Question

Prove the following equation by proving that the left hand side is equal to the right hand side.
\[\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\dfrac{1+\sin x}{\cos x}\]

Answer 1

Hint:In order to solve the above question containing various trigonometric identities we will just do multiplication by an expression to the numerator and the denominator which is a conjugate of the denominator. Also, we will use the identity given as follows:
\[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]

Complete step-by-step answer:
We have been given to prove that the left hand side of the following equation is equal to the right hand side of the following equation. So we will begin solving the equation by solving the left hand side of the equation and will try to reach the right hand side of the equation.
Solving the left hand side of the equation as follows:
\[\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\dfrac{1+\sin x}{\cos x}\]
On multiplying the left hand side of the equality by \[\left( 1+\operatorname{cosx}+sinx \right)\] to the numerator as well as denominator of the equation, we get the expression as follows,
\[\begin{align}
  & \Rightarrow \dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}\times \dfrac{\left( 1+\cos x \right)+\sin x}{\left( 1+\cos x \right)+\sin x} \\
 & =\dfrac{\left( 1+\cos x \right)+\sin x}{\left( 1+\cos x \right)-\sin x}\times \dfrac{\left( 1+\cos x \right)+\sin x}{\left( 1+\cos x \right)+\sin x} \\
 & =\dfrac{\left( 1+\cos x \right)\left( 1+\cos x \right)+{{\sin }^{2}}x+2\left( 1+\cos x \right)\sin x}{\left( 1+\cos x \right)\left( 1+\cos x \right)-{{\sin }^{2}}x} \\
 & =\dfrac{{{\left( 1+\cos x \right)}^{2}}+{{\sin }^{2}}x+2\sin x+2\sin x\cos x}{{{\left( 1+\cos x \right)}^{2}}-{{\sin }^{2}}x} \\
 & =\dfrac{1+{{\cos }^{2}}x+2\cos x+{{\sin }^{2}}x+2\sin x+2\sin x\cos x}{1+{{\cos }^{2}}x-{{\sin }^{2}}x+2\cos x} \\
 & =\dfrac{1+\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)+2\cos x+2\sin x+2\sin x\cos x}{\left( 1-{{\sin }^{2}}x \right)+{{\cos }^{2}}x+2\cos x} \\
\end{align}\]
Now by using the identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] in the above expression, we get as follows:
\[\begin{align}
  & =\dfrac{1+1+2\cos x+2\sin x+2\sin x\cos x}{{{\cos }^{2}}x+{{\cos }^{2}}x+2\cos x} \\
 & =\dfrac{2+2\cos x+2\sin x+2\sin x\cos x}{2{{\cos }^{2}}x+2\cos x} \\
 & {\text{By taking 2 as common in numerator}} \\
 & =\dfrac{2\left( 1+\cos x \right)+2\sin x\left( 1+\cos x \right)}{2{{\cos }^{2}}x+2\cos x} \\
 & =\dfrac{2\left( 1+\cos x \right)\left( 1+\sin x \right)}{2\cos x\left( 1+\cos x \right)} \\
 & =\dfrac{1+\sin x}{\cos x}=RHS \\
\end{align}\]
So, we have proved \[\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}\] = \[\dfrac{1+\sin x}{\cos x}\] which means that we have proved that the left hand side is equal to the right hand side.

Note: Be careful of the sign while multiplying the numerator and denominator of the given expression by \[\left[ \left( 1+\cos x \right)+\sin x \right]\]. Also, you have to rearrange all the terms by taking common terms and solving and then simplifying by using the identity of \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] after simplification of the equation.Students should remember important trigonometric identities and formulas for solving these types of questions.