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Question: Prove that \(\dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\...

Prove that tan⁡θ(1−cot⁡θ)+cot⁡θ(1−tan⁡θ)=1+tan⁡θ+cot⁡θ=sec⁡θ×csc⁡θ+1\dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}} = 1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1.

Explanation

Solution

Hint – In this question we have to prove that the left hand side of the given expression is equal to the right hand side. Use the trigonometric identities and algebraic formula likecot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }},[tan2θ+1=sec⁡2θ]\left[ {{\text{ta}}{{\text{n}}^2}\theta + 1 = {{\sec }^2}\theta } \right], (a3−b3)=(a−b)(a2+b2+ab)\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)to simplify the L.H.S part and prove it equal to the R.H.S part.

Complete step by step solution:
Given equation is
tan⁡θ(1−cot⁡θ)+cot⁡θ(1−tan⁡θ)=1+tan⁡θ+cot⁡θ=sec⁡θ×csc⁡θ+1\dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}} = 1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1
Consider L.H.S
⇒tan⁡θ(1−cot⁡θ)+cot⁡θ(1−tan⁡θ)\Rightarrow \dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}
Now as we know cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }} so, substitute this value in above equation we have,
tan⁡θ(1−1tan⁡θ)+cot⁡θ(1−tan⁡θ)\dfrac{{\tan \theta }}{{\left( {1 - \dfrac{1}{{\tan \theta }}} \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}
Now simplify the above equation we have,
⇒tan⁡2θ(tan⁡θ−1)+cot⁡θ(1−tan⁡θ)\Rightarrow \dfrac{{{{\tan }^2}\theta }}{{\left( {\tan \theta - 1} \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}
⇒tan⁡2θ(tan⁡θ−1)−cot⁡θ(tan⁡θ−1)=tan⁡2θ−1tan⁡θtan⁡θ−1=tan⁡3θ−1tan⁡θ(tan⁡θ−1)\Rightarrow \dfrac{{{{\tan }^2}\theta }}{{\left( {\tan \theta - 1} \right)}} - \dfrac{{\cot \theta }}{{\left( {\tan \theta - 1} \right)}} = \dfrac{{{{\tan }^2}\theta - \dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}} = \dfrac{{{{\tan }^3}\theta - 1}}{{\tan \theta \left( {\tan \theta - 1} \right)}}
Now as we know (a3−b3)=(a−b)(a2+b2+ab)\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right) so, use this property in above equation we have,
⇒(tan⁡θ−1)(tan⁡2θ+1+tan⁡θ)tan⁡θ(tan⁡θ−1)=(tan⁡2θ+1+tan⁡θ)tan⁡θ\Rightarrow \dfrac{{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta \left( {\tan \theta - 1} \right)}} = \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }}
Now divide by tan⁡θ\tan \theta in above equation we have,
⇒(tan⁡2θ+1+tan⁡θ)tan⁡θ=1+tan⁡θ+cot⁡θ\Rightarrow \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }} = 1 + \tan \theta + \cot \theta
= R.H.S
Now we have to again prove that
1+tan⁡θ+cot⁡θ=sec⁡θ×csc⁡θ+11 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1
Now consider L.H.S
⇒1+tan⁡θ+cot⁡θ\Rightarrow 1 + \tan \theta + \cot \theta
⇒1+tan⁡θ+cot⁡θ=(tan⁡2θ+1+tan⁡θ)tan⁡θ\Rightarrow 1 + \tan \theta + \cot \theta = \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }} (From above equation)
Now as we know that [tan2θ+1=sec⁡2θ]\left[ {{\text{ta}}{{\text{n}}^2}\theta + 1 = {{\sec }^2}\theta } \right] so, substitute this value in above equation we have,
⇒sec⁡2θ+tan⁡θtan⁡θ=sec⁡2θtan⁡θ+1\Rightarrow \dfrac{{{{\sec }^2}\theta + \tan \theta }}{{\tan \theta }} = \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} + 1
Now we know that secθ = 1cos⁡θ, tan⁡θ=sin⁡θcos⁡θ{\text{sec}}\theta {\text{ = }}\dfrac{1}{{\cos \theta }},{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} so, substitute this value in above equation we have,
⇒sec⁡2θtan⁡θ+1=cos⁡θcos⁡2θ×sin⁡θ+1=1cos⁡θ×sin⁡θ+1\Rightarrow \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} + 1 = \dfrac{{\cos \theta }}{{{{\cos }^2}\theta \times \sin \theta }} + 1 = \dfrac{1}{{\cos \theta \times \sin \theta }} + 1
Now we know 1sin⁡θ=csc⁡θ\dfrac{1}{{\sin \theta }} = \csc \theta so, substitute this value in above equation we have,
⇒1cos⁡θ×sin⁡θ+1=sec⁡θ×csc⁡θ+1\Rightarrow \dfrac{1}{{\cos \theta \times \sin \theta }} + 1 = \sec \theta \times \csc \theta + 1
= R.H.S
Hence proved

Note – Whenever we face such types of questions the key point is simply to have a good grasp over the trigonometric identities, some of them have been mentioned above while performing the solution. Start by simplifying any one side and application of these identities along with some algebraic identities will help you reach the right answer.