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Question: Prove the given expression: \(\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \thet...

Prove the given expression: tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcos⁡ecθ\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \cos ec\theta

Explanation

Solution

Hint: Start with taking LCM in LHS and simplify it using the identity a3−b3{{a}^{3}}-{{b}^{3}} , Now convert tan into sin and cos then use the fact that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 for all θ\theta to get the required answer.

Complete step-by-step answer:

An equality with sine, cosine or tangent in them is called a trigonometric equation. These are solved by interrelations known beforehand.
All the interrelations which relate sine, cosine, tangent, secant, cotangent, cosecant are called trigonometric identities. These trigonometric identities solve the equation and make them simpler to understand for a proof. These are the main and crucial steps to take us nearer to result.
Given equation in the question which we need to prove:
tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcos⁡ecθ\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \cos ec\theta
Take the left hand separately and simplify it
tan⁡θ1−tan⁡θ+1tan⁡θ1−tan⁡θ\dfrac{\tan \theta }{1-\tan \theta }+\dfrac{\dfrac{1}{\tan \theta }}{1-\tan \theta }
By taking the common term outside the expression we can turn into:
1tan⁡θ−1(tan⁡2θ−1tan⁡θ)=tan⁡3θ−1tan⁡θ(tan⁡θ−1)\dfrac{1}{\tan \theta -1}\left( {{\tan }^{2}}\theta -\dfrac{1}{\tan \theta } \right)=\dfrac{{{\tan }^{3}}\theta -1}{\tan \theta \left( \tan \theta -1 \right)}
By simplifying further, the expression turns into form of:
(tan⁡θ−1)(tan⁡2θ+1+tan⁡θ)tan⁡θ(tan⁡θ−1)\dfrac{\left( \tan \theta -1 \right)\left( {{\tan }^{2}}\theta +1+\tan \theta \right)}{\tan \theta \left( \tan \theta -1 \right)}
By cancelling the term and using relation between tan, cot, we get
tan⁡θ+cot⁡θ+1\tan \theta +\cot \theta +1
By converting all the terms into sine, cosines, we get that:
(sin⁡2θ+cos⁡2θ)sin⁡θcos⁡θ+1=1+sec⁡θcos⁡ecθ\dfrac{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}{\sin \theta \cos \theta }+1=1+\sec \theta \cos ec\theta by using the identity sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 we proved the required equation.
Hence proved.

Note: (i) Be careful while using the algebraic identity a3−b3{{a}^{3}}-{{b}^{3}}
(ii) After converting into sine, cosine takes the L.C.M properly.
(iii) You can directly take LCM in the first step and solve in a brute force method. But it will take some time to get results.