Question

Prove that:
$\left( 1+\cot A-\cos ecA \right)\left( 1+\tan A+\sec A \right)=2$

Answer 1

Hint: First, before proceeding for this, we must know the following conversions of trigonometry to solve this kind of problem as $\cot A=\dfrac{\cos A}{\sin A},\tan A=\dfrac{\sin A}{\cos A},\cos ecA=\dfrac{1}{\sin A},\sec A=\dfrac{1}{\cos A}$. Then, by solving the left hand side of the equation weather it gives the value 2 or not to prove the given expression. Then, by applying the identity as $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get the final result which is equal to the right hand side of expression which in turn proves the given expression is correct.

Complete step by step answer:
In this question, we are supposed to prove that $\left( 1+\cot A-\cos ecA \right)\left( 1+\tan A+\sec A \right)=2$.
So, before proceeding for this, we must know the following conversions of trigonometry to solve this kind of problem as:
$\begin{align}
  & \cot A=\dfrac{\cos A}{\sin A} \\
 & \tan A=\dfrac{\sin A}{\cos A} \\
 & \cos ecA=\dfrac{1}{\sin A} \\
 & \sec A=\dfrac{1}{\cos A} \\
\end{align}$
Then, by substituting the value of all the functions in the given expression, we get:
$\left( 1+\dfrac{\cos A}{\sin A}-\dfrac{1}{\sin A} \right)\left( 1+\dfrac{\sin A}{\cos A}+\dfrac{1}{\cos A} \right)=2$
Now, by solving the left hand side of the equation whether it gives the value 2 or not to prove the given expression.
So, by solving the left hand side of the expression, we get:
$\begin{align}
  & \left( \dfrac{\cos A+\sin A-1}{\sin A} \right)\left( \dfrac{\cos A+\sin A+1}{\cos A} \right) \\
 & \Rightarrow \dfrac{1}{\sin A\cos A}\left( \sin A+\cos A+1 \right)\left( \sin A+\cos A-1 \right) \\
\end{align}$
Then, by applying the identity as $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$, we get:
$\dfrac{1}{\sin A\cos A}\left[ {{\left( \sin A+\cos A \right)}^{2}}-{{1}^{2}} \right]$
Then, by solving the above expression, we get:
$\dfrac{1}{\sin A\cos A}\left[ {{\sin }^{2}}A+{{\cos }^{2}}A+2\sin A\cos A-1 \right]$
Then again by using another trigonometric identity as ${{\sin }^{2}}A+{{\cos }^{2}}A=1$, we get:
$\dfrac{1}{\sin A\cos A}\left[ 1+2\sin A\cos A-1 \right]$
Then, after solving further, we get:
$\dfrac{2\sin A\cos A}{\sin A\cos A}=2$
So, it gives the value as 2 which is equal to the right hand side of the equation as 2.
Hence, the given expression $\left( 1+\cot A-\cos ecA \right)\left( 1+\tan A+\sec A \right)=2$ is proved to be true.

Note:
Now, to solve these types of questions we need to follow the above mentioned pattern exactly the same. We can also multiply all the terms of the expression of the left hand side and then substitute all the values as stated above, then also we get the same result and we can prove the given expression.