Find zeros of the given polynomials
P(x) = 2x+2
P(y) = 3y+3
P(z) = 4x+1
P(x) = 8x+1
P(x) = 4x-8
P(x) = 3x-12
P(x) = 5x +15
P(x) = 6x-36
P(x) = 7x-35
P(x) = 8x+72
Please give the answer as soon as possible
Answers
Answer:
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}
csc
2
θ+sec
2
θ
csc
2
θ−sec
2
θ
=
4
3
\huge\bold{Solution :}Solution:
★══════════════════════★
\sf tan\ \theta=\dfrac{1}{\sqrt{7}}tan θ=
7
1
:\to \sf tan^2\theta=\dfrac{1}{(\sqrt{7})^2}:→tan
2
θ=
(
7
)
2
1
:\to \sf \textsf{\textbf{\pink{tan$^\text{2} \boldsymbol \theta\ $ =\ $\dfrac{\text{1}}{\text{7}}$}}}\ \; \bigstar:→tan
2
θ =
7
1
★
\sf \dfrac{1}{cot\ \theta}=\dfrac{1}{\sqrt{7}}
cot θ
1
=
7
1
:\to \sf cot\ \theta=\sqrt{7}:→cot θ=
7
:\to \sf cot^2\theta=(\sqrt{7})^2:→cot
2
θ=(
7
)
2
:\to \sf \textsf{\textbf{\green{cot$^\text{2}\ \boldsymbol \theta $\ =\ 7}}}\ \; \bigstar:→cot
2
θ = 7 ★
★══════════════════════★
LHS
:\to \bf \blue{\dfrac{csc^2\theta-sec^2\theta}{csc^2\theta+sec^2\theta}}:→
csc
2
θ+sec
2
θ
csc
2
θ−sec
2
θ
From Trigonometric identities ,
csc²θ = 1 + cot²θ
sec²θ = 1 + tan²θ
:\to \sf \dfrac{(1+cot^2\theta)-(1+tan^2\theta)}{(1+cot^2\theta)+(1+tan^2\theta)}:→
(1+cot
2
θ)+(1+tan
2
θ)
(1+cot
2
θ)−(1+tan
2
θ)
tan²θ = ¹/₇
cot²θ = 7
:\to \sf \dfrac{(1+7)-(1+\frac{1}{7})}{(1+7)+(1+\frac{1}{7})}:→
(1+7)+(1+
7
1
)
(1+7)−(1+
7
1
)
:\to\ \sf \dfrac{8-\frac{8}{7}}{8+\frac{8}{7}}:→
8+
7
8
8−
7
8
:\to\ \sf \dfrac{48}{64}:→
64
48
:\to\ \textsf{\textbf{\orange{$\dfrac{\text{3}}{\text{4}}$}}}\ \; \bigstar:→
4
3
★
4x, 6y, 4x, 9x, -4x, -9x, 20x, -30x, -27x, 80x