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Question

Question: How do you solve \(\dfrac{1+\cos x}{1+\sec x}\) ?...

How do you solve 1+cosx1+secx\dfrac{1+\cos x}{1+\sec x} ?

Explanation

Solution

To solve the given trigonometric expression i.e. 1+cosx1+secx\dfrac{1+\cos x}{1+\sec x}, we are going to use the property of secx\sec x which is equal to the reciprocal of cosx\cos x and mathematically we can write secx=1cosx\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:
1+cosx1+secx\dfrac{1+\cos x}{1+\sec x}
We know the relation between secx&cosx\sec x\And \cos x is equal to:
secx=1cosx\sec x=\dfrac{1}{\cos x}
Using the above relation in the given trigonometric expression we get,
1+cosx1+1cosx\Rightarrow \dfrac{1+\cos x}{1+\dfrac{1}{\cos x}}
Now, taking L.C.M of cosx\cos x in the denominator in the above expression we get,
1+cosxcosx+1cosx\Rightarrow \dfrac{1+\cos x}{\dfrac{\cos x+1}{\cos x}}
Rearranging the above trigonometric expression we get,
cosx(1+cosx)1+cosx\Rightarrow \dfrac{\cos x\left( 1+\cos x \right)}{1+\cos x}
In the above expression, (1+cosx)\left( 1+\cos x \right) is common in the numerator and denominator so (1+cosx)\left( 1+\cos x \right) will be cancelled out from the numerator and the denominator and we get,
cosx\Rightarrow \cos x
From the above simplification of the given trigonometric expression we get,
cosx\Rightarrow \cos x

Note: You might think how we know whether to use secx=1cosx\sec x=\dfrac{1}{\cos x}, the answer is in the numerator you can see there is a term cosx\cos x so if we can use this secx&cosx\sec x\And \cos x relation then there is a possibility in which cosx\cos x term might get cancelled and which you can see is the case in the above solution when we use this relation (1+cosx)\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:
1+sinx1+cosecx\dfrac{1+\sin x}{1+\text{cosec}x}
So, here we can use the relation between cosecx=1sinx\text{cosec}x=\dfrac{1}{\sin x} and then simplify and we get,
1+sinx1+1sinx\Rightarrow \dfrac{1+\sin x}{1+\dfrac{1}{\sin x}}
Taking sinx\sin x as L.C.M in the denominator of the above expression we get,
1+sinxsinx+1sinx =sinx(1+sinx)1+sinx \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, (1+sinx)\left( 1+\sin x \right) is common in numerator and the denominator so we can cancel this expression and we get,
sinx\Rightarrow \sin x