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Question: Prove that : \(\dfrac{1}{{\cos ecA\, - \,\cot A}}\, - \,\dfrac{1}{{\sin A}}\,\, = \,\,\,\dfrac{1}{{\...

Prove that : 1cos⁡ecA − cot⁡A − 1sin⁡A  =   1sin⁡A  −  1cos⁡ecA+cot⁡A\dfrac{1}{{\cos ecA\, - \,\cot A}}\, - \,\dfrac{1}{{\sin A}}\,\, = \,\,\,\dfrac{1}{{\sin A}}\,\, - \,\,\dfrac{1}{{\cos ecA + \cot A}}

Explanation

Solution

We will use the basic trigonometric formulas to solve the question. This type of trigonometric proofs are solved by taking both sides that are Left hand side(LHS) and right hand side(RHS) one by one and simplifying both to obtain the same simplified solution.

Complete step-by-step answer:
We will start the proof by taking LHS and RHS, one by one and then simplifying them.
Firstly we will take Left hand side (LHS) as
LHS =  1cos⁡ecA − cot⁡A − 1sin⁡ALHS\, = \,\,\dfrac{1}{{\cos ecA\, - \,\cot A}}\, - \,\dfrac{1}{{\sin A}}
Now, we know that cos⁡ecθ = 1sin⁡θ  \cos ec\theta \, = \,\dfrac{1}{{\sin \theta }}\,\, and cot⁡θ = 1tan⁡θ = 1sin⁡θcos⁡θ = cos⁡θsin⁡θ\cot \theta \, = \,\dfrac{1}{{\tan \theta }}\, = \,\dfrac{1}{{\dfrac{{\sin \theta }}{{\cos \theta }}}}\, = \,\dfrac{{\cos \theta }}{{\sin \theta }}
So, we will simplify the whole equation in terms of sin⁡\sin and cos⁡\cos , putting the values as
=  11sin⁡A−cos⁡Asin⁡A   −   1sin⁡A= \,\,\dfrac{1}{{\dfrac{1}{{\sin A}} - \dfrac{{\cos A}}{{\sin A}}}}\,\,\, - \,\,\,\dfrac{1}{{\sin A}}
Now, simplifying the above equation as
=   1(1−cos⁡A)sin⁡A  −  1sin⁡A= \,\,\,\dfrac{1}{{\dfrac{{\left( {1 - \cos A} \right)}}{{\sin A}}}}\,\, - \,\,\dfrac{1}{{\sin A}}
=   sin⁡A(1−cos⁡A)  −  1sin⁡A= \,\,\,\dfrac{{\sin A}}{{\left( {1 - \cos A} \right)}}\,\, - \,\,\dfrac{1}{{\sin A}}
Now, we will cross multiply for further simplification,
=   sin⁡2A−(1−cos⁡A)sin⁡A(1−cos⁡A)= \,\,\,\dfrac{{{{\sin }^2}A - \left( {1 - \cos A} \right)}}{{\sin A\left( {1 - \cos A} \right)}}
Now, we have the trigonometric identity as sin⁡2A + cos⁡2A = 1  ..................(i){\sin ^2}A\, + \,{\cos ^2}A\, = \,1\,\,..................\left( i \right)
From the identity we will calculate the value of cos⁡2A{\cos ^2}A as 1−sin⁡2A1 - {\sin ^2}A and substitute in above equation as
=  (1−cos⁡2A) − (1−cos⁡A)sin⁡A(1−cos⁡A)= \,\,\dfrac{{\left( {1 - {{\cos }^2}A} \right)\, - \,\left( {1 - \cos A} \right)}}{{\sin A\left( {1 - \cos A} \right)}}
Now we have the identity as (a2−b2) = (a+b)(a−b) .................(ii)\left( {{a^2} - {b^2}} \right)\, = \,\left( {a + b} \right)\left( {a - b} \right)\,.................\left( {ii} \right)
so we will use it for (1−cos⁡2A)\left( {1 - {{\cos }^2}A} \right)
=  (1−cos⁡A)(1+cosA⁡) − (1−cos⁡A)sin⁡A(1−cos⁡A)= \,\,\dfrac{{\left( {1 - \cos A} \right)\left( {1 + \operatorname{cosA} } \right)\, - \,\left( {1 - \cos A} \right)}}{{\sin A\left( {1 - \cos A} \right)}}
Now taking (1−cos⁡A)\left( {1 - \cos A} \right) common from numerator
=   (1−cos⁡A)[1+cos⁡A−1]sin⁡A(1−cos⁡A)= \,\,\,\dfrac{{\left( {1 - \cos A} \right)\left[ {1 + \cos A - 1} \right]}}{{\sin A\left( {1 - \cos A} \right)}}
Now cancelling the term (1−cos⁡A)\left( {1 - \cos A} \right) from both numerator and denominator
=  cos⁡AsinA⁡= \,\,\dfrac{{\cos A}}{{\operatorname{sinA} }}
=  cotA⁡= \,\,\operatorname{cotA}
Now, similarly we will solve for right hand side as
RHS  = 1sinA⁡  −  1cos⁡ecA+cot⁡ARHS\,\, = \,\dfrac{1}{{\operatorname{sinA} }}\,\, - \,\,\dfrac{1}{{\cos ecA + \cot A}}
Simplifying in terms of sin⁡\sin and cos⁡\cos as
=  1sin⁡A  −  11sin⁡A + cos⁡Asin⁡A= \,\,\dfrac{1}{{\sin A}}\,\, - \,\,\dfrac{1}{{\dfrac{1}{{\sin A}}\, + \,\dfrac{{\cos A}}{{\sin A}}}}
=  1sinA⁡ −  1(1+cos⁡A)sin⁡A= \,\,\dfrac{1}{{\operatorname{sinA} }}\, - \,\,\dfrac{1}{{\dfrac{{\left( {1 + \cos A} \right)}}{{\sin A}}}}
=  1sin⁡A − sin⁡A(1+cos⁡A)= \,\,\dfrac{1}{{\sin A}}\, - \,\dfrac{{\sin A}}{{\left( {1 + \cos A} \right)}}
Again cross multiplying and solving as
=  (1+cos⁡A)−sin⁡2Asin⁡A(1+cosA⁡)= \,\,\dfrac{{\left( {1 + \cos A} \right) - {{\sin }^2}A}}{{\sin A\left( {1 + \operatorname{cosA} } \right)}}
Now again substituting value of sin⁡2A{\sin ^2}A by using the identity (i)\left( i \right) as
=  (1+cos⁡A)−(1−cos⁡2A)sin⁡A(1+cos⁡A)= \,\,\dfrac{{\left( {1 + \cos A} \right) - \left( {1 - {{\cos }^2}A} \right)}}{{\sin A\left( {1 + \cos A} \right)}}
Now using identity (ii)\left( {ii} \right) as
=   (1+cos⁡A)−(1−cos⁡A)(1+cos⁡A)sin⁡A(1+cos⁡A)= \,\,\,\dfrac{{\left( {1 + \cos A} \right) - \left( {1 - \cos A} \right)\left( {1 + \cos A} \right)}}{{\sin A\left( {1 + \cos A} \right)}}
Taking (1+cos⁡A)\left( {1 + \cos A} \right) common and cancelling with the denominator we get
=  [1−1+cos⁡A]sin⁡A= \,\,\dfrac{{\left[ {1 - 1 + \cos A} \right]}}{{\sin A}}
=  cos⁡Asin⁡A= \,\,\dfrac{{\cos A}}{{\sin A}}
=  cotA⁡= \,\,\operatorname{cotA}
=  LHS= \,\,LHS
Thus from the above solution we can prove that LHS=RHS.

Note: The above proof can also be done by simplifying the equation in terms of cos⁡ec\cos ec and cot⁡\cot . For this we will multiply the numerator and denominator by a common term (cos⁡ecA+cot⁡A)\left( {\cos ecA + \cot A} \right), which will not change the value of terms in LHS or RHS but will help to simplify the equation.
Try to simplify the problem using simple sin⁡\sin and cos⁡\cos functions, because they are the basic functions and a variety of identities are based on them, which will help in further simplifications. If you try to solve using cos⁡ec\cos ec and cot⁡\cot then make sure you know all the identities related to them, otherwise the equation may become complex and hard to solve.