Question

Prove the following identity, where the angles involved are acute angles for which the expressions are defined.
\[\dfrac{\cos A-\sin A+1}{\cos A+\sin A+1}=\text{cosec}A+\cot A\]

Answer 1

Hint: To do this question, firstly we will multiply numerator and denominator by sinA. Then, we will solve the brackets and simplify the expression by the use of identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to simplify numerator. After that, we will use algebraic identity \[{{a}^{2}}-{{b}^{2}}=(a-b)\left( a+b \right)\] and we will take the common factors out from both numerator and denominator and at last we will left with only expression equals to RHS which is cosecA + cotA.

Complete step by step answer:
In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.
The below mentioned formulae may be used before solving, in the solution which is as follows
\[\begin{align}
  & \tan x=\dfrac{\sin x}{\cos x} \\
 & \cot x=\dfrac{\cos x}{\sin x} \\
 & \text{cosec}x=\dfrac{1}{\sin x} \\
 & \sec x=\dfrac{1}{\cos x} \\
\end{align}\]
Now, these are the results that would be used to prove the proof mentioned in this question as using these identities, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.
In this particular question, we will first convert all the trigonometric functions in terms of sin and cos function and then we can convert the expression in terms of tan and cot function and then we will try to make the L.H.S. and the R.H.S. equal.
As mentioned in the question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows
\[\Rightarrow LHS=\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\]
\[\Rightarrow LHS=\dfrac{\sin A\cdot \left( \cos A-\sin A+1 \right)}{\sin A\cdot \left( \cos A+\sin A-1 \right)}\]
\[\Rightarrow LHS=\dfrac{\sin A\cdot \cos A-{{\sin }^{2}}A+\sin A}{\sin A\cdot \left( \cos A+\sin A-1 \right)}\]
\[\begin{align}
  & we\text{ know that, }{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & \Rightarrow LHS=\dfrac{\sin A\cdot \cos A+\sin A-\left( 1-{{\cos }^{2}}A \right)}{\sin A\cdot \left( \cos A+\sin A-1 \right)} \\
 & also,\text{ we know that}\left( {{a}^{2}}-{{b}^{2}}=(a-b)\left( a+b \right) \right) \\
 &\Rightarrow LHS=\dfrac{\sin A\cdot \cos A+\sin A-\left( 1+\cos A \right)\cdot \left( 1-\cos A \right)}{\sin A\cdot \left( \cos A+\sin A-1 \right)} \\
\end{align}\]
Taking the common terms outside, we get
\[\begin{align}
  &\Rightarrow LHS =\dfrac{\sin A\cdot \left( \cos A+1 \right)-\left( 1+\cos A \right)\cdot \left( 1-\cos A \right)}{\sin A\cdot \left( \cos A+\sin A-1 \right)} \\
 & \Rightarrow LHS=\dfrac{\left( \cos A+1 \right)\left( \sin A-\left( 1-\cos A \right) \right)}{\sin A\cdot \left( \cos A+\sin A-1 \right)} \\
 & \Rightarrow LHS=\dfrac{\left( \cos A+1 \right)\left( \sin A+\cos A-1 \right)}{\sin A\cdot \left( \cos A+\sin A-1 \right)} \\
 &\Rightarrow LHS=\dfrac{\cos A+1}{\sin A} \\
 &\Rightarrow LHS =\dfrac{\cos A}{\sin A}+\dfrac{1}{\sin A} \\
 &\Rightarrow LHS =\cot A+\text{cosec}A \\
\end{align}\]
(Using the identity of trigonometry which is \[({{\sin }^{2}}A+{{\cos }^{2}}A=1)\] )
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.

Note: Always remember some of the trigonometric identities such as \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \],\[1+{{\cot }^{2}}\theta ={\text{cosec}^{2}}\theta \] also some algebraic identities such as \[{{a}^{2}}-{{b}^{2}}=(a-b)\left( a+b \right)\] and \[{{a}^{2}}-2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}\]. While rationalizing, remember that you do not misplace or forget any of the signs of terms as this will lead to wrong answers. Try not to make any calculation errors while doing the solution of the question.