Question

if 2 sin theta-1=0then prove that sec theta +Tan theta=root 3​

Answer 1

Question :

if 2sinθ -1 = 0

then , prove that secθ+ tanθ =√3

Formula's and values used :

$1) \: secx = \frac{1}{cosx}$

$2) \tan(x) = \frac{1}{ \cot(x) }$

$3) \csc(x) = \frac{1}{ \sin(x) }$

$4) \: \sec( \frac{\pi}{6} ) = \frac{2}{ \sqrt{3} }$

$5) \tan( \frac{\pi}{6} ) = \frac{1}{ \sqrt{3} }$

${\red{\huge{\underline{\mathbb{Answer:-}}}}}$

2sinθ -1 = 0

⇒sinθ =$\frac{1}{2}$

or θ = $\frac{\pi }{6}$

we have to prove ;secθ+ tanθ =√3

LHS = secθ+ tanθ

put the value of θ

$= \sec( \frac{\pi}{6} ) + \tan( \frac{\pi}{6} )$

$= \frac{2}{ \sqrt{3} } + \frac{1}{ \sqrt{3} }$

$= \frac{2 + 1}{ \sqrt{3} }$

$= \frac{3}{ \sqrt{3} } = \sqrt{3}$

RHS = √3

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⇒LHS = RHS

${\purple{\boxed{\large{\bold{secθ+ tanθ =\sqrt{3} }}}}}$

$\huge{\bold{ Hence\: Proved }}$

Answer 2

❥${\purple{\underline{\underline{\large{\mathtt{ANSWER:-}}}}}}$

sec∅+tan∅=√3 [proved]

❥${\purple{\underline{\underline{\large{\mathtt{EXPLANATION:-}}}}}}$

• Given:-

• 2sin∅ - 1 = 0

• To prove:-

• sec∅+tan∅=√3

• Solution:-

${\sf{\green{2sin\theta-1=0}}}$

${\sf{\implies{2sin\theta = 1}}}$

${\sf{\implies{sin\theta =\frac{1}{2}}}}$

${\sf{\implies{sin\theta =sin\: 30}}}$

${\sf{\implies{\theta=30}}}$

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${\sf{\red{sec\theta+tan\theta}}}$

${\sf{\implies{sec30\:+tan30}}}$

${\sf{\implies{\frac{2}{\sqrt{3}}\:+\frac{1}{\sqrt{3}}}}}$

${\sf{\implies{\frac{3}{\sqrt{3}}}}}$

${\sf{\implies{\frac{3\sqrt{3}}{3}}}}$

${\sf{\implies{\sqrt{3}(Proved)}}}$

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❥${\underline{\underline{\large{\mathtt{MORE INFORMATION:-}}}}}$

Some values related to trigonometry:-

• sin30° = 1/2
• sin45° =1/√2
• sin60° = √3/2

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• cos30° = √3/2
• cos45° = 1/√2
• cos60° =1/2

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• tan30° = 1/√3
• tan45°= 1
• tan60° = √3

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