Question

Prove that: \[\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right) = \sec \theta + \cos ec \theta \].

Answer 1

Hint:
Here, we will first consider the left hand side of the equation and convert all the tangent and cotangent functions into the sine and cosine functions. Then we will apply suitable trigonometric identity to simplify the equation further. We will then apply the reciprocal trigonometric identity to prove the given equation.

Formula Used:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]

Complete step by step solution:
In order to prove the given equation, we will try to prove that the given left hand side is equal to the right hand side.
Thus, we have,
LHS\[ = \sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right)\]
Now, using the relationship between trigonometric identities, we know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\].
Hence, substituting these values in the given LHS, we get,
LHS \[ = \sin \theta \left( {1 + \dfrac{{\sin \theta }}{{\cos \theta }}} \right) + \cos \theta \left( {1 + \dfrac{{\cos \theta }}{{\sin \theta }}} \right)\]
Now, multiplying the brackets by the respective trigonometric functions present outside the brackets, we get,
\[ \Rightarrow \] LHS \[ = \sin \theta + \dfrac{{\sin \theta \times \sin \theta }}{{\cos \theta }} + \cos \theta + \dfrac{{\cos \theta \times \cos \theta }}{{\sin \theta }}\]
\[ \Rightarrow \] LHS \[ = \sin \theta + \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }} + \cos \theta + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}\]
Rewriting the above equation, we get
\[ \Rightarrow \] LHS \[ = \left( {\sin \theta + \dfrac{{{{\cos }^2}\theta }}{{\sin \theta }}} \right) + \left( {\cos \theta + \dfrac{{{{\sin }^2}\theta }}{{\cos \theta }}} \right)\]
Now, taking LCM in the two brackets respectively, we get,
\[ \Rightarrow \] LHS \[ = \left( {\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{\sin \theta }}} \right) + \left( {\dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{\cos \theta }}} \right)\]………………………..\[\left( 1 \right)\]
As we know that the sum of squares of sine and cosine functions of the same angle is equal to 1 i.e. \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Hence, substituting this value in \[\left( 1 \right)\], we get,
\[ \Rightarrow \] LHS \[ = \dfrac{1}{{\sin \theta }} + \dfrac{1}{{\cos \theta }}\]
Here, substituting the values \[\cos ec \theta = \dfrac{1}{{\sin \theta }}\] and \[\sec \theta = \dfrac{1}{{\cos \theta }}\], we get
\[ \Rightarrow \] LHS \[ = \cos ec \theta + \sec \theta = \] RHS
Therefore, \[\sin \theta \left( {1 + \tan \theta } \right) + \cos \theta \left( {1 + \cot \theta } \right) = \sec \theta + \cos ec \theta \]
Hence, proved

Note:
In this question, we have used trigonometry. Trigonometry is a branch of mathematics which helps us to study the relationship between the sides and the angles of a triangle. In practical life, trigonometry is used by cartographers (to make maps). It is also used by the aviation and naval industries. In fact, trigonometry is even used by Astronomers to find the distance between two stars. Hence, it has an important role to play in everyday life. The three most common trigonometric functions are the tangent function, the sine and the cosine function. In simple terms, they are written as ‘sin’, ‘cos’ and ‘tan’. Hence, trigonometry is not just a chapter to study, in fact, it is being used in everyday life.