Question

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

Answer 1

Hint:Consider the LHS of the equation given in the question and in that use \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta }\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to simplify it. Now, consider the RHS and use \[\tan \theta =\dfrac{1}{\cot \theta }=\dfrac{\sin \theta }{\cos \theta }\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] 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,
\[\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}\]
Let us consider the LHS of the given equation.
\[LHS=\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)\]
We know that \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\]. By using this in the above expression, we get,
\[LHS=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \sec A-\cos A \right)\]
We also know that, \[\sec \theta =\dfrac{1}{\cos \theta }\]. By using this in the above expression, we get,
\[LHS=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]
By simplifying the above expression, we get,
\[LHS=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]
Now, we know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] or \[{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \] and \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]. By using these in the above equation, we get,
\[LHS=\dfrac{{{\cos }^{2}}A}{\sin A}.\dfrac{{{\sin }^{2}}A}{\cos A}\]
Now, by canceling the like terms from the numerator and denominator of the above expression, we get,
\[LHS=\cos A\sin A....\left( i \right)\]
Now, let us consider the RHS of the equation given in the question.
\[RHS=\dfrac{1}{\tan A+\cot A}\]
We know that \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] and \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]. By using this in the above expression, we get,
\[RHS=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}\]
By simplifying the above expression, we get,
\[RHS=\dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\sin A}}\]
\[RHS=\dfrac{\cos A\sin A}{{{\sin }^{2}}A+{{\cos }^{2}}A}\]
We know that,
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
By using this, we get,
\[RHS=\cos A\sin A....\left( ii \right)\]
From equation (i) and (ii), we get,
LHS = RHS
Hence proved
Therefore, we have proved that
\[\left( \operatorname{cosec}A-\sin A \right)\left( \sec A-\cos A \right)=\dfrac{1}{\tan A+\cot A}\]

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