Question

CosA/(SinA+CosB)+CosB/(SinB-CosA) =CosA/(SinA-CosB)+CosB/(SinB+CosA)

Answer 1

Step-by-step explanation:

CosA/(SinA+CosB)+CosB/(SinB-CosA) =CosA/(SinA-CosB)+CosB/(SinB+CosA)

CosA/(SinA+CosB)+CosB/(SinB-CosA) =CosA/(SinA-CosB)+CosB/(SinB+CosA)

CosA/(SinA+CosB)+CosB/(SinB-CosA) =CosA/(SinA-CosB)+CosB/(SinB+CosA)

Answer 2

Appropriate Question :-

Prove that,

$\rm \: \frac{cosA}{sinA + cosB} + \frac{cosB}{sinB - cosA} = \frac{cosA}{sinA - cosB} + \frac{cosB}{sinB + cosA}$

$\large\underline{\sf{Solution-}}$

To solve this question, Let we Consider

$\rm \: \frac{cosA}{sinA + cosB} - \frac{cosA}{sinA - cosB}$

$\rm \: = cosA\bigg( \frac{1}{sinA + cosB} - \frac{1}{sinA - cosB}\bigg)$

$\rm \: = cosA\bigg( \frac{sinA - cosB - sinA - cosB}{(sinA + cosB)(sinA - cosB)}\bigg)$

$\rm \: = cosA\bigg( \frac{-2 cosB}{ {sin}^{2}A - {cos}^{2}B}\bigg)$

$\rm \: = \frac{-2 \: cosA \: cosB}{ {sin}^{2}A - {cos}^{2}B}$

$\color{green}\rm\implies \: \frac{cosA}{sinA + cosB} - \frac{cosA}{sinA - cosB} = \frac{-2 \: cosA \: cosB}{ {sin}^{2}A - {cos}^{2}B} - - (1)$

Now, Consider

$\rm \:\frac{cosB}{sinB + cosA} - \frac{cosB}{sinB - cosA}$

$\rm \: = \: cosB\bigg(\frac{1}{sinB + cosA} - \frac{1}{sinB - cosA}\bigg)$

$\rm \: = \: cosB\bigg(\frac{sinB - cosA - sinB - cosA}{(sinB + cosA)(sinB - cosA)}\bigg)$

$\rm \: = \: cosB\bigg(\frac{- 2cosA}{ {sin}^{2} B - {cos}^{2}A}\bigg)$

$\rm \: = \: \frac{- 2 \: cosA \: cosB}{ {sin}^{2} B - {cos}^{2}A}$

$\rm \: = \: \frac{- 2 \: cosA \: cosB}{(1 - {cos}^{2} B) - (1 - {sin}^{2}A)}$

$\rm \: = \: \frac{- 2 \: cosA \: cosB}{1 - {cos}^{2} B - 1 + {sin}^{2}A}$

$\rm \: = \: \frac{- 2 \: cosA \: cosB}{ - {cos}^{2} B + {sin}^{2}A}$

$\rm \: = \frac{-2 \: cosA \: cosB}{ {sin}^{2}A - {cos}^{2}B}$

$\color{green}\rm\implies \:\frac{cosB}{sinB + cosA} - \frac{cosB}{sinB - cosA} = \frac{-2 \: cosA \: cosB}{ {sin}^{2}A - {cos}^{2}B} - - (2)$

So, from equation (1) and (2), we get

$\rm \: \frac{cosA}{sinA + cosB} - \frac{cosA}{sinA - cosB} = \frac{cosB}{sinB + cosA} - \frac{cosB}{sinB - cosA}$

which can be further rewritten as

$\rm \: \frac{cosA}{sinA + cosB} + \frac{cosB}{sinB - cosA} = \frac{cosA}{sinA - cosB} + \frac{cosB}{sinB + cosA}$

Hence, Proved

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(90 \degree - x) = cosx}\\ \\ \bigstar \: \bf{cos(90 \degree - x) = sinx}\\ \\ \bigstar \: \bf{tan(90 \degree - x) = cotx}\\ \\ \bigstar \: \bf{cot(90 \degree - x) = tanx}\\ \\ \bigstar \: \bf{cosec(90 \degree - x) = secx}\\ \\ \bigstar \: \bf{sec(90 \degree - x) = cosecx}\\ \\ \bigstar \: \bf{ {sin}^{2}x + {cos}^{2}x = 1 } \\ \\ \bigstar \: \bf{ {sec}^{2}x - {tan}^{2}x = 1 }\\ \\ \bigstar \: \bf{ {cosec}^{2}x - {cot}^{2}x = 1 }\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$