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Question: Prove the following \[\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfr...

Prove the following
(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}

Explanation

Solution

Hint:Consider the LHS of the equation given in the question and in that use cosec⁡θ=1sin⁡θ,sec⁡θ=1cos⁡θ\operatorname{cosec}\theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta } and sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 to simplify it. Now, consider the RHS and use tan⁡θ=1cot⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta } and sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 to simplify it. From this get, LHS = RHS and prove the desired result.

Complete step-by-step answer:
In this question, we have to prove that,
(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}
Let us consider the LHS of the given equation.
LHS=(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)LHS=\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)
We know that cosec⁡θ=1sin⁡θ\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }. By using this in the above expression, we get,
LHS=(1sin⁡A−sin⁡A)(sec⁡A−cos⁡A)LHS=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \sec A-\cos A \right)
We also know that, sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta }. By using this in the above expression, we get,
LHS=(1sin⁡A−sin⁡A)(1cos⁡A−cos⁡A)LHS=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)
By simplifying the above expression, we get,
LHS=(1−sin⁡2Asin⁡A)(1−cos⁡2Acos⁡A)LHS=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)
Now, we know that, sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 or cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta and sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta . By using these in the above equation, we get,
LHS=cos⁡2Asin⁡A.sin⁡2Acos⁡ALHS=\dfrac{{{\cos }^{2}}A}{\sin A}.\dfrac{{{\sin }^{2}}A}{\cos A}
Now, by canceling the like terms from the numerator and denominator of the above expression, we get,
LHS=cos⁡Asin⁡A....(i)LHS=\cos A\sin A....\left( i \right)
Now, let us consider the RHS of the equation given in the question.
RHS=1tan⁡A+cot⁡ARHS=\dfrac{1}{\tan A+\cot 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 this in the above expression, we get,
RHS=1sin⁡Acos⁡A+cos⁡Asin⁡ARHS=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}
By simplifying the above expression, we get,
RHS=1sin⁡2A+cos⁡2Acos⁡Asin⁡ARHS=\dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\sin A}}
RHS=cos⁡Asin⁡Asin⁡2A+cos⁡2ARHS=\dfrac{\cos A\sin A}{{{\sin }^{2}}A+{{\cos }^{2}}A}
We know that,
sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1
By using this, we get,
RHS=cos⁡Asin⁡A....(ii)RHS=\cos A\sin A....\left( ii \right)
From equation (i) and (ii), we get,
LHS = RHS
Hence proved
Therefore, we have proved that
(cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡A+cot⁡A\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}

Note: In this question, after getting LHS = sin A cos A, students can write it as, sin⁡Acos⁡A1\dfrac{\sin A\cos A}{1} and replace 1 by sin⁡2A+cos⁡2A{{\sin }^{2}}A+{{\cos }^{2}}A. Now divide the numerator and denominator by sin A cos A to get the RHS by using tan⁡θ=1cot⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta }. But we generally don’t use this because it is very tough for students to get this idea of replacing 1 by sin⁡2A+cos⁡2A{{\sin }^{2}}A+{{\cos }^{2}}A though this technique is useful in many questions.