Question

Prove the identity \[\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\dfrac{\sec A}{{{\csc }^{2}}A}-\dfrac{\csc A}{{{\sec }^{2}}A}\] .

Answer 1

Hint: To the identity \[\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\dfrac{\sec A}{{{\csc }^{2}}A}-\dfrac{\csc A}{{{\sec }^{2}}A}\] , we have to consider the LHS. We have to first multiply the terms. Then, we have to apply $\cot A=\dfrac{\cos A}{\sin A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ . We have to simplify the terms and apply $\csc A=\dfrac{1}{\sin A}$ and $\sec A=\dfrac{1}{\cos A}$ . After a few rearrangements, the LHS will be equal to the RHS.

Complete step by step solution:
We have to prove the identity \[\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)=\dfrac{\sec A}{{{\csc }^{2}}A}-\dfrac{\csc A}{{{\sec }^{2}}A}\] . Let us consider the LHS.
\[\Rightarrow LHS=\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)\]
Let us multiply the terms.
\[\Rightarrow LHS=\sin A-\cos A+\cot A\sin A-\cot A\cos A+\tan A\sin A-\tan A\cos A\]
We know that $\cot A=\dfrac{\cos A}{\sin A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ . Hence, the above equation becomes
\[\Rightarrow LHS=\sin A-\cos A+\dfrac{\cos A}{\sin A}\times \sin A-\dfrac{\cos A}{\sin A}\times \cos A+\dfrac{\sin A}{\cos A}\times \sin A-\dfrac{\sin A}{\cos A}\times \cos A\]
Let us cancel the common terms from the numerator and denominator and simplify.
\[\Rightarrow LHS=\sin A-\cos A+\cos A-\dfrac{{{\cos }^{2}}A}{\sin A}+\dfrac{{{\sin }^{2}}A}{\cos A}-\sin A\]
We can further cancel the common terms.
\[\Rightarrow LHS=-\dfrac{{{\cos }^{2}}A}{\sin A}+\dfrac{{{\sin }^{2}}A}{\cos A}\]
We know that $\csc A=\dfrac{1}{\sin A}$ and $\sec A=\dfrac{1}{\cos A}$ . Hence, we can write the above form as
\[\begin{align}
  & \Rightarrow LHS=\left( -\dfrac{1}{\dfrac{1}{{{\cos }^{2}}A}}\times \dfrac{1}{\sin A} \right)+\left( \dfrac{1}{\dfrac{1}{{{\sin }^{2}}A}}\times \dfrac{1}{\cos A} \right) \\
 & \Rightarrow LHS=-\dfrac{1}{{{\sec }^{2}}A}\times \csc A+\dfrac{1}{\csc {{\,}^{2}}A}\times \sec A \\
 & \Rightarrow LHS=-\dfrac{\csc A}{{{\sec }^{2}}A}+\dfrac{\sec A}{\csc {{\,}^{2}}A}=RHS \\
\end{align}\]
Therefore, LHS = RHS.
Hence proved.

Note: Students must know the functions thoroughly and how to apply them. They must know that $\csc A=\dfrac{1}{\sin A}$ , $\sec A=\dfrac{1}{\cos A}$ , $\cot A=\dfrac{\cos A}{\sin A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ . They must begin the proof always from the LHS. We can prove the given identity in an alternate method.
Let us consider the LHS.
\[\Rightarrow LHS=\left( 1+\cot A+\tan A \right)\left( \sin A-\cos A \right)\]
Let us multiply the terms.
\[\Rightarrow LHS=\sin A-\cos A+\cot A\sin A-\cot A\cos A+\tan A\sin A-\tan A\cos A\]
We know that $\cot A=\dfrac{\cos A}{\sin A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ . Hence, the above equation becomes
\[\Rightarrow LHS=\sin A-\cos A+\dfrac{\cos A}{\sin A}\times \sin A-\cot A\times \cos A+\tan A\times \sin A-\dfrac{\sin A}{\cos A}\times \cos A\]
Let us cancel the common terms from the numerator and denominator and simplify.
\[\begin{align}
  & \Rightarrow LHS=\sin A-\cos A+\cos A-\cot A\times \cos A+\tan A\times \sin A-\sin A \\
 & \Rightarrow LHS=-\cot A\cos A+\tan A\sin A...\left( i \right) \\
\end{align}\]
Now, let us consider the RHS.
\[\Rightarrow RHS=-\dfrac{\csc A}{{{\sec }^{2}}A}+\dfrac{\sec A}{\csc {{\,}^{2}}A}\]
We know that $\csc A=\dfrac{1}{\sin A}$ and $\sec A=\dfrac{1}{\cos A}$ . Hence, we can write the above form as
\[\begin{align}
  & \Rightarrow RHS=-\dfrac{\dfrac{1}{\sin A}}{\dfrac{1}{{{\cos }^{2}}A}}+\dfrac{\dfrac{1}{\cos A}}{\dfrac{1}{{{\sin }^{2}}A}} \\
 & \Rightarrow RHS=-\dfrac{{{\cos }^{2}}A}{\sin A}+\dfrac{{{\sin }^{2}}A}{\cos A} \\
\end{align}\]
Let us take the LCM.
\[\Rightarrow RHS=-\dfrac{{{\cos }^{3}}A}{\sin A\cos A}+\dfrac{{{\sin }^{3}}A}{\sin A\cos A}\]
Let us take cos A outside from the first term and sin A from the second term.
\[\Rightarrow RHS=-\cos A\times \dfrac{{{\cos }^{2}}A}{\sin A\cos A}+\sin A\times \dfrac{{{\sin }^{2}}A}{\sin A\cos A}\]
Let us cancel the common terms.
\[\Rightarrow RHS=-\cos A\times \dfrac{\cos A}{\sin A}+\sin A\times \dfrac{\sin A}{\cos A}\]
We know that $\cot A=\dfrac{\cos A}{\sin A}$ and $\tan A=\dfrac{\sin A}{\cos A}$ . Hence, the above equation becomes
\[\Rightarrow RHS=-\cos A\cot A+\sin A\tan A...\left( ii \right)\]
From (i) and (ii), we can see that LHS = RHS.
Hence proved.