Question

How do you solve $\dfrac{1+\cos x}{1+\sec x}$ ?

Answer 1

To solve the given trigonometric expression i.e. $\dfrac{1+\cos x}{1+\sec x}$, we are going to use the property of $\sec x$ which is equal to the reciprocal of $\cos x$ and mathematically we can write $\sec x=\dfrac{1}{\cos x}$ in the above expression and then simplify it.

Complete step by step answer:
The trigonometric expression given in the above problem is as follows:
$\dfrac{1+\cos x}{1+\sec x}$
We know the relation between $\sec x\And \cos x$ is equal to:
$\sec x=\dfrac{1}{\cos x}$
Using the above relation in the given trigonometric expression we get,
$\Rightarrow \dfrac{1+\cos x}{1+\dfrac{1}{\cos x}}$
Now, taking L.C.M of $\cos x$ in the denominator in the above expression we get,
$\Rightarrow \dfrac{1+\cos x}{\dfrac{\cos x+1}{\cos x}}$
Rearranging the above trigonometric expression we get,
$\Rightarrow \dfrac{\cos x\left( 1+\cos x \right)}{1+\cos x}$
In the above expression, $\left( 1+\cos x \right)$ is common in the numerator and denominator so $\left( 1+\cos x \right)$ will be cancelled out from the numerator and the denominator and we get,
$\Rightarrow \cos x$
From the above simplification of the given trigonometric expression we get,
$\Rightarrow \cos x$

Note: You might think how we know whether to use $\sec x=\dfrac{1}{\cos x}$, the answer is in the numerator you can see there is a term $\cos x$ so if we can use this $\sec x\And \cos x$ relation then there is a possibility in which $\cos x$ term might get cancelled and which you can see is the case in the above solution when we use this relation $\left( 1+\cos x \right)$ in the numerator and the denominator will be cancelled out.
The similar problem which can be possible is as follows:
$\dfrac{1+\sin x}{1+\text{cosec}x}$
So, here we can use the relation between $\text{cosec}x=\dfrac{1}{\sin x}$ and then simplify and we get,
$\Rightarrow \dfrac{1+\sin x}{1+\dfrac{1}{\sin x}}$
Taking $\sin x$ as L.C.M in the denominator of the above expression we get,
$\begin{aligned} & \Rightarrow \dfrac{1+\sin x}{\dfrac{\sin x+1}{\sin x}} \\\ & =\dfrac{\sin x\left( 1+\sin x \right)}{1+\sin x} \\\ \end{aligned}$
In the above expression, $\left( 1+\sin x \right)$ is common in numerator and the denominator so we can cancel this expression and we get,
$\Rightarrow \sin x$