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Question: How do you verify \(\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)}={{\cos }^{2}}x\) ?...

How do you verify sin⁡2(−x)tan⁡2(x)=cos⁡2x\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)}={{\cos }^{2}}x ?

Explanation

Solution

First analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. This can be done by using the suitable trigonometric identities (formulae) and simplify any one side of the equation.

Complete step by step solution:
Sine, cosine and tangent of an angle are trigonometric ratios. These ratios are also called trigonometric functions. When we plot a graph of the trigonometric ratios with respect to all the real values of an angle we get a graph that has a periodic property. This means that the graph repeats itself after equal intervals of the angle. Other than the trigonometric ratios sine, cosine and tangent we have other trigonometric ratios called cosecant, secant and cotangent.

All the above six trigonometric ratios (functions) are dependent on each other. There are different properties and identities that related the trigonometric ratios.The equation that has to be verified is sin⁡2(−x)tan⁡2(x)=cos⁡2x\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)}={{\cos }^{2}}x. Let us first analyse the left hand side of the equation and check whether it is equal to the right hand side of the equation. The left hand side of the equation is,
sin⁡2(−x)tan⁡2(x)\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)} ….. (i)
We know that sin⁡(−x)=sin⁡x\sin (-x)=\sin x. Then,
sin⁡2(−x)=(−sin⁡x)2=sin⁡2x{{\sin }^{2}}(-x)={{\left( -\sin x \right)}^{2}}={{\sin }^{2}}x

We also know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}. Then,
tan⁡2x=sin⁡2xcos⁡2x{{\tan }^{2}}x=\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}
Substitute the known values in (i). Then,
sin⁡2(−x)tan⁡2(x)=sin⁡2(x)sin⁡2(x)cos⁡2(x)\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)}=\dfrac{{{\sin }^{2}}(x)}{\dfrac{{{\sin }^{2}}(x)}{{{\cos }^{2}}(x)}}
∴sin⁡2(−x)tan⁡2(x)=sin⁡2(x)×cos⁡2(x)sin⁡2(x)=cos⁡2x\therefore\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)}={{\sin }^{2}}(x)\times \dfrac{{{\cos }^{2}}(x)}{{{\sin }^{2}}(x)}={{\cos }^{2}}x
Then this means that sin⁡2(−x)tan⁡2(x)=cos⁡2x\dfrac{{{\sin }^{2}}(-x)}{{{\tan }^{2}}(x)}={{\cos }^{2}}x.Therefore, the left hand side of the given equation is equal to the right hand side of the equation. Hence, the given equation is verified and is correct.

Note: There are more than one way to answer the given question. You may use other trigonometric identities like,
tan⁡2x=1−sec⁡2x{{\tan }^{2}}x=1-{{\sec }^{2}}x and sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x
Then you can use the relation between cosine and secant function, which is,
sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}
Further, if you substitute these values in the left hand side of the equation then the equation will be verified.