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sukhopar [10]
3 years ago
10

Prove tan(θ / 2) = sin θ / (1 + cos θ) for θ in quadrant 1 .

Mathematics
2 answers:
Leona [35]3 years ago
3 0
We are asked to prove tan(θ / 2) = sin θ / (1 + cos θ). In this case, tan θ is equal to sin θ / cos θ. we can apply this to the equality. sin θ is equal to square root of (1-cos θ)/2 while cos θ is equal to <span>square root of (1 + cos θ)/2.
Hence, when we replace cos </span><span>θ with </span>square root of (1-cos θ)/2, we can prove already.
Novay_Z [31]3 years ago
3 0

Answer:

Step-by-step explanation:

We have to prove the identitytan(\frac{\Theta }{2})=\frac{sin\Theta}{1+cos\Theta }

We will take right hand side of the identity

\frac{sin\Theta}{1+cos\Theta}=\frac{2sin(\frac{\Theta }{2})cos(\frac{\Theta }{2})}{1+[2cos^{2}(\frac{\Theta }{2})-1]}

=\frac{2sin(\frac{\Theta }{2})cos(\frac{\Theta }{2})}{2cos^{2}(\frac{\Theta }{2})}=\frac{sin(\frac{\Theta }{2})}{cos(\frac{\Theta }{2})}

=tan(\frac{\Theta }{2}) [ Tan θ will be positive since θ lies in 1st quadrant ]

= L. H. S.

Hence proved.

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I will answer this question using the attached triangle

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Sin \theta =\frac{Opposite}{Hypotenuse} and

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Substitute values for BC and BA

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Sin \angle A =\frac{8}{10}

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Substitute values for AC and BA

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Solving (b): Sine and Cosine B

In trigonometry:

Sin \theta =\frac{Opposite}{Hypotenuse} and

Cos \theta =\frac{Adjacent}{Hypotenuse}

So:

Sin \angle B =\frac{AC}{BA}

Substitute values for AC and BA

Sin \angle B =\frac{6cm}{10cm}

Sin \angle B =\frac{6}{10}

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Cos \angle B=\frac{BC}{BA}

Substitute values for BC and BA

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