Question

Prove the following equation: $\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\dfrac{1+\sin x}{\cos x}$

Answer 1

Hint: Assuming we have to prove the following relation that the LHS=RHS, we multiply the numerator and denominator of the expression on the left-hand side of the equation with the conjugate of the denominator. We then simplify the expression obtained and prove that the LHS=RHS.

Complete step by step solution:
Consider the left-hand side of the equation,
$\Rightarrow \dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}$
The given question can be simplified by considering the conjugate of the denominator of the left-hand side of the equation. This conjugate of the denominator is given by,
$\Rightarrow 1+\cos x+\sin x$
We multiply this with the numerator and denominator of the expression given above,
$\Rightarrow \dfrac{\left( 1+\cos x+\sin x \right)\left( 1+\cos x+\sin x \right)}{\left( 1+\cos x-\sin x \right)\left( 1+\cos x+\sin x \right)}$
The numerator has the same two terms multiplied; hence it will be squared. The denominator is of the form $\left( a-b \right).\left( a+b \right)={{a}^{2}}-{{b}^{2}}.$ Here, a is taken as $\left( 1+\cos x \right)$ and b as $\sin x.$
$\Rightarrow \dfrac{{{\left( 1+\cos x+\sin x \right)}^{2}}}{{{\left( 1+\cos x \right)}^{2}}-{{\sin }^{2}}x}$
Expanding the numerator by considering $1$ as a and $\cos x+\sin x$ as b and using in the formula for ${{\left( a+b \right)}^{2}},$
$\Rightarrow \dfrac{{{1}^{2}}+{{\left( \cos x+\sin x \right)}^{2}}+2.\left( \cos x+\sin x \right)}{{{\left( 1+\cos x \right)}^{2}}-{{\sin }^{2}}x}$
Expanding the expressions in the numerator and denominator using the same ${{\left( a+b \right)}^{2}}$ formula,
$\Rightarrow \dfrac{{{1}^{2}}+{{\cos }^{2}}x+{{\sin }^{2}}x+2.\cos x\sin x+2.\left( \cos x+\sin x \right)}{1+{{\cos }^{2}}x+2\cos x-{{\sin }^{2}}x}$
We know the identity that,
${{\sin }^{2}}x+{{\cos }^{2}}x=1\ldots \ldots \left( 1 \right)$
Substituting 1 for this ${{\sin }^{2}}x+{{\cos }^{2}}x$ terms in the numerator, and grouping the $1-{{\sin }^{2}}x$ terms together in the denominator,
$\Rightarrow \dfrac{1+1+2.\cos x\sin x+2.\left( \cos x+\sin x \right)}{\left( 1-{{\sin }^{2}}x \right)+{{\cos }^{2}}x+2\cos x}$
Rearranging equation (1), we get ${{\cos }^{2}}x=1-{{\sin }^{2}}x.$ Substituting this and adding the constants in the numerator,
$\Rightarrow \dfrac{2+2.\cos x\sin x+2.\left( \cos x+\sin x \right)}{{{\cos }^{2}}x+{{\cos }^{2}}x+2\cos x}$
Adding the two ${{\cos }^{2}}x$ terms in the denominator,
$\Rightarrow \dfrac{2+2.\cos x\sin x+2.\left( \cos x+\sin x \right)}{2{{\cos }^{2}}x+2\cos x}$
Taking the 2 common out from the numerator and denominator,
$\Rightarrow \dfrac{2.\left( 1+\cos x\sin x+\cos x+\sin x \right)}{2.\left( {{\cos }^{2}}x+\cos x \right)}$
Cancelling the 2 from the numerator and denominator,
$\Rightarrow \dfrac{1+\cos x\sin x+\cos x+\sin x}{{{\cos }^{2}}x+\cos x}$
Taking the $\sin x$ term in the numerator common and the $\cos x$ term common from the denominator,
$\Rightarrow \dfrac{\left( 1+\cos x \right)+\sin x\left( 1+\cos x \right)}{\cos x\left( 1+\cos x \right)}$
Taking the $1+\cos x$ term common out,
$\Rightarrow \dfrac{\left( 1+\cos x \right)\left( 1+\sin x \right)}{\left( 1+\cos x \right).\cos x}$
Cancelling the $1+\cos x$ terms,
$\Rightarrow \dfrac{1+\sin x}{\cos x}$
This is nothing but the term on the RHS.
Hence, we have proved that LHS=RHS, that is, $\dfrac{1+\cos x+\sin x}{1+\cos x-\sin x}=\dfrac{1+\sin x}{\cos x}.$

Note: Students need to know the trigonometric identities and expansions as these are the basic formulae required to solve such questions. Care must be taken to consider only the conjugate of the denominator because if we take the conjugate of the numerator and solve, the equation becomes more and more complex and we do not get a solution.