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Lady bird [3.3K]
3 years ago
14

Root 3 cosec140° - sec140°=4prove that​

Mathematics
1 answer:
Lorico [155]3 years ago
3 0

Answer:

Step-by-step explanation:

We are to show that \sqrt{3} cosec140^{0} - sec140^{0} = 4\\

<u>Proof:</u>

From trigonometry identity;

cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}

\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\

From trigonometry, 2sinAcosA = Sin2A

= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\=  \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\=  \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2  \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}

Also note that sin(B-C) = sinBcosC - cosBsinC

sin420cos140 - cos420sin140 = sin(420-140)

The resulting equation becomes;

\frac{4(sin(420-140)) }{sin280}

= \frac{4sin280}{sin280}\\ = 4 \ Proved!

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<em>Comment on the figure</em>

The usual geometry here is that the outside legs (opposite the vertical angles) are parallel, meaning that the angle indicators are the correct marks. It is possible, but unusual, for the red hash marks to be correct and the angle indicators to be mismarked. The red hash marks seem tentatively drawn, so seem like they're more likely to be the incorrect marks.

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