Question

If \[\sin \theta + \cos \theta = \sqrt 3 \] then prove that \[\tan \theta + \cot \theta = 1\].

Answer 1

Hint: In this question we use the knowledge of trigonometric ratio \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] to prove \[\tan \theta + \cot \theta = 1\] by substation of the values.
* Trigonometric identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]

Complete step-by-step solution:
Given, \[\sin \theta + \cos \theta = \sqrt 3 \]
Squaring both sides of the equation
\[{\left( {\sin \theta + \cos \theta } \right)^2} = {\left( {\sqrt 3 } \right)^2}\]
Use the formula \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] where \[a = \sin \theta \] and \[b = \cos \theta \] to simplify.
\[{\sin ^2}\theta + {\cos ^2}\theta + 2\sin \theta \cos \theta = 3\]
Use the Trigonometric identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[1 + 2\sin \theta \cos \theta = 3\]
Subtract 1 from both sides of the equation and simplify the equation
\[
  1 + 2\sin \theta \cos \theta - 1 = 3 - 1 \\
  2\sin \theta \cos \theta = 2 \\
 \]
Divide both sides by 2 we get
\[\sin \theta \cos \theta = 1\,\,\, \ldots \left( 1 \right)\]
Now, substitute \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\] in the equation \[\tan \theta + \cot \theta \]
\[\tan \theta + \cot \theta = \dfrac{{\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }}\]
Take the LCM on the right hand side of the equation and combine into a single fraction.
\[\tan \theta + \cot \theta = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }}\]
Use the Trigonometric identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]and substitute it in the equation.
\[\tan \theta + \cot \theta = \dfrac{1}{{\cos \theta \sin \theta }}\]
From equation (1) substitute \[\sin \theta \cos \theta = 1\] in the equation.
\[\tan \theta + \cot \theta = \dfrac{1}{1} = 1\]
It can be seen that the LHS = RHS

Hence proved.

Note: In questions which require proving use the given equations to deduce a result and apply the result on the equation to prove. These types of questions always convert all the trigonometric forms into \[\sin \] and \[\cos \] form and then solve the equation.
Otherwise, if we solve directly and don’t substitute \[\sin \] and \[\cos \], then the solution becomes more complicated and we will be using another formula of trigonometry which might be hard to remember than these formulas.

Alternative method:
Since we know \[\cot \theta = \dfrac{1}{{\tan \theta }}\] , therefore we substitute the value of \[\cot \theta \] in the equation
\[\tan \theta + \cot \theta = \tan \theta + \dfrac{1}{{\tan \theta }}\]
Taking the LCM on right hand side of the equation
\[\tan \theta + \cot \theta = \dfrac{{{{\tan }^2}\theta + 1}}{{\tan \theta }}\]
Now we substitute \[1 + {\tan ^2}\theta = {\sec ^2}\theta \] in the numerator of right hand side of the equation
\[\tan \theta + \cot \theta = \dfrac{{{{\sec }^2}\theta }}{{\tan \theta }}\]
Now we know \[{\sec ^2}\theta = \dfrac{1}{{{{\cos }^2}\theta }}\] and \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Therefore we get \[\tan \theta + \cot \theta = \dfrac{1}{{\cos \theta .\cos \theta }} \times \dfrac{{\cos \theta }}{{\sin \theta }} = \dfrac{1}{{\sin \theta \cos \theta }}\]
And again using equation \[(1)\]we have \[\sin \theta \cos \theta = 1\]
Therefore substituting the value of \[\sin \theta \cos \theta = 1\] in the equation we get
\[\tan \theta + \cot \theta = \dfrac{1}{1} = 1\]

Hence proved.