Question

solve plz plz solve​

Answer 1

$\sf{Given \ : \ {\dfrac{tan \theta}{1 - cot \theta}} + {\dfrac{cot \theta}{1 - tan \theta}} = 1 + sec \theta . cosec \theta }$




$\sf{L.H.S. = {\dfrac{tan \theta}{1 - cot \theta}} + {\dfrac{cot \theta}{1 - tan \theta}}}$



${\boxed{\sf{tan \theta = {\dfrac{sin \theta}{cos \theta}} }}}$



${\boxed{\sf{cot \theta = {\dfrac{cos \theta}{sin \theta}} }}}$



$\sf{= {\dfrac{ {\dfrac{sin \theta}{cos \theta}} }{ 1 - {\dfrac{cos \theta}{sin \theta}} }} + {\dfrac{ {\dfrac{cos \theta}{sin \theta}} }{ 1 - {\dfrac{sin \theta}{cos \theta}} }} }$



$\sf{= {\dfrac{ {\dfrac{sin \theta}{cos \theta}} }{ {\dfrac{sin \theta - cos \theta}{sin \theta}} }} + {\dfrac{ {\dfrac{cos \theta}{sin \theta}} }{ {\dfrac{cos \theta - sin \theta}{cos \theta}} }} }$



$\sf{= {\dfrac{sin \theta}{cos \theta}} × {\dfrac{sin \theta}{sin \theta - cos \theta}} + {\dfrac{cos \theta}{sin \theta}} × {\dfrac{cos \theta}{cos \theta - sin \theta}} }$



$\sf{= {\dfrac{sin^2 \theta}{cos \theta (sin \theta - cos \theta)}} + {\dfrac{cos^2 \theta}{sin \theta (cos \theta - sin \theta)}} }$



$\sf{Multiplying \ the \ numerator \ and \ denominator \ of \ {\dfrac{cos^2 \theta}{sin \theta (cos \theta - sin \theta)}} \ by \ - 1, \ we \ get}$



$\sf{= {\dfrac{sin^2 \theta}{cos \theta (sin \theta - cos \theta)}} - {\dfrac{cos^2 \theta}{sin \theta (sin \theta - cos \theta)}} }$



$\sf{= {\dfrac{sin^3 \theta - cos^3 \theta}{cos \theta sin \theta (sin \theta - cos \theta)}}}$



${\boxed{\sf{a^3 - b^3 = (a - b)(a^2 + b^2 + ab)}}}$



$\sf{Here, \ a = sin \theta, \ b = cos \theta}$



$\sf{= {\dfrac{(sin \theta - cos \theta)(sin^2 \theta + cos^2 \theta + sin \theta cos \theta)}{cos \theta sin \theta (sin \theta - cos \theta)}}}$



$\sf{= {\dfrac{ {\cancel{(sin \theta - cos \theta)}} (sin^2 \theta + cos^2 \theta + sin \theta cos \theta)}{cos \theta sin \theta {\cancel{(sin \theta - cos \theta)}}}}}$



${\boxed{\sf{sin^2 \theta + cos^2 \theta = 1}}}$



$\sf{= {\dfrac{1 + sin \theta cos \theta}{cos \theta sin \theta}}}$



$\sf{= {\dfrac{1}{cos \theta sin \theta}} + {\dfrac{sin \theta cos \theta}{cos \theta sin \theta}}}$



$\sf{= {\dfrac{1}{cos \theta sin \theta}} + {\dfrac{ {\cancel{sin \theta cos \theta}}}{{\cancel{cos \theta sin \theta}}}}}$



$\sf{= {\dfrac{1}{cos \theta}} . {\dfrac{1}{sin \theta}} + 1}$



${\boxed{\sf{ {\dfrac{1}{cos \theta}} = sec \theta}}}$



${\boxed{\sf{ {\dfrac{1}{sin \theta}} = cosec \theta}}}$



$\sf{= sec \theta . cosec \theta + 1}$



$\sf{= R.H.S.}$



$\sf{Hence, \ proved \ !!}$

Answer 2

Answer:

R.H.S

Step-by-step explanation: