a+b)(a−b)a^{2}-b^{2}=\left( a+b\right) \left( a-b\right)
So by this identity we can write the above expression (1) as, [where a=cos⁡2x and b=sin⁡2xa=\cos^{2} x\ and\ b=\sin^{2} x]
(cos⁡2x)2−(sin⁡2x)2sin⁡2xcos⁡2x\dfrac{\left( \cos^{2} x\right)^{2} -\left( \sin^{2} x\right)^{2} }{\sin^{2} x\cos^{2} x}
=(cos⁡2x+sin⁡2x)(cos⁡2x−sin⁡2x)sin⁡2xcos⁡2x\dfrac{\left( \cos^{2} x+\sin^{2} x\right) \left( \cos^{2} x-\sin^{2} x\right) }{\sin^{2} x\cos^{2} x}
=(cos⁡2x−sin⁡2x)sin⁡2xcos⁡2x\dfrac{\left( \cos^{2} x-\sin^{2} x\right) }{\sin^{2} x\cos^{2} x} [∵sin⁡2x+cos⁡2x=1\because \sin^{2} x+\cos^{2} x=1]
=cos⁡2xsin⁡2xcos⁡2x\dfrac{\cos 2x}{\sin^{2} x\cos^{2} x} [∵cos⁡2x−sin⁡2x=cos⁡2x\because \cos^{2} x-\sin^{2} x=\cos 2x]
=4cos⁡2x4sin⁡2xcos⁡2x\dfrac{4\cos 2x}{4\sin^{2} x\cos^{2} x} [multiplying numerator and denominator by 4]
=4cos⁡2x22×sin⁡2xcos⁡2x\dfrac{4\cos 2x}{2^{2}\times \sin^{2} x\cos^{2} x}
=4cos⁡2x(2sin⁡xcos⁡x)2\dfrac{4\cos 2x}{\left( 2\sin x\cos x\right)^{2} }
=4cos⁡2x(sin⁡2x)2\dfrac{4\cos 2x}{\left( \sin 2x\right)^{2} } [∵2sin⁡xcos⁡x=sin⁡2x\because 2\sin x\cos x=\sin 2x]
=4×cos⁡2xsin⁡2x×1sin⁡2x4\times \dfrac{\cos 2x}{\sin 2x} \times \dfrac{1}{\sin 2x}
=4cot⁡2xcsc⁡2x4\cot 2x\csc 2x
[Since as we know that cot⁡θ=cos⁡θsin⁡θ and csc⁡θ=1sin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } \ and\ \csc \theta =\dfrac{1}{\sin \theta }]
So, therefore,
cot⁡2x−tan⁡2x=4cot⁡2xcsc⁡2x\cot^{2} x-\tan^{2} x=4\cot 2x\csc 2x
Hence proved.
Note: In this question we have proved it by taking the LHS part, but you can also start the solution from the RHS part, and by using appropriate trigonometric formula you will get the LHS part. But in each an every way you have transform any given trigonometric function into sin⁡x,cos⁡x\sin x,\cos x form, like-cot⁡x=cos⁡xsin⁡x , csc⁡x=1sin⁡x\cot x=\dfrac{\cos x}{\sin x} \ ,\ \csc x=\dfrac{1}{\sin x}.