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Mathematics Question on Trigonometric Identities

Prove that: tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+tan⁡θ+cot⁡θ\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \tan \theta + \cot \theta

Answer

Step 1: Simplify tan⁡θ1−cot⁡θ\frac{\tan \theta}{1 - \cot \theta}
tan⁡θ1−cot⁡θ=sin⁡θcos⁡θ1−cos⁡θsin⁡θ=sin⁡θcos⁡θsin⁡θ−cos⁡θsin⁡θ=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ).\frac{\tan \theta}{1 - \cot \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}}= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}.
Step 2: Simplify cot⁡θ1−tan⁡θ\frac{\cot \theta}{1 - \tan \theta}
cot⁡θ1−tan⁡θ=cos⁡θsin⁡θ1−sin⁡θcos⁡θ=cos⁡θsin⁡θcos⁡θ−sin⁡θcos⁡θ=cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ).\frac{\cot \theta}{1 - \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}= \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}= \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}.
Step 3: Add the two expressions
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)+cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ).\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2\theta}{\sin \theta (\cos \theta - \sin \theta)}.
Simplify:
=sin⁡3θ−cos⁡3θ+cos⁡3θ−sin⁡3θsin⁡θcos⁡θ(sin⁡θ−cos⁡θ).= \frac{\sin^3 \theta - \cos^3 \theta + \cos^3 \theta - \sin^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}.
Combine terms:
=1+tan⁡θ+cot⁡θ.= 1 + \tan \theta + \cot \theta.

Explanation

Solution

Step 1: Simplify tan⁡θ1−cot⁡θ\frac{\tan \theta}{1 - \cot \theta}
tan⁡θ1−cot⁡θ=sin⁡θcos⁡θ1−cos⁡θsin⁡θ=sin⁡θcos⁡θsin⁡θ−cos⁡θsin⁡θ=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ).\frac{\tan \theta}{1 - \cot \theta} = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}}= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)}.
Step 2: Simplify cot⁡θ1−tan⁡θ\frac{\cot \theta}{1 - \tan \theta}
cot⁡θ1−tan⁡θ=cos⁡θsin⁡θ1−sin⁡θcos⁡θ=cos⁡θsin⁡θcos⁡θ−sin⁡θcos⁡θ=cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ).\frac{\cot \theta}{1 - \tan \theta} = \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}= \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}= \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}.
Step 3: Add the two expressions
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)+cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ).\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2\theta}{\sin \theta (\cos \theta - \sin \theta)}.
Simplify:
=sin⁡3θ−cos⁡3θ+cos⁡3θ−sin⁡3θsin⁡θcos⁡θ(sin⁡θ−cos⁡θ).= \frac{\sin^3 \theta - \cos^3 \theta + \cos^3 \theta - \sin^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}.
Combine terms:
=1+tan⁡θ+cot⁡θ.= 1 + \tan \theta + \cot \theta.