Question

Faça as relações com os coeficientes das equações do 2° grau completas abaixo.

-4x^2-8x-10=0


ME AJUDEMMMMMMMMMMMM PORRRRRRRRRRRRRR FAVORRR ​

Answer 1

\huge\bold{Question :}Question:

If tan θ = ¹/√7 then , show that \sf \dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}=\dfrac{3}{4}csc2θ+sec2θcsc2θ−sec2θ​=43​

\huge\bold{Solution :}Solution:

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\sf tan\ \theta=\dfrac{1}{\sqrt{7}}tan θ=7​1​

:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}:→tan2θ=(7​)21​

:\to \sf \textsf{\textbf{\pink{tan$^\text{2} \boldsymbol \theta\ $ =\ $\dfrac{\text{1}}{\text{7}}$}}}\ \; \bigstar:→tan2θ  = 71​ ★

\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}cot θ1​=7​1​

:\to \sf cot\ \theta=\sqrt{7}:→cot θ=7​

:\to \sf cot^2\theta=(\sqrt{7})^2:→cot2θ=(7​)2

:\to \sf \textsf{\textbf{\green{cot$^\text{2}\ \boldsymbol \theta $\ =\ 7}}}\ \; \bigstar:→cot2 θ = 7 ★

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LHS

:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}:→csc2θ+sec2θcsc2θ−sec2θ​

From Trigonometric identities ,

csc²θ = 1 + cot²θ

sec²θ = 1 + tan²θ

:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}:→(1+cot2θ)+(1+tan2θ)(1+cot2θ)−(1+tan2θ)​

tan²θ = ¹/₇

cot²θ = 7

:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}:→(1+7)+(1+71​)(1+7)−(1+71​)​

:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}:→ 8+78​8−78​​

:\to\ \sf \dfrac{48}{64}:→ 6448​

:\to\ \textsf{\textbf{\orange{$\dfrac{\text{3}}{\text{4}}$}}}\ \; \bigstar:→ 43​ ★