Question

pl solve 12 question on a note book and attach pic of that pl​

Answer 1

$\mathsf{Given :\;\;\dfrac{sec\theta + tan\theta - 1}{tan\theta - sec\theta + 1}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{sec^2\theta - tan^2\theta = 1}}}$

$\mathsf{\implies \dfrac{(sec\theta + tan\theta) - [sec^2\theta - tan^2\theta]}{tan\theta - sec\theta + 1}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^2 - b^2 = (a + b)(a - b)}}}$

$\mathsf{\implies \dfrac{(sec\theta + tan\theta) - [(sec\theta + tan\theta)(sec\theta - tan\theta)]}{tan\theta - sec\theta + 1}}$

$\mathsf{Taking\;(sec\theta + tan\theta)\;common\;in\;the\;numerator,\;We\;get :}$

$\mathsf{\implies \dfrac{(sec\theta + tan\theta)[1 - (sec\theta - tan\theta)]}{tan\theta - sec\theta + 1}}$

$\mathsf{\implies \dfrac{(sec\theta + tan\theta)[1 - sec\theta + tan\theta]}{tan\theta - sec\theta + 1}}$

$\mathsf{\implies sec\theta + tan\theta}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{sec\theta = \dfrac{1}{cos\theta}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies \dfrac{1}{cos\theta} + \dfrac{sin\theta}{cos\theta}}$

$\mathsf{\implies \dfrac{1 + sin\theta}{cos\theta}}$

$\mathsf{Multiplying\;both\;Numerator\;and\;Denominator\;with\;(1 - sin\theta),\;We\;get :}$

$\mathsf{\implies \dfrac{(1 + sin\theta)(1 - sin\theta)}{cos\theta(1 - sin\theta)}}$

$\mathsf{\implies \dfrac{1 - sin^2\theta}{cos\theta(1 - sin\theta)}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{1 - sin^2\theta = cos^2\theta}}}$

$\mathsf{\implies \dfrac{cos^2\theta}{cos\theta(1 - sin\theta)}}$

$\mathsf{\implies \dfrac{cos\theta}{1 - sin\theta}}$