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Question: How do I show that \[\dfrac{{\cos \theta }}{{1 + \sin \theta }} + \dfrac{{1 + \sin \theta }}{{\cos \...

How do I show that cos⁡θ1+sin⁡θ+1+sin⁡θcos⁡θ=2sec⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }} + \dfrac{{1 + \sin \theta }}{{\cos \theta }} = 2\sec \theta

Explanation

Solution

Hint : We can see that the terms are written as in ab+ba\dfrac{a}{b} + \dfrac{b}{a} . We will take the LCM of the terms. Then we will simplify the terms. That will lead to the standard trigonometric identity sin⁡2t+cos⁡2t{\sin ^2}t + {\cos ^2}t that is equal to 1. Then again rearranging the terms we will reach the answer. Only remember that we cannot directly get to the answer. Or we do not need to add or remove any new term in the expression. Simply LCM will help us in getting the answer.

Complete step-by-step answer :
Given that
cos⁡θ1+sin⁡θ+1+sin⁡θcos⁡θ\dfrac{{\cos \theta }}{{1 + \sin \theta }} + \dfrac{{1 + \sin \theta }}{{\cos \theta }}
Taking LCM,
=cos⁡θ.cos⁡θ+(1+sin⁡θ)(1+sin⁡θ)(1+sin⁡θ)cos⁡θ= \dfrac{{\cos \theta .\cos \theta + \left( {1 + \sin \theta } \right)\left( {1 + \sin \theta } \right)}}{{\left( {1 + \sin \theta } \right)\cos \theta }}
Now multiplying the terms as well as the brackets in the numerator,
=cos⁡2θ+1+sin⁡θ+sin⁡θ+sin⁡2θ(1+sin⁡θ)cos⁡θ= \dfrac{{{{\cos }^2}\theta + 1 + \sin \theta + \sin \theta + {{\sin }^2}\theta }}{{\left( {1 + \sin \theta } \right)\cos \theta }}
Adding the two same terms sin⁡θ\sin \theta ,
=cos⁡2θ+1+2sin⁡θ+sin⁡2θ(1+sin⁡θ)cos⁡θ= \dfrac{{{{\cos }^2}\theta + 1 + 2\sin \theta + {{\sin }^2}\theta }}{{\left( {1 + \sin \theta } \right)\cos \theta }}
Now we will take standard identity terms on one side,
=sin⁡2θ+cos⁡2θ+1+2sin⁡θ(1+sin⁡θ)cos⁡θ= \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta + 1 + 2\sin \theta }}{{\left( {1 + \sin \theta } \right)\cos \theta }}
We know that sin⁡2t+cos⁡2t=1{\sin ^2}t + {\cos ^2}t = 1 so we will substitute the value,
=1+1+2sin⁡θ(1+sin⁡θ)cos⁡θ= \dfrac{{1 + 1 + 2\sin \theta }}{{\left( {1 + \sin \theta } \right)\cos \theta }}
Adding the numbers,
=2+2sin⁡θ(1+sin⁡θ)cos⁡θ= \dfrac{{2 + 2\sin \theta }}{{\left( {1 + \sin \theta } \right)\cos \theta }}
Taking 2 common from the terms in numerator,
=2(1+sin⁡θ)(1+sin⁡θ)cos⁡θ= \dfrac{{2\left( {1 + \sin \theta } \right)}}{{\left( {1 + \sin \theta } \right)\cos \theta }}
Cancelling the common terms,
=2cos⁡θ= \dfrac{2}{{\cos \theta }}
Now we know that 1cos⁡θ=secθ\dfrac{1}{{\cos \theta }} = sec\theta
=2secθ= 2sec\theta
This is our final answer.
So, the correct answer is “2secθ2sec\theta”.

Note : Here note that the trigonometric functions are sine and cosine only used. But we need not to reshuffle any functions. We will just take the LCM in order to simplify the ratios. Also note that simplify the answer as much as we can. Such that =2cos⁡θ = \dfrac{2}{{\cos \theta }} can also be written as =2secθ= 2sec\theta so both steps are correct but try to write near to simplified answers.