Question

C = sec theta + tan theta

Question : If C = root 3,what is the value of theta?​

Answer 1

$\small\pink{\mathcal{\underline{\:\:Given\:that\:\:}}}$

• C=secϴ+tanϴ

Here, C = √3

$\small\pink{\mathcal{\underline{\:\:Accroding\:to\:question,\:\:}}}$

• secϴ + tanϴ = √3

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;\frac{1}{cosϴ}+\frac{sinϴ}{cosϴ}\;=\;\bf{3}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;\frac{1+sinϴ}{cosϴ}\;=\;\bf{3}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;1+sinϴ\;=\;\bf{\;√3\:cosϴ\;}}}\end{gathered}$

$\small\pink{\mathcal{\underline{\:\:By\:squaring\:both\:side,\:}}}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;(1+sinϴ)^2\;=\;\bf{(\;√3\:cosϴ)^2\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;1+2sinϴ+sin^2ϴ\;=\;\bf{\;3cos^2ϴ\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;1+2sinϴ+sin^2ϴ\;=\;\bf{\;3(1-sin^2ϴ)\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;sin^2ϴ+2sinϴ+1\;=\;\bf{\;3-3sin^2ϴ\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;sin^2ϴ+3sin^2ϴ+2sinϴ+1-3\;=\;\bf{\;0\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;4sin^2ϴ+2sinϴ-2\;=\;\bf{\;0\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;2sin^2ϴ+2sinϴ-1\;=\;\bf{\;0\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;2sin^2ϴ+2sinϴ-sinϴ-1\;=\;\bf{\;0\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;2sinϴ(sinϴ+1)-1(sinϴ+1)\;=\;\bf{\;0\;}}}\end{gathered}$

$\begin{gathered}\:\:\:\:\:\displaystyle{\sf{\leadsto\;(2sinϴ-1)(sinϴ+1)\;=\;\bf{\;0\;}}}\end{gathered}$

______________________________

Either, 2sinϴ - 1 = 0

⇒ 2sinϴ = 0

⇒ sinϴ = $\frac{1}{2}$

⇒ ϴ = 30°

Or, sinϴ+1 = 0

⇒ sinϴ = -1

This is not possibe as ϴ is an acute angle.

$\large\pink{\boxed{\underline{\mathscr{\blue{So,the\:value\:of\:ϴ=30°}}}}}$

Answer 2

GIVEN THAT

• C = secϴ + tanϴ
• C = √3

SOLUTION

According To Question

• secϴ + tanϴ = √3
• 1/{cosϴ} + {sinϴ}/{cosϴ} = 3
• {1+sinϴ}/{cosϴ} = 3
• 1+sinϴ=√3cosϴ

Squaring Both Side

• (1+sinϴ)² = (√3cosϴ)²
• 1+2sinϴ+sin²ϴ=3cos²ϴ
• 1+2sinϴ+sin²ϴ=3(1−sin²ϴ)
• sin²ϴ+2sinϴ+1=3−3sin²ϴ
• sin²ϴ+3sin²ϴ+2sinϴ+1−3=0
• 4sin²ϴ+2sinϴ−2=0
• 2sin²ϴ+2sinϴ−1=0
• 2sin²ϴ+2sinϴ−sinϴ−1=0
• 2sinϴ(sinϴ+1)−1(sinϴ+1)=0
• (2sinϴ−1)(sinϴ+1)=0

$\boxed{\begin{array}{c|c}2sinϴ - 1 = 0& sinϴ+1 = 0 \: \\ sinϴ = \frac{1}{2} \: &sinϴ = -1 \end{array}}\\ \bf \: since \: ϴ \: is \: an \: acute \: angle.\\ \cr \: Hence, the \: value \: of \: \boxed{ \bf ϴ=30°}$