# Prove that $\dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}} = 1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1$.

Last updated date: 30th Mar 2023

•

Total views: 306.3k

•

Views today: 2.83k

Answer

Verified

306.3k+ views

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 like\[\cot \theta = \dfrac{1}{{\tan \theta }}\],\[\left[ {{\text{ta}}{{\text{n}}^2}\theta + 1 = {{\sec }^2}\theta } \right]\], $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$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

$\dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}} = 1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1$

Consider L.H.S

$ \Rightarrow \dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}$

Now as we know \[\cot \theta = \dfrac{1}{{\tan \theta }}\] so, substitute this value in above equation we have,

\[\dfrac{{\tan \theta }}{{\left( {1 - \dfrac{1}{{\tan \theta }}} \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}\]

Now simplify the above equation we have,

\[ \Rightarrow \dfrac{{{{\tan }^2}\theta }}{{\left( {\tan \theta - 1} \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}\]

\[ \Rightarrow \dfrac{{{{\tan }^2}\theta }}{{\left( {\tan \theta - 1} \right)}} - \dfrac{{\cot \theta }}{{\left( {\tan \theta - 1} \right)}} = \dfrac{{{{\tan }^2}\theta - \dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}} = \dfrac{{{{\tan }^3}\theta - 1}}{{\tan \theta \left( {\tan \theta - 1} \right)}}\]

Now as we know $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ so, use this property in above equation we have,

\[ \Rightarrow \dfrac{{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta \left( {\tan \theta - 1} \right)}} = \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }}\]

Now divide by $\tan \theta $ in above equation we have,

\[ \Rightarrow \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }} = 1 + \tan \theta + \cot \theta \]

= R.H.S

Now we have to again prove that

$1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1$

Now consider L.H.S

\[ \Rightarrow 1 + \tan \theta + \cot \theta \]

\[ \Rightarrow 1 + \tan \theta + \cot \theta = \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }}\] (From above equation)

Now as we know that \[\left[ {{\text{ta}}{{\text{n}}^2}\theta + 1 = {{\sec }^2}\theta } \right]\] so, substitute this value in above equation we have,

\[ \Rightarrow \dfrac{{{{\sec }^2}\theta + \tan \theta }}{{\tan \theta }} = \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} + 1\]

Now we know that \[{\text{sec}}\theta {\text{ = }}\dfrac{1}{{\cos \theta }},{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] so, substitute this value in above equation we have,

\[ \Rightarrow \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} + 1 = \dfrac{{\cos \theta }}{{{{\cos }^2}\theta \times \sin \theta }} + 1 = \dfrac{1}{{\cos \theta \times \sin \theta }} + 1\]

Now we know $\dfrac{1}{{\sin \theta }} = \csc \theta $ so, substitute this value in above equation we have,

\[ \Rightarrow \dfrac{1}{{\cos \theta \times \sin \theta }} + 1 = \sec \theta \times \csc \theta + 1\]

= 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.

Complete step by step solution:

Given equation is

$\dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}} = 1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1$

Consider L.H.S

$ \Rightarrow \dfrac{{\tan \theta }}{{\left( {1 - \cot \theta } \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}$

Now as we know \[\cot \theta = \dfrac{1}{{\tan \theta }}\] so, substitute this value in above equation we have,

\[\dfrac{{\tan \theta }}{{\left( {1 - \dfrac{1}{{\tan \theta }}} \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}\]

Now simplify the above equation we have,

\[ \Rightarrow \dfrac{{{{\tan }^2}\theta }}{{\left( {\tan \theta - 1} \right)}} + \dfrac{{\cot \theta }}{{\left( {1 - \tan \theta } \right)}}\]

\[ \Rightarrow \dfrac{{{{\tan }^2}\theta }}{{\left( {\tan \theta - 1} \right)}} - \dfrac{{\cot \theta }}{{\left( {\tan \theta - 1} \right)}} = \dfrac{{{{\tan }^2}\theta - \dfrac{1}{{\tan \theta }}}}{{\tan \theta - 1}} = \dfrac{{{{\tan }^3}\theta - 1}}{{\tan \theta \left( {\tan \theta - 1} \right)}}\]

Now as we know $\left( {{a^3} - {b^3}} \right) = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)$ so, use this property in above equation we have,

\[ \Rightarrow \dfrac{{\left( {\tan \theta - 1} \right)\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta \left( {\tan \theta - 1} \right)}} = \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }}\]

Now divide by $\tan \theta $ in above equation we have,

\[ \Rightarrow \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }} = 1 + \tan \theta + \cot \theta \]

= R.H.S

Now we have to again prove that

$1 + \tan \theta + \cot \theta = \sec \theta \times \csc \theta + 1$

Now consider L.H.S

\[ \Rightarrow 1 + \tan \theta + \cot \theta \]

\[ \Rightarrow 1 + \tan \theta + \cot \theta = \dfrac{{\left( {{{\tan }^2}\theta + 1 + \tan \theta } \right)}}{{\tan \theta }}\] (From above equation)

Now as we know that \[\left[ {{\text{ta}}{{\text{n}}^2}\theta + 1 = {{\sec }^2}\theta } \right]\] so, substitute this value in above equation we have,

\[ \Rightarrow \dfrac{{{{\sec }^2}\theta + \tan \theta }}{{\tan \theta }} = \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} + 1\]

Now we know that \[{\text{sec}}\theta {\text{ = }}\dfrac{1}{{\cos \theta }},{\text{ }}\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] so, substitute this value in above equation we have,

\[ \Rightarrow \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }} + 1 = \dfrac{{\cos \theta }}{{{{\cos }^2}\theta \times \sin \theta }} + 1 = \dfrac{1}{{\cos \theta \times \sin \theta }} + 1\]

Now we know $\dfrac{1}{{\sin \theta }} = \csc \theta $ so, substitute this value in above equation we have,

\[ \Rightarrow \dfrac{1}{{\cos \theta \times \sin \theta }} + 1 = \sec \theta \times \csc \theta + 1\]

= 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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE