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Question: Prove the following statement: \[\dfrac{\operatorname{cosec}A}{\cot A+\tan A}=\cos A\]...

Prove the following statement:
cosec⁡Acot⁡A+tan⁡A=cos⁡A\dfrac{\operatorname{cosec}A}{\cot A+\tan A}=\cos A

Explanation

Solution

Hint: First of all, consider the LHS of the given equation and convert the whole expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta by using the formulas cosec⁡θ=1sin⁡θ,tan⁡θ=1cot⁡θ=sin⁡θcos⁡θ\operatorname{cosec}\theta =\dfrac{1}{\sin \theta },\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta }. Now, simplify the given expression to prove the desired result.
Complete step-by-step answer:
In this question, we have to prove that cosec⁡Acot⁡A+tan⁡A=cos⁡A\dfrac{\operatorname{cosec}A}{\cot A+\tan A}=\cos A. Let us consider the LHS of the equation given in the question.
E=cosec⁡Acot⁡A+tan⁡AE=\dfrac{\operatorname{cosec}A}{\cot A+\tan A}
We know that cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }. By using this in the numerator of the above expression, we get,
E=1sin⁡Acot⁡A+tan⁡AE=\dfrac{\dfrac{1}{\sin A}}{\cot A+\tan A}
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta }. By using these in the denominator of the above expression, we get,
E=1sin⁡Acos⁡Asin⁡A+sin⁡Acos⁡AE=\dfrac{\dfrac{1}{\sin A}}{\dfrac{\cos A}{\sin A}+\dfrac{\sin A}{\cos A}}
By taking sin⁡Acos⁡A\sin A\cos A as LCM in the denominator and simplifying the expression, we get,
E=1sin⁡Acos⁡A.cos⁡A+sin⁡A.sin⁡Asin⁡Acos⁡AE=\dfrac{\dfrac{1}{\sin A}}{\dfrac{\cos A.\cos A+\sin A.\sin A}{\sin A\cos A}}
E=1sin⁡Acos⁡2A+sin⁡2Asin⁡Acos⁡AE=\dfrac{\dfrac{1}{\sin A}}{\dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A}{\sin A\cos A}}
We know that cos⁡2θ+sin⁡2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1. By using this in the above expression, we get,
E=1sin⁡A1sin⁡Acos⁡AE=\dfrac{\dfrac{1}{\sin A}}{\dfrac{1}{\sin A\cos A}}
By simplifying the above expression, we get,
E=1sin⁡A.sin⁡Acos⁡A1E=\dfrac{1}{\sin A}.\dfrac{\sin A\cos A}{1}
By canceling the like terms in the above expression, we get,
E=cos⁡AE=\cos A
E = RHS
So, we get, LHS = RHS
Hence proved.
So, we have proved that cosec⁡Acot⁡A+tan⁡A=cos⁡A\dfrac{\operatorname{cosec}A}{\cot A+\tan A}=\cos A.

Note: In these types of questions, it is always better to convert the whole equation in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta and then simplify it to get the desired result. Also, students often make mistakes while taking the LCM in questions related to trigonometry. So, this must be taken care of. Finally, students must remember the general trigonometric formulas to solve the question easily.