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Question: How do you simplify the expression \(\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{1 + \sin x}}\) ?...

How do you simplify the expression sinxcosx+cosx1+sinx\dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{1 + \sin x}} ?

Explanation

Solution

In order to solve this sum, we need to simplify the given trigonometric equation. First we take the LCM of the denominators, and then we multiply the numerators with respective denominators accordingly. We simply sum and solve it further to get our required answer.

Complete step-by-step solution:
In the given question, we are asked to simplify the given trigonometric expression which is
sinxcosx+cosx1+sinx\Rightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\cos x}}{{1 + \sin x}}
Let us take the LCM of the denominators, thus we have:
sinx(1+sinx)+cosx(cosx)(cosx)(1+sinx)\Rightarrow \dfrac{{\sin x\left( {1 + \sin x} \right) + \cos x\left( {\cos x} \right)}}{{\left( {\cos x} \right)\left( {1 + \sin x} \right)}}
Let us simplify the numerators by multiplying the respective terms:
sinx+sin2x+cos2x(cosx)(1+sinx)\Rightarrow \dfrac{{\sin x + {{\sin }^2}x + {{\cos }^2}x}}{{\left( {\cos x} \right)\left( {1 + \sin x} \right)}}
Now as we know that: sin2x+cos2x=1{\sin ^2}x + {\cos ^2}x = 1
sinx+1(cosx)(1+sinx)\Rightarrow \dfrac{{\sin x + 1}}{{\left( {\cos x} \right)\left( {1 + \sin x} \right)}}
On solving it further, we find that:
̸(1+sinx)(cosx)̸(1+sinx)\Rightarrow \dfrac{{\not{{\left( {1 + \sin x} \right)}}}}{{\left( {\cos x} \right)\not{{\left( {1 + \sin x} \right)}}}}
1cosx=secx\Rightarrow \dfrac{1}{{\cos x}} = \sec x → as it is the inverse of cosine.

secx\sec x is the answer for the given question.

Note: Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot. Some common formulas of trigonometric identities are:
sinθ=perpendicularhypotenuse{\text{sin}}\theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}} , where perpendicular is the side containing the right angle in a right angled triangle and hypotenuse is the side opposite to the perpendicular.
cosθ=basehypotenuse{\text{cos}}\theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}} , where base is the side containing the perpendicular and hypotenuse
tanθ=perpendicularbase{\text{tan}}\theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}
Now in order to solve questions like these, one should be careful with the sign of the trigonometric functions. And we should also remember the different inverse of the main trigonometric functions like:
Cosec is the inverse of sin 1sinθ \to \dfrac{1}{{\sin \theta }}
Sec is the inverse of cos 1cosθ \to \dfrac{1}{{\cos \theta }}
Cot is the inverse of tan 1tanθ \to \dfrac{1}{{\tan \theta }}