blahh trig... q (1 Viewer)

[shkspeare]

ok basically this is coming out of "part" of a question

how does [ 1+tan(x/2) ] / [ 1 - tan(x/2) ] = secx + tanx

in simplest form

i would have thought it would be tan[pi/4 + x/2]

anywayz some1 help me get secx + tanx ?

 

[Heinz]

[ 1+tan(x/2) ] / [ 1 - tan(x/2) ] = secx + tanx
LHS = [1 + 2tan(x/2) + tan²(x/2)]/[1-tan²(x/2)]
= [1 + tan²(x/2)]/[1-tan²(x/2)] + [ 2tan(x/2)]/[1-tan²(x/2)]
= [sec²(x/2)]/[1-tan²(x/2)] + [tan(x)] since [2tan(x/2)]/[1-tan²(x/2)] is the expansion of tan(x/2 + x/2)
=[1]/[cos²(x/2) -sin²(x/2)] + [tan(x)] multiplying the numerator and demoninator of the first fraction by cos²(x/2)
= [1]/[cos(x)] + [tan(x)] double angle formula for cos(x)
= sec(x) + tan(x)
= RHS

 

[shkspeare]

thx

just wondering.. which is more simplified? tan[pi/4 + x/2] or secx + tanx

if the question said to simplify it...

 

[nike33]

hehe let t = tanx/2

lhs = (1 + t) / (1 - t)
rhs = (1 + t^2 + 2t)/(1 - t^2)
= (1 + t)^2 / (1+t)(1-t)
= (1 + t) / (1 - t)
= lhs

 

[Heinz]

Originally posted by nike33
hehe let t = tanx/2

lhs = (1 + t) / (1 - t)
rhs = (1 + t^2 + 2t)/(1 - t^2)
= (1 + t)^2 / (1+t)(1-t)
= (1 + t) / (1 - t)
= lhs

Ahh.. I forgot about t formula. Much faster

 

[~GrOoVy~]

woopsi double post

 

[~GrOoVy~]

<p>hey i am having trouble with a trig question as well..<br>
<br>
find θ<br>
<br>
√6cosθ + √2sinθ + √3cotθ + 1=0<br>
<br>
20cotθ + 15cotθcosecθ- 4cosecθ= 3(1+cot^2&#952)<br>
<br>
thanx</p>

 

[Estel]

You haven't specified domains, so I've just left each at the final step.

The key to both is to factorise.

Question 1

rt6cosA + rt2sinA + rt3cotA + 1=0
Factorising:
rt3cosA(rt2+1/sinA)+sinA(rt2+1/sinA)=0
(rt2+1/sinA)(rt3cosA+sinA)=0

(rt2+1/sinA)=0
sinA= -1/rt2

(rt3cosA+sinA)=0
2cos(A-30deg)=0


Question 2

20cotA + 15cotAcosecA - 4cosecA= 3(1+cot^2A)
20cotA + 15cotAcosecA - 4cosecA - 3 - 3cot^2A = 0
20cotA + 15cotAcosecA - 4cosecA - (3cosec^2A - 3cot^2A)
- 3cot^2A = 0
20cotA + 15cotAcosecA - 4cosecA - 3cosec^2A = 0
5cotA(4 + 3 cosecA) - cosecA(4 + 3cosecA) = 0
(4 + 3cosecA)(5cotA - cosecA) = 0

3cosecA = -4
cosecA = -4/3
sinA = -3/4

5cotA - cosecA = 0
5cosA/sinA - 1/sinA=0
cosA = 1/5

 

[~GrOoVy~]

thanx estel

btw the domain was <p class="MsoNormal">0≤x≤360</p>

 

[nike33]

can you think of a quicker way now without having to rely on factorisation? (probably the same speed for qn1, faster for qn2) but you dont need to be able to factorise, generally more reliable