Question

Prove LHS=PHS for the following trigonometric equation
\[\dfrac{{\cos t}}{{1 + \sin t}} + \dfrac{{1 + \sin t}}{{\cos t}} = 2\sec t\]

Answer 1

Hint: In order to prove LHS=RHS for the given question, we will at first divide both numerator and denominator of the LHS by such a trigonometric function so that all of the trigonometric function in the given expression on LHS become same as the ones on the RHS.

Complete step by step solution:
We have,
\[\dfrac{{\cos t}}{{1 + \sin t}} + \dfrac{{1 + \sin t}}{{\cos t}} = 2\sec t\]
Taking the complicates side fist, that is LHS for the given question,
Therefore,
\[LHS = \dfrac{{\cos t}}{{1 + \sin t}} + \dfrac{{1 + \sin t}}{{\cos t}} - - - - - \left( 1 \right)\]
Here, we see that the LHS is divided into two parts which are separated by the plus sign in the middle. Taking first part of the LHS we get,
\[ \Rightarrow \dfrac{{\cos t}}{{1 + \sin t}}\]
To simplify the above expression we multiply both its numerator and denominator by its numerator.
Thus, we get
\[ \Rightarrow \dfrac{{\cos t \times \cos t}}{{(1 + \sin t)\cos t}} - - - - - \left( a \right)\]
Taking the second part of the LHS, we get,
\[ \Rightarrow \dfrac{{1 + \sin t}}{{\cos t}}\]
To simplify the above expression we will again use the same technique that we used for first part of LHS. We multiply both the numerator and the denominator by the numerator itself. We get,
\[ \Rightarrow \dfrac{{(1 + \sin t)(1 + \sin t)}}{{\cos t(1 + \sin t)}} - - - - - \left( b \right)\]
Substituting both (a) and (b) in the equation (1)
Therefore we get,
\[ \Rightarrow \dfrac{{\cos t \times \cos t}}{{(1 + \sin t)\cos t}} + \dfrac{{(1 + \sin t)(1 + \sin t)}}{{\cos t(1 + \sin t)}}\]
Further,
\[ \Rightarrow \dfrac{{{{\cos }^2}t}}{{\cos t(1 + \sin t)}} + \dfrac{{{{(1 + \sin t)}^2}}}{{\cos t(1 + \sin t)}}\]
Taking L.C.M.,
\[ \Rightarrow \dfrac{{{{\cos }^2}t + {{\left( {1 + \sin t} \right)}^2}}}{{\cos t(1 + \sin t)}}\]
Comparing and putting the value in \[{\left( {1 + \sin t} \right)^2}\]by using algebraic identity, which is,
\[{\left( {a + b} \right)^2} = {a^{}} + 2ab + {b^2}\]
Therefore, we get,
\[ \Rightarrow \dfrac{{{{\cos }^2}t + 1 + 2\sin t + {{\sin }^2}t}}{{\cos t(1 + \sin t)}}\]
\[ \Rightarrow \dfrac{{2 + 2\sin t}}{{\cos t(1 + \sin t)}}\]
Thus,
\[ \Rightarrow \dfrac{{2(1 + \sin t)}}{{\cos t(1 + \sin t)}}\]
Eliminating the common factors,
\[ \Rightarrow \dfrac{2}{{\cos t}}\]
Or
\[ \Rightarrow 2 \times \dfrac{1}{{\cos t}}\]
We know that, \[\dfrac{1}{{\cos t}} = \sec t\]
Thus, we get,
\[ \Rightarrow 2 \times \dfrac{1}{{\cos t}} = 2\sec t = RHS\]
Hence, proved.

Note: The reciprocal and quotient identities of trigonometry are always very important while solving a trigonometric equation. However, the algebraic identities are no less important. The nature of the identities will depend on the nature of the question. It is always helpful to start solving from the complicated side of the equation.