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Question: How do you simplify \(\dfrac{{\sin \theta }}{{1 + \cos \theta }} + \dfrac{{1 + \cos \theta }}{{\sin ...

How do you simplify sinθ1+cosθ+1+cosθsinθ\dfrac{{\sin \theta }}{{1 + \cos \theta }} + \dfrac{{1 + \cos \theta }}{{\sin \theta }}?

Explanation

Solution

In this question we will try to take the Lowest common multiple to get a common denominator and then we will use the trigonometric identities to simplify the equation and get the final answer.

Complete step-by-step solution:
We have the equation:
sinθ1+cosθ+1+cosθsinθ\Rightarrow \dfrac{{\sin \theta }}{{1 + \cos \theta }} + \dfrac{{1 + \cos \theta }}{{\sin \theta }}
Now since it is in the form of addition of two fractions, we will find the L.C.M as:
sinθ×(sinθ)1+cosθ×(sinθ)+(1+cosθ)×(1+cosθ)sinθ×(1+cosθ)\Rightarrow \dfrac{{\sin \theta \times (\sin \theta )}}{{1 + \cos \theta \times (\sin \theta )}} + \dfrac{{(1 + \cos \theta ) \times (1 + \cos \theta )}}{{\sin \theta \times (1 + \cos \theta )}}
Now since the denominator is same in both the fractions and since they are addition, we can write the equation as:
sinθ×(sinθ)+(1+cosθ)×(1+cosθ)(1+cosθ)×sinθ\Rightarrow \dfrac{{\sin \theta \times (\sin \theta ) + (1 + \cos \theta ) \times (1 + \cos \theta )}}{{(1 + \cos \theta ) \times \sin \theta }}
On simplifying the numerator, we get:
sin2θ+1+2cosθ+cos2θ(1+cosθ)×sinθ\Rightarrow \dfrac{{{{\sin }^2}\theta + 1 + 2\cos \theta + {{\cos }^2}\theta }}{{(1 + \cos \theta ) \times \sin \theta }}
Now since we know that cos2θ+sin2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1 the numerator can be simplified and written as:
1+1+2cosθ(1+cosθ)×sinθ\Rightarrow \dfrac{{1 + 1 + 2\cos \theta }}{{(1 + \cos \theta ) \times \sin \theta }}
On simplifying we get:
2+2cosθ(1+cosθ)×sinθ\Rightarrow \dfrac{{2 + 2\cos \theta }}{{(1 + \cos \theta ) \times \sin \theta }}
Now since the number 22 is common in both the terms in the numerator, we can take it out and write it as:
2(1+cosθ)(1+cosθ)×sinθ\Rightarrow \dfrac{{2(1 + \cos \theta )}}{{(1 + \cos \theta ) \times \sin \theta }}
Now since 1+cosθ1 + \cos \theta is common in both the numerator and denominator we can cancel and write it as:
2sinθ\Rightarrow \dfrac{2}{{\sin \theta }}
Now we know that 1sinθ=cscθ\dfrac{1}{{\sin \theta }} = \csc \theta

Therefore, on substituting we get:
2cscθ\Rightarrow 2\csc \theta, which is the final answer.

Note: It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken.
The various trigonometric identities and formulae should be remembered while doing these types of sums. The various Pythagorean identities should also be remembered while doing these types of questions.
To simplify any given equation, it is good practice to convert all the identities into sin\sin and cos\cos for simplifying.
If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.