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Question: Find the following identity is true/false \(\dfrac{2\sin x\cos x-\cos x}{1-\sin x-{{\sin }^{2}}x-{{\...

Find the following identity is true/false 2sin⁡xcos⁡x−cos⁡x1−sin⁡x−sin⁡2x−cos⁡2x=cot⁡x\dfrac{2\sin x\cos x-\cos x}{1-\sin x-{{\sin }^{2}}x-{{\cos }^{2}}x}=\cot x. $$$$

Explanation

Solution

We proceed from left hand side of equation and convert cosine into sine using Pythagorean identity for any acute angle θ\theta as sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 and then we convert sine and cosine into co-tangent using the identitycot⁡θ=sin⁡θcos⁡θ\cot \theta =\dfrac{\sin \theta }{\cos \theta }. We equate the expression and check if the identity holds true for all values of xx.$$$$

Complete step-by-step solution:
We know that identity is an equality relation defined on parameters is true when the equality relation is true for all real values of parameter, for example the identity of square of sum two numbers with parameters a,ba,b
(a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab
A equality relation is not always an identity, for example the following statement may not rue for all a,ba,b
(a+b)2=a2+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}
We know from Pythagorean identity of trigonometric ratios that for any angle θ\theta we have,
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
We can co-tangent of the angle θ\theta in terms of sine and cosine of the angle as
cot⁡θ=sin⁡θcos⁡θ\cot \theta =\dfrac{\sin \theta }{\cos \theta }
We are given in the question to find the truth value of following identity
2sin⁡xcos⁡x−cos⁡x1−sin⁡x−sin⁡2x−cos⁡2x=cot⁡x.......(1)\dfrac{2\sin x\cos x-\cos x}{1-\sin x-{{\sin }^{2}}x-{{\cos }^{2}}x}=\cot x.......\left( 1 \right)
Let us proceed from the left hand side of the given identity and try to express terms of cot⁡x\cot x. If we want to get cot⁡x\cot x we have to keep sin⁡x\sin x in the denominator. We have,
2sin⁡xcos⁡x−cos⁡x1−sin⁡x−sin⁡2x−cos⁡2x=2sin⁡xcos⁡x−cos⁡x1−cos⁡2x−sin⁡x−sin⁡2x\dfrac{2\sin x\cos x-\cos x}{1-\sin x-{{\sin }^{2}}x-{{\cos }^{2}}x}=\dfrac{2\sin x\cos x-\cos x}{1-{{\cos }^{2}}x-\sin x-{{\sin }^{2}}x}
Let us convert to cosine into sine using Pythagorean trigonometric identity of sine-cosine for angle θ=x\theta =x and have 1−cos⁡2x=sin⁡2x1-{{\cos }^{2}}x={{\sin }^{2}}x. So we have,

& =\dfrac{2\sin x\cos x-\cos x}{{{\sin }^{2}}x-\sin x-{{\sin }^{2}}x} \\\ & =\dfrac{2\sin x\cos x-\cos x}{-\sin x} \\\ & =\dfrac{2\sin x\cos x}{-\sin x}-\left( \dfrac{\cos x}{-\sin x} \right) \\\ & =2\cos x\left( \dfrac{\sin x}{-\sin x} \right)-\left( \dfrac{\cos x}{-\sin x} \right) \\\ \end{aligned}$$ Let us convert the sine and cosine in terms of cotangent using the trigonometric identity $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ for$x=\theta $. We have, $$=-2\cos x+\cot x$$ We have $\cot x$ on the right-hand side of the equation (1). Let us check if $$-2\cos x+\cot x=\cos x$$ The above expression on the left-hand side is not equal to $\cot x$ for all values$x$. So the if equation (1) is an identity with parameter $x$ as given in the question then the identity is false.$$$$ **Note:** We note that the identity is defined for $x=n\pi $ as the denominator cannot be zero. We note that if the question would have asked to find the truth value of relation$\dfrac{2\sin x\cos x-\cos x}{1-\sin x-{{\sin }^{2}}x-{{\cos }^{2}}x}=\cot x$, we would have solved for $x$ from $-2\cos x+\cot x=\cos x$ and found the relation holds true only for $x=\left( 2n+1 \right)\dfrac{\pi }{2}$ some integer$n$.