Answer:
Step-by-step explanation:
We are to show that [tex]\sqrt{3} cosec140^{0} - sec140^{0} = 4\\[/tex]
Proof:
From trigonometry identity;
[tex]cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}[/tex]
[tex]\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\[/tex]
From trigonometry, 2sinAcosA = Sin2A
[tex]= \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}[/tex]
Also note that sin(B-C) = sinBcosC - cosBsinC
sin420cos140 - cos420sin140 = sin(420-140)
The resulting equation becomes;
[tex]\frac{4(sin(420-140)) }{sin280}[/tex]
= [tex]\frac{4sin280}{sin280}\\ = 4 \ Proved![/tex]
