Question

if 270°→theta→360 and costheta =1/4 find tan theta/2

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Answer 1

Given :- if 270° < A < 360° and cosA = (1/4) . find tan(A/2) = ?

Solution :-

we know that,

• cos2A = (1 - tan²A)/(1 + tan²A)

So,

• cosA = {1 - tan²(A/2)} / {1 + tan²(A/2)}

Putting value we get,

→ (1/4) = {1 - tan²(A/2)} / {1 + tan²(A/2)}

Let tan(A/2),

→ (1/4) = {1 - x²} / {1 + x²}

→ 1 + x² = 4(1 - x²)

→ 1 + x² = 4 - 4x²

→ 4x² + x² + 1 - 4 = 0

→ 5x² - 3 = 0

→ 5x² = 3

→ x² = 3/5

→ x = ±√(3/5)

therefore,

→ tan(A/2) = ±√(3/5)

Now, since 270° < A < 360° .

• A = in 4th quadrant .
• in 4th quadrant cosA is positive .

Hence,

→ tan(A/2) = (-√(3/5) (Ans.)

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Answer 2

SOLUTION :-

GIVEN :-

$\displaystyle \sf{ {270}^{ \circ} < \theta < {360}^{ \circ} \: \: \: and \: \cos \theta = \frac{1}{4} }$

TO DETERMINE :-

$\displaystyle \sf{ \tan \frac{ \theta}{2} }$

FORMULA TO BE IMPLEMENTED :-

We are aware of the Trigonometric identity that

$\displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{1 - \cos \theta}{1 + \cos \theta} }$

EVALUATION :-

Here it is given that

$\displaystyle \sf{ {270}^{ \circ} < \theta < {360}^{ \circ} \: \: \: and \: \cos \theta = \frac{1}{4} }$

Now

$\displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{1 - \cos \theta}{1 + \cos \theta} }$

$\implies \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{1 - \frac{1}{4} }{1 + \frac{1}{4} } }$

$\implies \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{ \frac{4 - 1}{4} }{ \frac{4 + 1}{4} } }$

$\implies \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{3 }{5} }$

$\because \: \: \displaystyle \sf{ {270}^{ \circ} < \theta < {360}^{ \circ} \: \: }$

$\therefore \: \: \displaystyle \sf{ {135}^{ \circ} < \frac{ \theta}{2} < {180}^{ \circ} \: \: }$

$\therefore \: \: \displaystyle \sf{ \frac{ \theta}{2} \: \: lies \: in \: second \: quadrant }$

$\therefore \: \: \displaystyle \sf{ \tan \frac{ \theta}{2} \: \: \: is \: negative }$

$\therefore \: \: \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{3 }{5} } \: \: gives$

$\displaystyle \sf{ {\tan} \frac{ \theta}{2} = - \sqrt{ \frac{3 }{5} }}$

FINAL ANSWER :-

$\boxed{\displaystyle \sf{ \: \: {\tan} \frac{ \theta}{2} = - \sqrt{ \frac{3 }{5} }} \: \: \: }$

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