Question

ans plz any one out of this

Answer 1

$\sf{\underline {\Large {QUESTION}}}$ :

( sin $\theta$ + 1 + cos $\theta$ ) ( sin $\theta$ - 1 + cos $\theta$). Sec $\theta$ cosec $\theta$ = 2

$\sf{\underline {\Large {PROVING}}}$ :

L. H. S. = ( sin $\theta$ + 1 + cos $\theta$) ( sin $\theta$ - 1 + cos $\theta$). Sec $\theta$ cosec $\theta$

L. H. S. = [ ( sin $\theta$ + cos $\theta$)² - ( 1 )² ]. Sec $\theta$ cosec $\theta$

$\fbox{Identity \:Used\:, \:a² \:- \:b² \:= \:( \:a\: + \:b\:)\: (\: a\: -\: b\: )}$

$\tt{\underline {Here \:, a\:=\:(\:sin theta \:+\: cos theta) \:and \:b\:= \: 1 }}$

L. H. S. = ( Sin²$\theta$ + cos²$\theta$ + 2 sin $\theta$ cos$\theta$ - 1 ) × Sec $\theta$ cosec $\theta$

L. H. S. = ( 1 + 2 sin $\theta$ cos $\theta$ - 1 ) × $\frac{1} {sin theta cos theta}$

L. H. S. = $\frac{( 2 sin theta cos theta)} {sin theta cos theta}$

L. H. S. = 2

$\sf{\large {\underline {L.\: H. \:S. \:= \:R. \:H.\: S.}}}$

Hence, Proved.

Answer 2

6. First Problem :

$\mathsf{Given : (sin\theta + cos\theta + 1)(sin\theta + cos\theta - 1).({sec}\theta.co{sec}\theta)}$

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

$\mathsf{:\implies [(sin\theta + cos\theta)^2 - (1)^2].[{sec}\theta.co{sec}\theta]}$

$\mathsf{:\implies [sin^2\theta + cos^2\theta + 2.sin\theta.cos\theta - 1].[{sec}\theta.co{sec}\theta]}$

★  We know that : sin²θ + cos²θ = 1

$\mathsf{:\implies [1 + 2.sin\theta.cos\theta - 1].[{sec}\theta.co{sec}\theta]}$

$\mathsf{:\implies [2.sin\theta.cos\theta].[{sec}\theta.co{sec}\theta]}$

$\mathsf{:\implies [2.sin\theta.co{sec}\theta].[{cos}\theta.{sec}\theta]}$

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

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

$\mathsf{:\implies 2\bigg(sin\theta \times \dfrac{1}{sin\theta}\bigg).\bigg(cos\theta \times \dfrac{1}{cos\theta}\bigg)}$

$\mathsf{:\implies 2}$

6. Second Problem :

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

$\mathsf{:\implies {\dfrac{\sqrt{{sec}\theta - 1}}{\sqrt{{sec}\theta + 1}}} + \dfrac{\sqrt{{sec}\theta + 1}}{\sqrt{{sec}\theta - 1}}}}$

Taking LCM, We get :

$\mathsf{:\implies \dfrac{(\sqrt{{sec}\theta - 1})(\sqrt{{sec}\theta - 1}) + (\sqrt{{sec}\theta + 1})(\sqrt{{sec}\theta + 1})}{(\sqrt{{sec}\theta + 1})(\sqrt{{sec}\theta - 1})}}$

$\mathsf{:\implies \dfrac{\sqrt{({sec}\theta - 1)({{sec}\theta - 1})} + \sqrt{({sec}\theta + 1)({{sec}\theta + 1})} }{\sqrt{({sec}\theta + 1)({{sec}\theta - 1})}}}$

$\mathsf{:\implies \dfrac{\sqrt{({sec}\theta - 1)^2} + \sqrt{({sec}\theta + 1)^2} }{\sqrt{({{sec}^2\theta - 1})}}}$

$\mathsf{:\implies \dfrac{{sec}\theta - 1 + {sec}\theta + 1}{\sqrt{({{sec}^2\theta - 1})}}}$

$\mathsf{:\implies \dfrac{2.{sec}\theta}{\sqrt{({{sec}^2\theta - 1})}}}$

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

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

$\mathsf{:\implies \dfrac{2.{sec}\theta}{\sqrt{(tan^2\theta)}}}$

$\mathsf{:\implies \dfrac{2.{sec}\theta}{tan\theta}}}$

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

$\mathsf{:\implies \dfrac{2.{sec}\theta}{sin\theta.{sec}\theta}}}$

$\mathsf{:\implies \dfrac{2}{sin\theta}}}$

$\mathsf{:\implies 2.co{sec}\theta}$