if270°<theta<theta<360°and cos theta=1/4find tan theta/2
Answers
Step-by-step explanation:
SOLUTION :-
GIVEN :-
\displaystyle \sf{ {270}^{ \circ} < \theta < {360}^{ \circ} \: \: \: and \: \cos \theta = \frac{1}{4} }270
∘
<θ<360
∘
andcosθ=
4
1
TO DETERMINE :-
\displaystyle \sf{ \tan \frac{ \theta}{2} }tan
2
θ
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} }tan
2
2
θ
=
1+cosθ
1−cosθ
EVALUATION :-
Here it is given that
\displaystyle \sf{ {270}^{ \circ} < \theta < {360}^{ \circ} \: \: \: and \: \cos \theta = \frac{1}{4} }270
∘
<θ<360
∘
andcosθ=
4
1
Now
\displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{1 - \cos \theta}{1 + \cos \theta} }tan
2
2
θ
=
1+cosθ
1−cosθ
\implies \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{1 - \frac{1}{4} }{1 + \frac{1}{4} } }⟹tan
2
2
θ
=
1+
4
1
1−
4
1
\implies \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{ \frac{4 - 1}{4} }{ \frac{4 + 1}{4} } }⟹tan
2
2
θ
=
4
4+1
4
4−1
\implies \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{3 }{5} }⟹tan
2
2
θ
=
5
3
\because \: \: \displaystyle \sf{ {270}^{ \circ} < \theta < {360}^{ \circ} \: \: }∵270
∘
<θ<360
∘
\therefore \: \: \displaystyle \sf{ {135}^{ \circ} < \frac{ \theta}{2} < {180}^{ \circ} \: \: }∴135
∘
<
2
θ
<180
∘
\therefore \: \: \displaystyle \sf{ \frac{ \theta}{2} \: \: lies \: in \: second \: quadrant }∴
2
θ
liesinsecondquadrant
\therefore \: \: \displaystyle \sf{ \tan \frac{ \theta}{2} \: \: \: is \: negative }∴tan
2
θ
isnegative
\therefore \: \: \displaystyle \sf{ {\tan}^{2} \frac{ \theta}{2} = \frac{3 }{5} } \: \: gives∴tan
2
2
θ
=
5
3
gives
\displaystyle \sf{ {\tan} \frac{ \theta}{2} = - \sqrt{ \frac{3 }{5} }}tan
2
θ
=−
5
3
FINAL ANSWER :-
\boxed{\displaystyle \sf{ \: \: {\tan} \frac{ \theta}{2} = - \sqrt{ \frac{3 }{5} }} \: \: \: }
tan
2
θ
=−
5
3
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