Question

Show that $\dfrac{1}{{\left( {\csc A - \cot A} \right)}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\left( {\csc A + \cot A} \right)}}$

Answer 1

Hint : Here we need to prove the given problem. We simplify the left hand side of the equation and the right hand side of the equation and we show that both are equal. To solve this we use the concept of reciprocal and definition of cotangent. We know that cosec reciprocal is sine function and cotangent is ratio of cosine to sine function.

Complete step by step solution:
Now take
$L.H.S = \dfrac{1}{{\left( {\csc A - \cot A} \right)}} - \dfrac{1}{{\sin A}}$ and $R.H.S = \dfrac{1}{{\sin A}} - \dfrac{1}{{\left( {\csc A + \cot A} \right)}}$.
Now take LHS
$L.H.S = \dfrac{1}{{\left( {\csc A - \cot A} \right)}} - \dfrac{1}{{\sin A}}$.
We know $\csc A = \dfrac{1}{{\sin A}}$ and $\cot A = \dfrac{{\cos A}}{{\sin A}}$,
$L.H.S = \dfrac{1}{{\left( {\dfrac{1}{{\sin A}} - \dfrac{{\cos A}}{{\sin A}}} \right)}} - \dfrac{1}{{\sin A}}$
$L.H.S = \dfrac{1}{{\left( {\dfrac{{1 - \cos A}}{{\sin A}}} \right)}} - \dfrac{1}{{\sin A}}$
$L.H.S = \dfrac{{\sin A}}{{\left( {1 - \cos A} \right)}} - \dfrac{1}{{\sin A}}$
Again taking LCM and simplifying we have,
$L.H.S = \dfrac{{\sin A.\sin A - \left( {1 - \cos A} \right)}}{{\sin A.\left( {1 - \cos A} \right)}}$
$L.H.S = \dfrac{{{{\sin }^2}A - \left( {1 - \cos A} \right)}}{{\sin A.\left( {1 - \cos A} \right)}}$
We know the trigonometric identity, ${\sin ^2}A + {\cos ^2}A = 1$ using this we have,
$L.H.S = \dfrac{{\left( {1 - {{\cos }^2}A} \right) - \left( {1 - \cos A} \right)}}{{\sin A.\left( {1 - \cos A} \right)}}$
$L.H.S = \dfrac{{\left( {{1^2} - {{\cos }^2}A} \right) - \left( {1 - \cos A} \right)}}{{\sin A.\left( {1 - \cos A} \right)}}$
Now Applying the identity ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.
$L.H.S = \dfrac{{\left( {\left( {1 - \cos A} \right)\left( {1 + \cos A} \right)} \right) - \left( {1 - \cos A} \right)}}{{\sin A.\left( {1 - \cos A} \right)}}$
Taking $1 - \cos A$ common,
$L.H.S = \dfrac{{\left( {1 - \cos A} \right)\left( {\left( {1 + \cos A} \right) - 1} \right)}}{{\sin A.\left( {1 - \cos A} \right)}}$
$L.H.S = \dfrac{{\cos A}}{{\sin A.}}$
We know that cotangent is ratio of cosine to tangent,
$\Rightarrow L.H.S = \cot A - - (1)$
Now take RHS,
$R.H.S = \dfrac{1}{{\sin A}} - \dfrac{1}{{\left( {\csc A + \cot A} \right)}}$
We are going simplify this easily using the identity ${\csc ^2}A - {\cot ^2}A = 1$
$R.H.S = \dfrac{1}{{\sin A}} - \dfrac{{{{\csc }^2}A - {{\cot }^2}A}}{{\left( {\csc A + \cot A} \right)}}$
$R.H.S = \dfrac{1}{{\sin A}} - \dfrac{{\left( {\csc A - \cot A} \right)\left( {\csc A + \cot A} \right)}}{{\left( {\csc A + \cot A} \right)}}$
$R.H.S = \dfrac{1}{{\sin A}} - \csc A + \cot A$
We also have $\csc A = \dfrac{1}{{\sin A}}$ , then
$R.H.S = \csc A - \csc A + \cot A$
$\Rightarrow R.H.S = \cot A - - (2)$
From (1) and (2) we have
$\Rightarrow \dfrac{1}{{\left( {\csc A - \cot A} \right)}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\left( {\csc A + \cot A} \right)}}$.
We can also simplify the LHS as we did in RHS.

Note : Sine, cosine, tangent, cosecant, secant and cotangent are the six types of trigonometric functions; sine, cosine and tangent are the main functions while cosecant, secant and cotangent are the reciprocal of sine, cosine and tangent respectively.