if-tan-sec-3-than-find-out-the-value-of-where-0-o-2pi-

Question Number 69953 by Shamim last updated on 29/Sep/19

if \tan θ + \sec θ = \sqrt{3} than find out the

value of \underset{}{} θ where 0^{o} ⩽ θ ⩽ 2 π .

Commented by Shamim last updated on 29/Sep/19

ans ki only x = \frac{π}{6} ? ? ?

Commented by mathmax by abdo last updated on 29/Sep/19

t a n θ + \frac{1}{c o s θ} = \sqrt{3} \Rightarrow c o s θ t a n θ + 1 = \sqrt{3} c o s θ \Rightarrow

s i n θ + 1 = \sqrt{3} c o s θ \Rightarrow \sqrt{3} c o s θ - s i n θ = 1 (w i t h θ \neq \frac{π}{2} + k π)

2 (\frac{\sqrt{3}}{2} c o s θ - \frac{1}{2} s i n θ) = 1 \Rightarrow c o s θ c o s (\frac{π}{6}) - s i n θ s i n (\frac{π}{6}) = \frac{1}{2} \Rightarrow

c o s (θ + \frac{π}{6}) = \frac{1}{2} \Rightarrow c o s (θ + \frac{π}{6}) = c o s (\frac{π}{3}) \Rightarrow θ + \frac{π}{6} = \frac{π}{3} + 2 k π o r

θ + \frac{π}{6} = - \frac{π}{3} + 2 k π \Rightarrow θ = \frac{π}{6} + 2 k π o r θ = - \frac{π}{2} + 2 k π (k \in Z)

s o l u t i o n s o n [0, 2 π]

0 ⩽ \frac{π}{6} + 2 k π ⩽ 2 π \Rightarrow 0 ⩽ \frac{1}{6} + 2 k ⩽ 2 \Rightarrow - \frac{1}{6} ⩽ 2 k ⩽ \frac{11}{6} \Rightarrow

- \frac{1}{12} ⩽ k ⩽ \frac{11}{12} \Rightarrow k = 0 \Rightarrow x = \frac{π}{6}

c_{2} \to 0 ⩽ - \frac{π}{2} + 2 k π ⩽ 2 π \Rightarrow 0 ⩽ - \frac{1}{2} + 2 k ⩽ 2 \Rightarrow \frac{1}{2} ⩽ 2 k ⩽ \frac{5}{2} \Rightarrow

\frac{1}{4} ⩽ k ⩽ \frac{5}{4} \Rightarrow k = 1 \Rightarrow x = \frac{3 π}{2} b u t t h i s i s n o t s o < u t i o n!

Commented by Prithwish sen last updated on 29/Sep/19

\tan θ + \sec θ = \sqrt{3} \dots . (i)

\Rightarrow \sec θ - \tan θ = \frac{1}{\sqrt{3}} \dots (ii)

Adding (i) and (ii)

2 \sec θ = \frac{4}{\sqrt{3}}

\cos θ = \frac{\sqrt{3}}{2} = \cos \frac{π}{6}

\Rightarrow θ = n π \pm \frac{π}{6}

Commented by mathmax by abdo last updated on 29/Sep/19

y e s .

Commented by Shamim last updated on 29/Sep/19

okk .

all kind of like this math maintain

same rule ? ? ?

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