Question

Prove that LHS=RHS
$\sin {40}^{o} - \cos {70}^{o} = \sqrt{3} \cos {80}^{o}$
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Answer 1

Answer:

x2−3xx2−9=1−3x−9x2−9.

and that I have to prove that the LHS = RHS; How would I go about doing so?

What I've done:

I've simply brought (3x−9)/(x2−9) over to the LHS, resulting in the equation (x2−3x)/(x2−9)+(3x−9)/(x2−9)=1 , which gives me (x2−9)/(x2−9)=1.(LHS = RHS)

However, I do not think that what I've done is the correct method of proving. Should I have instead transformed the LHS into RHS without bringing anything from LHS to RHS and vice versa?

Furthermore , after which I'm integrating (x2−3x)/(x2−9) by doing the following:

What I first did was factorise the denominator into (x+3)(x−3). I then equated (x2−3x)=A(x+3)+B(x−3).

Given that x=3⟹A=0 and given that x=−3⟹B=−3.

This would then give me the following equation to integrate:

[(−3)/(x+3)+0(x−3)]dx

which led me to the answer −3ln|x+3|+C.

However, when I checked my answer, the correct answer should be

x−3ln|x+3|+C.

Where did the missing x come from?

Any assistance would be greatly appreciated

Thanks!

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Step-by-step explanation:

Answer 2

Answer:-

LHS=

$\sin {40}^{o} - \cos {70}^{o}$

$= \sin( {90}^{o} - {50}^{o} ) - \cos {70}^{o}$

$= \cos {50}^{o} - \cos {70}^{o}$

$= - 2 \sin {60}^{o} \sin( { -10 }^{o} )$

$= 2 \sin {60}^{o} \sin {10}^{o}$

$= 2 \times \frac{ \sqrt{3} }{2} \cos80$

$= \sqrt{3} \cos80$

=RHS

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