Question

How do you simplify \[\dfrac{{1 + \cos y}}{{1 + \sec y}}\] ?

Answer 1

Hint: Trigonometric functions are those functions that tell us the relation between the three sides of a right-angled triangle. Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively. After using the reciprocal relation and taking the LCM we can solve this.

Complete step-by-step answer:
Given, \[\dfrac{{1 + \cos y}}{{1 + \sec y}}\]
We know that reciprocal of cosine is secant. That is,
 \[\sec y = \dfrac{1}{{\cos y}}\] .
Substituting this in the given problem we have,
 \[ \Rightarrow \dfrac{{1 + \cos y}}{{1 + \sec y}} = \dfrac{{\left( {1 + \cos y} \right)}}{{\left( {1 + \dfrac{1}{{\cos y}}} \right)}}\]
In the denominator of the right hand side of the equation we take LCM and simplifying we have,
 \[ = \dfrac{{\left( {1 + \cos y} \right)}}{{\left( {\dfrac{{\cos y + 1}}{{\cos y}}} \right)}}\] .
Further simplification we have,
 \[ = \dfrac{{\left( {1 + \cos y} \right)}}{{\left( {\cos y + 1} \right)}} \times \cos y\]
 \[ = \dfrac{{\left( {1 + \cos y} \right)}}{{\left( {1 + \cos y} \right)}} \times \cos y\]
Cancelling the terms we have,
 \[ = \cos y\]
That is we have \[ \Rightarrow \dfrac{{1 + \cos y}}{{1 + \sec y}} = \cos y\] .
So, the correct answer is “\[ \dfrac{{1 + \cos y}}{{1 + \sec y}} = \cos y\]”.

Note: In the above problem we can also apply the reciprocal of cosine and after simplification we get the same answer. That is
 \[\cos y = \dfrac{1}{{\sec y}}\]
Substituting this in \[\dfrac{{1 + \cos y}}{{1 + \sec y}}\] we have,
 \[ = \dfrac{{\left( {1 + \dfrac{1}{{\sec y}}} \right)}}{{\left( {1 + \sec y} \right)}}\]
Taking LCM in the numerator and simplifying we have,
 \[ = \dfrac{{\left( {\dfrac{{\sec y + 1}}{{\sec y}}} \right)}}{{\left( {1 + \sec y} \right)}}\]
Rearranging we have
 \[ = \dfrac{{\left( {1 + \sec y} \right)}}{{\sec y\left( {1 + \sec y} \right)}}\]
Cancelling the terms we have,
 \[ = \dfrac{1}{{\sec y}}\]
We know the reciprocal of secant is cosine then we have
 \[ = \cos y\] .
That is we have \[\dfrac{{1 + \cos y}}{{1 + \sec y}} = \cos y\] . In both the cases we have the same answer. We can do anyone one of the methods. Also remember A graph is divided into four quadrants, all the trigonometric functions are positive in the first quadrant, all the trigonometric functions are negative in the second quadrant except sine and cosine functions, tangent and cotangent are positive in the third quadrant while all others are negative and similarly all the trigonometric functions are negative in the fourth quadrant except cosine and secant.