Question

solve the above problem of trigonometric chapter.....​

Answer 1

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


${\boxed{\tt{tanA = {\dfrac{sinA}{cosA}}}}}$


${\boxed{\tt{cotA = {\dfrac{cosA}{sinA}}}}}$


$\tt{\implies {\dfrac{cosA}{1 - {\dfrac{sinA}{cosA}} }} + {\dfrac{sinA}{1 - {\dfrac{cosA}{sinA}} }} }$


$\tt{\implies {\dfrac{cosA}{ {\dfrac{cosA - sinA}{cosA}} }} + {\dfrac{sinA}{ {\dfrac{sinA - cosA}{sinA}} }} }$


$\tt{\implies {\dfrac{cos^2A}{cosA - sinA}} + {\dfrac{sin^2A}{sinA - cosA}} }$


$\tt{Multiply \ {\dfrac{sin^2A}{sinA - cosA}} \ by \ -1, \ we \ get}$


$\tt{\implies {\dfrac{cos^2}{cosA - sinA}} - {\dfrac{sin^2A}{cosA - sinA}} }$


$\tt{\implies {\dfrac{cos^2A - sin^2A}{cosA - sinA}} }$


${\boxed{\tt{a^2 - b^2 = (a + b)(a - b)}}}$


$\tt{Here, \ a = cosA, \ b = sinA}$


$\tt{\implies {\dfrac{(cosA + sinA)(cosA - sinA)}{cosA - sinA}} }$


$\tt{\implies {\dfrac{(cosA + sinA)( {\cancel{cosA - sinA}} )}{ {\cancel{cosA - sinA}} }}}$


$\tt{\implies sinA + cosA}$


$\tt{\implies R.H.S.}$


$\tt{Hence, \ proved.}$