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Question

Question: What is \(\cot A + \cos ecA\) equal to?...

What is cotA+cosecA\cot A + \cos ecA equal to?

Explanation

Solution

The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as cosec(x)=1sin(x)\cos ec(x) = \dfrac{1}{{\sin (x)}} and cot(x)=cos(x)sin(x)\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}} . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem. We must know the half angle formulae of trigonometric functions in order to solve the question.

Complete step by step answer:
In the given problem, we have to simplify the trigonometric expression cotA+cosecA\cot A + \cos ecA.
So, we have, cotA+cosecA\cot A + \cos ecA
Now, using the trigonometric formulae cosec(x)=1sin(x)\cos ec(x) = \dfrac{1}{{\sin (x)}} and cot(x)=cos(x)sin(x)\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}}, we get,
cosAsinA+1sinA\Rightarrow \dfrac{{\cos A}}{{\sin A}} + \dfrac{1}{{\sin A}}
Since the denominators are equal, adding up the numerators, we get,
cosA+1sinA\Rightarrow \dfrac{{\cos A + 1}}{{\sin A}}
Using the half angle formula of cosine cosx=2cos2(x2)1\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1 in numerator, we get,
(2cos2(A2)1)+1sinA\Rightarrow \dfrac{{\left( {2{{\cos }^2}\left( {\dfrac{A}{2}} \right) - 1} \right) + 1}}{{\sin A}}
Now, using the half angle formula of sine sinx=2sinx2cosx2\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}, we get,
2cos2(A2)1+12sinA2cosA2\Rightarrow \dfrac{{2{{\cos }^2}\left( {\dfrac{A}{2}} \right) - 1 + 1}}{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}
Cancelling the like terms with opposite signs, we get,
2cos2(A2)2sinA2cosA2\Rightarrow \dfrac{{2{{\cos }^2}\left( {\dfrac{A}{2}} \right)}}{{2\sin \dfrac{A}{2}\cos \dfrac{A}{2}}}
Cancelling the common factors in numerator and denominator, we get,
cos(A2)sin(A2)\Rightarrow \dfrac{{\cos \left( {\dfrac{A}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}
Now, using the trigonometric formula of cotangent as cotx=cosxsinx\cot x = \dfrac{{\cos x}}{{\sin x}}, we get,
cot(A2)\Rightarrow \cot \left( {\dfrac{A}{2}} \right)
Hence, cotA+cosecA=cotA2\cot A + \cos ecA = \cot \dfrac{A}{2}

Note:
Given problem deals with trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart such as: cosec(x)=1sin(x)\cos ec(x) = \dfrac{1}{{\sin (x)}} and cot(x)=cos(x)sin(x)\cot (x) = \dfrac{{\cos (x)}}{{\sin (x)}} . Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations. We must remember the half angle formulae of sine and cosine as cosx=2cos2(x2)1\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1 and sinx=2sinx2cosx2\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} to solve the given problem. However, questions involving this type of simplification of trigonometric ratios may also have multiple interconvertible answers.