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3 years ago

8

claire was able to verify that x=3 was a solution to her teacher's linear equation, but the equation got erasted from thw board.

what might the equation have been? identify as many equations as you can with a solution of x=3.

1 answer:

svet-max [94.6K]3 years ago

7 0

X+5=8
x+(-7)=-4
4+x=7
7-x=4
3-x=0
-9×x= -27
those are just a few options

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Show that √(1-cos A/1+cos A) =cosec A - cot A

Gennadij [26K]

Hi there!

$\sqrt{\frac{1-cosA}{1+cosA}} =$

We can begin by multiplying by its conjugate:

$\sqrt{\frac{1-cosA}{1+cosA}} * \sqrt{\frac{1+cosA}{1+cosA}} = \\\\\sqrt{\frac{(1-cosA)(1 + cosA)}{(1+cosA)(1 + cosA)}} =$

Simplify using the identity:

$1 - cos^2A = sin^2A$

$\sqrt{\frac{(1-cos^2A)}{(1+cosA)^2}} =\\\\\sqrt{\frac{(sin^2A)}{(1+cosA)^2}} =$

Take the square root of the expression:

${\frac{sinA}{1+cosA} =$

Multiply again by the conjugate to get a SINGLE term in the denominator:

${\frac{sinA}{1+cosA} * {\frac{1-cosA}{1-cosA} =\\$

Simplify:

${\frac{sinA(1-cosA)}{1-cos^2A} =$

Use the above trig identity one more:

${\frac{sinA(1-cosA)}{sin^2A} =$

Cancel out sinA:

${\frac{(1-cosA)}{sinA} =$

Split the fraction into two:

${\frac{1}{sinA} - \frac{cosA}{sinA} =$

Recall:

$1/sinA = cscA\\\\cosA/sinA = cotA$

Simplify:

$\frac{1}{sinA} + \frac{cosA}{sinA} = \boxed{cscA - cotA}$

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