Question

pls help and give answer I am 8th class boy​

Answer 1

$\mathsf{Given :\; \sqrt{2}.{sec}\theta + tan\theta = 1}$

$\mathsf{\implies \sqrt{2}.{sec}\theta = 1 - tan\theta}$

$\textsf{Squaring on both sides, We get :}$

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

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

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

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

$\mathsf{\implies 2(1 + tan^2\theta) - (1 + tan^2\theta) + 2tan\theta = 0}$

$\mathsf{\implies 1 + tan^2\theta + 2tan\theta = 0}$

$\mathsf{\implies (1 + tan\theta)^2 = 0}$

$\implies \mathsf{tan\theta + 1 = 0}$

$\implies \mathsf{tan\theta = -1}$

$\implies \mathsf{tan\theta = tan\bigg(\pi - \dfrac{\pi}{4}\bigg)}$

$\implies \mathsf{tan\theta = tan\bigg(\dfrac{4\pi -\pi}{4}\bigg)}$

$\implies \mathsf{tan\theta = tan\bigg(\dfrac{3\pi}{4}\bigg)}$

$\textsf{We know that :}$

$\mathsf{\bigstar\;\;If\;TanA = TanB \implies A = n\pi + B\;\;where : (n\;\epsilon\;Z)}$

$\bigstar \;\;\mathsf{Above\;Condition\;is\;valid\;only\;when\;A\;and\;B\;are\;not\;odd\;multiple\;of\;\dfrac{\pi}{2}}$

$\mathsf{\implies \theta = n\pi + \dfrac{3\pi}{4}\;\;where : (n\;\epsilon\;Z)}$