$A$ is an angle of the right angled triangle. Trigonometric functions $\sin{A}$ and $\cos{A}$ formed following special trigonometric function.

$\dfrac{\sin^2{A}}{\cos{A}(\sin{A}-\cos{A})}$ $+$ $\dfrac{\cos^2{A}}{\sin{A}(\cos{A}-\sin{A})}$

It can be simplified in trigonometric mathematics by using trigonometric identities.

Observe the denominator of each term of the trigonometric expression. Part of the denominator of each term contains similar factors. So, try to adjust the denominator to make both common factors should be appeared same.

$=\,\,\,$ $\dfrac{\sin^2{A}}{\cos{A}(\sin{A}-\cos{A})}$ $-$ $\dfrac{\cos^2{A}}{\sin{A}(\sin{A}-\cos{A})}$

$\sin{A}-\cos{A}$ is the common factor in the denominator of each term of the expression. Therefore, take out the common factor from both terms.

$=\,\,\,$ $\dfrac{1}{(\sin{A}-\cos{A})}$ $\Bigg[\dfrac{\sin^2{A}}{\cos{A}}$ $-$ $\dfrac{\cos^2{A}}{\sin{A}}\Bigg]$

Simplify the expression by using subtraction rule of the fractions.

$=\,\,\,$ $\dfrac{1}{(\sin{A}-\cos{A})}$ $\Bigg[\dfrac{\sin^2{A} \times \sin{A} – \cos^2{A} \times \cos{A}}{\cos{A}\sin{A}}\Bigg]$

$=\,\,\,$ $\dfrac{1}{(\sin{A}-\cos{A})}$ $\Bigg[\dfrac{\sin^3{A}-\cos^3{A}}{\cos{A}\sin{A}}\Bigg]$

The numerator of the multiplying factor is in cube form. It can be simplified by using a-b whole cube formula.

${(a-b)}^3$ $=$ $a^3-b^3-3ab(a-b)$

$\implies {(a-b)}^3+3ab(a-b)$ $=$ $a^3-b^3$

$\implies a^3-b^3$ $=$ ${(a-b)}^3+3ab(a-b)$

Take $a = \sin{A}$ and $b = \cos{A}$

Therefore, $\sin^3{A}-\cos^3{A}$ $=$ ${(\sin{A}-\cos{A})}^3+3\sin{A}\cos{A}{(\sin{A}-\cos{A})}$

$\sin^3{A}-\cos^3{A}$ can be expanded as displayed above.

$=\,\,\,$ $\dfrac{1}{(\sin{A}-\cos{A})}$ $\Bigg[\dfrac{{(\sin{A}-\cos{A})}^3+3\sin{A}\cos{A}{(\sin{A}-\cos{A})}}{\cos{A}\sin{A}}\Bigg]$

$\sin{A}-\cos{A}$ is common multiplying factor in the numerator. Take it out common from them.

$=\,\,\,$ $\dfrac{(\sin{A}-\cos{A})}{(\sin{A}-\cos{A})}$ $\Bigg[\dfrac{{(\sin{A}-\cos{A})}^2+3\sin{A}\cos{A}}{\cos{A}\sin{A}}\Bigg]$

$=\,\,\,$ $\require{cancel} \dfrac{\cancel{(\sin{A}-\cos{A})}}{\cancel{(\sin{A}-\cos{A})}}$ $\Bigg[\dfrac{{(\sin{A}-\cos{A})}^2+3\sin{A}\cos{A}}{\cos{A}\sin{A}}\Bigg]$

$=\,\,\,$ $\dfrac{{(\sin{A}-\cos{A})}^2+3\sin{A}\cos{A}}{\cos{A}\sin{A}}$

Part of the numerator is in the form of a-b whole square. Expand it as per ${(a-b)}^2$ formula.

$=\,\,\,$ $\dfrac{\sin^2{A}+\cos^2{A}-2\sin{A}\cos{A}+3\sin{A}\cos{A}}{\cos{A}\sin{A}}$

According to Pythagorean identity of sin and cos functions, the sum of the squares of them is equal to one.

$=\,\,\,$ $\dfrac{1+\sin{A}\cos{A}}{\cos{A}\sin{A}}$

$=\,\,\,$ $\dfrac{1+\sin{A}\cos{A}}{\sin{A}\cos{A}}$

$=\,\,\,$ $\dfrac{1}{\sin{A}\cos{A}}+\dfrac{\sin{A}\cos{A}}{\sin{A}\cos{A}}$

$=\,\,\,$ $\require{cancel} \dfrac{1}{\sin{A}\cos{A}}+\dfrac{\cancel{\sin{A}\cos{A}}}{\cancel{\sin{A}\cos{A}}}$

$=\,\,\,$ $\dfrac{1}{\sin{A}\cos{A}}+1$

The denominator of the first term of the expression is similar to the expansion of the sin double angle rule. Make an adjustment to transform the denominator as sin of double angle. It can be done by multiplying the term by $2$ and dividing it by $2$.

$=\,\,\,$ $\dfrac{2}{2 \times \sin{A}\cos{A}}+1$

$=\,\,\,$ $\dfrac{2}{2\sin{A}\cos{A}}+1$

$=\,\,\,$ $\dfrac{2}{\sin{2A}}+1$

As per reciprocal identity of cosec function, the reciprocal of cosecant is written as sin function.

$=\,\,\,$ $2\csc{2A}+1$

It is also written as follows

$=\,\,\,$ $2\operatorname{cosec}{2A}+1$

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved