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Question: Prove that: \[\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{...

Prove that: 1csc⁡A−cot⁡A−1sin⁡A=1sin⁡A−1csc⁡A+cot⁡A.\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}.

Explanation

Solution

To prove this, we simplify the both sides separately by writing all trigonometric ratios in terms of sin⁡A\sin A and cos⁡A\cos A, then rationalize the term in the fraction which has complex denominator and solve to simplest form. We show that both sides give the same values.

  • csc⁡A=1sin⁡A;cot⁡A=cos⁡Asin⁡A\csc A = \dfrac{1}{{\sin A}};\cot A = \dfrac{{\cos A}}{{\sin A}}

Complete step-by-step answer:
First we solve the LHS of the equation.
Write all trigonometric ratios in 1csc⁡A−cot⁡A−1sin⁡A\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} in terms of sin⁡A\sin A and cos⁡A\cos A.
1csc⁡A−cot⁡A−1sin⁡A=11sin⁡A−cos⁡Asin⁡A−1sin⁡A\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\dfrac{1}{{\sin A}} - \dfrac{{\cos A}}{{\sin A}}}} - \dfrac{1}{{\sin A}}
Taking LCM in the denominator of the first fraction.
=11−cos⁡Asin⁡A−1sin⁡A= \dfrac{1}{{\dfrac{{1 - \cos A}}{{\sin A}}}} - \dfrac{1}{{\sin A}}
=sin⁡A1−cos⁡A−1sin⁡A= \dfrac{{\sin A}}{{1 - \cos A}} - \dfrac{1}{{\sin A}}
Now we rationalize the first fraction by multiplying both numerator and denominator by (1+cos⁡θ)(1 + \cos \theta ).
=sin⁡A1−cos⁡A×1+cos⁡A1+cos⁡A−1sin⁡A= \dfrac{{\sin A}}{{1 - \cos A}} \times \dfrac{{1 + \cos A}}{{1 + \cos A}} - \dfrac{1}{{\sin A}}

                                     $$ = \dfrac{{\sin A(1 + \cos A)}}{{(1 + \cos A)(1 - \cos A)}} - \dfrac{1}{{\sin A}}$$

Now we know from the property that (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2}
Using the formula solve denominator of first fraction where a=1,b=cos⁡Aa = 1,b = \cos A
=sin⁡A(1+cos⁡A)(1−cos⁡2A)−1sin⁡A= \dfrac{{\sin A(1 + \cos A)}}{{(1 - {{\cos }^2}A)}} - \dfrac{1}{{\sin A}}
Now from the property sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 we can write 1−cos⁡2θ=sin⁡2θ1 - {\cos ^2}\theta = {\sin ^2}\theta
So, we substitute the value of 1−cos⁡2A=sin⁡2A1 - {\cos ^2}A = {\sin ^2}A in the denominator of the first fraction.
=sin⁡A(1+cos⁡A)sin⁡2A−1sin⁡A= \dfrac{{\sin A(1 + \cos A)}}{{{{\sin }^2}A}} - \dfrac{1}{{\sin A}}

Cancel out the same factors from numerator and denominator.

       $$ = \dfrac{{(1 + \cos A)}}{{\sin A}} - \dfrac{1}{{\sin A}}$$

Taking LCM of both the fractions.

                                       $$

= \dfrac{{1 + \cos A - 1}}{{\sin A}} \\
= \dfrac{{\cos A}}{{\sin A}} \\

$$ = \cot A$$ {since $$\cot A = \dfrac{{\cos A}}{{\sin A}}$$ } $$ \Rightarrow \dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \cot A$$ $...(1)$ Now we solve the RHS of the equation. Write all trigonometric ratios in $$\dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}$$ in terms of $\sin A$ and $\cos A$. $$\dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\dfrac{1}{{\sin A}} + \dfrac{{\cos A}}{{\sin A}}}}$$ Take LCM in the denominator of the second fraction. $ = \dfrac{1}{{\sin A}} - \dfrac{1}{{\dfrac{{1 + \cos A}}{{\sin A}}}}$ $ = \dfrac{1}{{\sin A}} - \dfrac{{\sin A}}{{1 + \cos A}}$ Now we rationalize the second fraction by multiplying both numerator and denominator by $$(1 - \cos A)$$ $ = \dfrac{1}{{\sin A}} - \dfrac{{\sin A}}{{1 + \cos A}} \times \dfrac{{1 - \cos A}}{{1 - \cos A}}$ $ = \dfrac{1}{{\sin A}} - \dfrac{{\sin A(1 - \cos A)}}{{(1 - \cos A)(1 + \cos A)}}$ Now we know from the property that $$(a + b)(a - b) = {a^2} - {b^2}$$ Using the formula solve denominator of second fraction where $$a = 1,b = \cos A$$ $ = \dfrac{1}{{\sin A}} - \dfrac{{\sin A(1 - \cos A)}}{{1 - {{\cos }^2}A}}$ Now from the property $${\sin ^2}\theta + {\cos ^2}\theta = 1$$ we can write $$1 - {\cos ^2}\theta = {\sin ^2}\theta $$ So, we substitute the value of $$1 - {\cos ^2}A = {\sin ^2}A$$ in the denominator of the second fraction. $ = \dfrac{1}{{\sin A}} - \dfrac{{\sin A(1 - \cos A)}}{{{{\sin }^2}A}}$ Cancel out factors from the numerator and denominator of the second fraction. $ = \dfrac{1}{{\sin A}} - \dfrac{{(1 - \cos A)}}{{\sin A}}$ Taking LCM of both fractions. $ = \dfrac{{1 - 1 + \cos A}}{{\sin A}} \\\ = \dfrac{{\cos A}}{{\sin A}} \\\ $ $$ = \cot A$$ {since $$\cot A = \dfrac{{\cos A}}{{\sin A}}$$ } $$ \Rightarrow \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}} = \cot A$$ $...(2)$ Now we check if LHS is equal to RHS Equating values of $\cot A$ from equation $(1)$ and $(2)$ we get $$\cot A = \cot A$$ LHS = RHS Hence, $$\dfrac{1}{{\csc A - \cot A}} - \dfrac{1}{{\sin A}} = \dfrac{1}{{\sin A}} - \dfrac{1}{{\csc A + \cot A}}.$$ **Note:** Students make mistakes of rationalizing with wrong factors, keep in mind we multiply with such a term that makes our denominator easy. Many students make the mistake of solving the question without converting into sin and cos, which makes our solution complex.