Question

Prove that \[\left( \csc A-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1}{\tan A+\cot A} \right)\].

Answer 1

Hint: Use the conversion: - \[\csc A=\dfrac{1}{\sin A}\] and \[\sec A=\dfrac{1}{\cos A}\] to simplify the terms in L.H.S. Now, take L.C.M inside the bracket and use the identity: - \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] for further simplification. Write \[\sin A\cos A\] in the numerator as \[\dfrac{1}{\left( \dfrac{1}{\sin A\cos A} \right)}\] and replace 1 with \[{{\sin }^{2}}A+{{\cos }^{2}}A\] in the denominator. Separate the terms to get the proof.

Complete step-by-step solution
Here, we have to prove the given expression: -
\[\Rightarrow \left( \csc A-\sin A \right)\left( \sec A-\cos A \right)=\left( \dfrac{1}{\tan A+\cot A} \right)\]
Here, we have to use some trigonometric identities, so let us see what are trigonometric identities.
In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving angles and side lengths of a triangle.
Now, let us come to the question. We have been given: -
L.H.S = \[\left( \csc A-\sin A \right)\left( \sec A-\cos A \right)\]
Using the conversion: - \[\csc A=\dfrac{1}{\sin A}\] and \[\sec A=\dfrac{1}{\cos A}\], we get,
\[\Rightarrow \] L.H.S = \[\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]
\[\Rightarrow \] L.H.S = \[\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]
Now, using the trigonometric identity: - \[1-{{\sin }^{2}}A={{\cos }^{2}}A\] and \[1-{{\cos }^{2}}A={{\sin }^{2}}A\] in the numerator of first and second term respectively, we get,
\[\Rightarrow \] L.H.S = \[\dfrac{{{\cos }^{2}}A}{\sin A}\times \dfrac{{{\sin }^{2}}A}{\cos A}\]
\[\Rightarrow \] L.H.S = \[\cos A\times \sin A\]
This can be written as: -
\[\Rightarrow \] L.H.S = \[\dfrac{1}{\left( \dfrac{1}{\cos A\sin A} \right)}\]
Using the identity: - \[1={{\sin }^{2}}A+{{\cos }^{2}}A\] to replace 1 in the denominator, we get,
\[\Rightarrow \] L.H.S = \[\dfrac{1}{\left( \dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\sin A} \right)}\]
Separating the terms, we get,
\[\Rightarrow \] L.H.S = \[\dfrac{1}{\left( \dfrac{{{\sin }^{2}}A}{\cos A\sin A}+\dfrac{{{\cos }^{2}}A}{\cos A\sin A} \right)}\]
\[\Rightarrow \] L.H.S = \[\dfrac{1}{\left( \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} \right)}\]
Using the conversion: - \[\dfrac{\sin A}{\cos A}=\tan A\] and \[\dfrac{\cos A}{\sin A}=\cot A\], we get,
\[\Rightarrow \] L.H.S = \[\dfrac{1}{\tan A+\cot A}\]
\[\Rightarrow \] L.H.S = R.H.S
Hence, proved

Note: One may note that we can also apply a reverse process to solve this question. That means we can start with the right-hand side to simplify the equation and reach towards the left-hand side. There can be a third process also. What we can do is, we start solving with L.H.S and come to a certain step and stop. Then we start solving the R.H.S side and try to get to that step where we had stopped the L.H.S.