From Omilili

### Show that (cos18+sin18)/(cos18-sin18)=tan63?

)/(1-tan18*(1)) the**numerator**and denominator by cos18 = (tan45+tan18)/(1-tan18*(tan45)) tan45 = 1 (cos18+sin18)/(cos18-sin18) divide

**numerator**and denominator by cos(18) we get (1+ tan(18))/(1-tan ) - Sin18) { because Sin(90+A) = Cos A } =(Sin(108) + Sin18)/(Sin(108) - Sin18) =2Sin((108+18)/

**2**).Cos

### Prove : (**1**+cosA - sinA)/(**1**+cosA + sinA) = secA - tanA

TanA = sinA/cosA cotA = cosA/sinA **1**+ cot^2A = cosec^2A

**tan**^2A +

**1**=

**sec**^2A cosecA = 1/sinA ... Prove that (

**1**+cosA - sinA)/(

**1**+cosA + sinA) = secA - tanA 2. Relevant equations

**sin**^2A + cos^2A = ... Prove that (1+cosA - sinA)/(1+cosA + sinA) = secA - tanA 2. Relevant equations sin^2A + cos^2A = 1 tanA = sinA/cosA cotA...

From Yahoo Answers

**Question:**A) 3 sqr(2)/2 B) 2 sqr(2)/3 C) sqr(3)/2 D) 1/3

**Answers:**cos A = 1/3 a = 2 sin A = [1 - (1/3) ] = (8/9) = ( 8)/3 = a/c c = a / ( 8)/3 = 3a/ 8 = 3(2)/2 2 = 3/ 2 = (3 2 )/2 A) 3 sqr(2)/2 ............. correct answer B) 2 sqr(2)/3 C) sqr(3)/2 D) 1/3

**Question:**Thank you, I have 2 more I'm stuck on, this one, I don't see the substitution: Need to Integrate sec[x]tan[x]/(1+sin[x]) -- answer is ln|1+sec[x]|+C.

**Answers:**There is a typo error in the question. It should have been secx tanx / (1 + secx) dx in which case, numerator = differentiation of denominator => integral = ln l denominator l + c = ln l 1 + secx l + c.

**Question:**for 0

**Answers:**Draw a circle with radius 1. Then draw any line from the center to the perimeter (which is the radius) and make a perpendicular line to the x-axis from the point where the radius touches the perimeter. You will then get a right angled triangle inside a circle. Take the point where the radius touches the perimeter as (x,y). You will get cos A = x and sin A = y. Pythagoras theorem x2 + y2 =1 so cos2 A + sin2 A = 1. To solve the equations just substitute cos x and sin x into the equation then expand them to get the y value.

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