Question

Prove that :
$\sf{\dfrac{sinA}{1\:+\:cotA}}\:-\:\sf{\dfrac{cosA}{1\:+\:tanA}}\:\sf\:=\:sinA\:-\:cosA$​

Answer 1

$\large \bf \clubs \:  To \: Find :-$

$\small\bf\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} =sinA - {cosA}$

-----------------------

$\large \bf \clubs \:  Formulae \: Used :-$

-----------------------

$\bf\maltese\:\:\: tanA=\dfrac{sinA}{cosA} \\ \\ \bf\maltese\:\:\:cotA=\dfrac{cosA}{sinA} \\ \\\bf\maltese\:\:\:a^{2} -b^{2} =(a-b) (a+b)$

-----------------------

$\large \bf \clubs \:  Proof :-$

$\LARGE\bf{LHS} : -$

$\sf\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} \\ \\ \sf=\dfrac{sinA}{1 + \dfrac{cosA}{sinA} } - \dfrac{cosA}{1 + \dfrac{sinA}{cosA} }\\ \\\sf=\dfrac{sin^2A}{sinA + cosA} - \dfrac{cos^{2}A }{cosA + sinA} \\ \\ \sf=\dfrac{sin^2A-cos^2A}{sinA + cosA} \\ \\ \sf = \frac{ \cancel{(sinA + cosA)}(sinA - cosA)}{ \cancel{sinA + cosA}} \\ \\ \bf= sinA + cosA$

$\LARGE=\bf{RHS}$

$\Large \purple{ \bf\bigstar \: Hence \: Proved \: \bigstar}$

-----------------------

$\large \bf \clubs \:  Additional \: Info :-$

$\qquad \pink{\bigstar \: \bf Fundamental \: Trigonometric\:\bigstar} \\ \qquad\bf\pink{\bigstar\:Identities\:\bigstar}$

$\large\qquad \boxed{\boxed{\begin{array}{cc} \maltese \: \sf \: { \sin }^{2} \theta + { \cos}^{2} \theta = 1 \\ \\ \maltese\sf \: \: { \sec}^{2} \theta = 1 + { \tan}^{2} \theta \\ \\ \maltese \: \sf \: { \cosec}^{2} \theta = 1 + { \cot}^{2} \theta \end{array}}}$

-----------------------

Answer 2

Correct Question :-

sin A/1 + cotA - cosA/1 + tanA = sinA - cosA

Required Proof :-

$\sf\dfrac{sinA}{1 + cotA} - \dfrac{cosA}{1 + tanA} = sinA + cosA$

$\sf \bullet cotA = \dfrac{cosA}{sinA} \;\;\& \;\;\;\bullet tanA=\dfrac{sinA}{cosA}$

$\sf\dfrac{sinA}{1+\dfrac{cosA}{sinA}} - \dfrac{cosA}{1 + \dfrac{sinA}{cosA}}=sinA+cosA$

$\sf\dfrac{cos^2A}{cosA-sinA}+\dfrac{sin^2A}{sinA-cosA}=sinA+cosA$

$\sf\dfrac{cos^2A}{cosA-sinA}-\dfrac{sin^2A}{cosA-sinA}=sinA+cosA$

$\sf \dfrac{cos^2A-sin^2A}{cosA-sinA}=sinA+cosA$

Apply identity

a² - b² = (a + b)(a - b)

$\sf\dfrac{(cosA+ sinA)(cosA-sinA)}{cosA-sinA}=sinA+cosA$

$\sf sinA+cosA=sinA+cosA$