Question

Prove LHS=RHS for the following equation
$\dfrac{{{{\cot }^2}x}}{{1 + cosecx}} = \dfrac{{1 - \sin x}}{{\sin x}}$

Answer 1

In order to solve the given question, we will take one side of the equation and simplify it using reciprocal and quotient identities. After successful application of the identities, we will solve the equation using algebraic identity until we get the same expression as on the other side.

Formula used: One of the most important identity that we have used in the above question is:
${x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)$
The other trigonometric identities we have used are:
${\cot ^2}x = \cos e{c^2}x - 1$
And
The standard reciprocal identity of sine function, that is,
$\cos ecx = \dfrac{1}{{\sin x}}$

Complete step-by-step solution:
Given,
$\dfrac{{{{\cot }^2}x}}{{1 + cosecx}} = \dfrac{{1 - \sin x}}{{\sin x}}$
Here,
LHS= $\dfrac{{{{\cot }^2}x}}{{1 + cosecx}} - - - - - \left( 1 \right)$
Now, according to trigonometric identities
We know that,
${\cot ^2}x = cose{c^2}x - 1 - - - - - \left( 2 \right)$
Thus, substituting the values from equation (2) to (1)
We get,
$\Rightarrow \dfrac{{\cos e{c^2}x - 1}}{{1 + cosecx}} - - - - - \left( 3 \right)$
We will rewrite equation (3) in the following form to proceed further,
$\Rightarrow \dfrac{{cose{c^2}x - {{\left( 1 \right)}^2}}}{{1 + cosecx}} - - - - - \left( 4 \right)$
Now,
We know that, ${x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)$
Thus, the numerator of equation (4) can be rewritten as
$\Rightarrow \cos e{c^2}x - {\left( 1 \right)^2} = \left( {\cos ecx - 1} \right)\left( {\cos ecx + 1} \right) - - - - - \left( 5 \right)$
Therefore, after substituting the values of equation (5) in equation (4) we get,
$\Rightarrow \dfrac{{\left( {\cos ecx - 1} \right)\left( {\cos ecx + 1} \right)}}{{\left( {\cos ecx + 1} \right)}}$
After eliminating the common factor from the above equation, we get,
$\Rightarrow \left( {\cos ecx - 1} \right) - - - - - \left( 6 \right)$
Here, according to the reciprocal identities of trigonometry, we know that
$\Rightarrow \cos ecx = \dfrac{1}{{\sin x}} - - - - - \left( 7 \right)$
After substituting the value of equation (7) in equation (6)
We get,
$\Rightarrow \dfrac{1}{{\sin x}} - 1$
Further, taking L.C.M. of the denominator.
We get,
$\Rightarrow \dfrac{{1 - \sin x}}{{\sin x}} = RHS$
Hence, proved.

Note: Here we have to remember the basic algebraic and trigonometric identities to solve this problem like we have used ${x^2} - {y^2} = \left( {x - y} \right)\left( {x + y} \right)$ and ${\cot ^2}x = \cos e{c^2}x - 1$ to solve this problem. General, algebraic identities are very helpful in simplifying complicated trigonometric equations. Basic formulas of algebra, such as difference of squares formula and perfect square formula are almost always useful in solving trigonometry. The other reciprocal identities and quotient identities of trigonometry also help in solving complicated trigonometric expressions.