Question

TRIGONOMETRY. only step by step answer no fake answer plz ​.......​

Answer 1

$\mathsf{Given :\;\dfrac{Cot^2A({Sec}A - 1)}{1 + SinA} + \dfrac{{Sec}^2A(SinA - 1)}{1 + {Sec}A}}$

Taking LCM, We get :

$\mathsf{\implies \dfrac{Cot^2A({Sec}A - 1)({Sec}A + 1) + {Sec}^2A(SinA - 1)(SinA + 1)}{(1 + {Sec}A)(1 + SinA)}}$

★  We know that : (a + b)(a - b) = a² - b²

$\mathsf{\implies \dfrac{Cot^2A({Sec}^2A - 1) + {Sec}^2A(Sin^2A - 1)}{(1 + {Sec}A)(1 + SinA)}}$

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

$\mathsf{\implies ({sec}^2\theta - 1) = tan^2\theta}$

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

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

$\mathsf{\implies \dfrac{Cot^2A.Tan^2A + {Sec}^2A.(-Cos^2A)}{(1 + {Sec}A)(1 + SinA)}}$

$\mathsf{\implies \dfrac{Cot^2A.Tan^2A - {Sec}^2A.Cos^2A}{(1 + {Sec}A)(1 + SinA)}}$

$\mathsf{\implies \dfrac{(CotA.TanA)^2 - ({Sec}A.CosA)^2}{(1 + {Sec}A)(1 + SinA)}}$

$\bigstar\;\;\mathsf{We\;know\;that : \boxed{\mathsf{tan\theta.cot\theta = 1}}\;\;\;\bigg(since\;cot\theta = \dfrac{1}{tan\theta}\bigg)}$

$\bigstar\;\;\mathsf{We\;know\;that : \boxed{\mathsf{cos\theta.{sec}\theta = 1}}\;\;\;\bigg(since\;{sec}\theta = \dfrac{1}{cos\theta}\bigg)}$

$\mathsf{\implies \dfrac{(1)^2 - (1)^2}{(1 + {Sec}A)(1 + SinA)}}$

$\mathsf{\implies \dfrac{(1 - 1)}{(1 + {Sec}A)(1 + SinA)}}$

$\mathsf{\implies \dfrac{0}{(1 + {Sec}A)(1 + SinA)}}$

$\mathsf{\implies 0}$