Please someone who knows how to solve it this should help. It's Trigonometry year 1 mathematics

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don't know if these is correct ..looks like ss3 syllabus.

coming for the next Q

Riele:
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Intelligent person

Riele:
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don't know if these is correct ..looks like ss3 syllabus.

coming for the next Q

Thanks bro.. We waiting

Modified..

Sorry for the bro. I never knew you were a girl... Just checked your profile and you're a fine girl. From my state too.

Riele:
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don't know if these is correct ..looks like ss3 syllabus.

coming for the next Q

You made a mistake in your first step by dividing both sides by 2.

Riele:
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don't know if these is correct ..looks like ss3 syllabus.

coming for the next Q

FOR once I thought someone has stolen a book of mine. You write like me.

CsRockefeller:


FOR once I thought someone has stolen a book of mine. You write like me.

Lol.

Darivie04:

You made a mistake in your first step by dividing both sides by 2.

You made another mistake too...

Please solve and post the corrections. Please

I hv a bad camera wud have helped out..
but i'll try to provide you some clues

well for 2b(i)
i'll give my explanations

since we are asked to prove..

I hope you know by now you start from the more complex side, and from the above question the L.H.S(left) is more complex....
let's take a glimpse at some derived trig formulas to be applicable in this question...
cos^2x + sin^2x=1
1+tan^2x=sec^2x
1 + cot^2x=cosec^2x

1/cosec^2(x)=sin^2(x)
1/sec^2(x)=cos^2(x)

with this above ;
step 2 gives
1/(sec^2x)^2 + cot^2x/(cosec^2x)^2



I believe you can proceed from here..
just keep substituting, there surely will be light in the tunnel

Darivie04:

You made a mistake in your first step by dividing both sides by 2.

You made another mistake too...

Yeah. She ought to find arc tan 5, then divide the resulting answer by 2 to get the angle and find the tan of the angle consequently.

@burnaboss, below is the correction to 1a.


tan 2@ = 5
2@ = arc tan 5
2@= 78.69
@ =78.69/2
@=39.345° (app 39.3°)

Therefore, tan @= tan 39.3°
=0.8185

I guess 2b(ii) is the most simple aspect of your question

2a) the binomial expansion jinx
remember the trend... you should have the question equals to (1+x)^1/2
expanding that gives
1+x + n(n-1)/2! *x^n-1 + n(n-1)(n-2)/3! *x^n-2...... the trend increases...

for no 1a

you can apply two distinctive methods


One, as queenara has solved;
and two;
the use of double Angles formula, it makes you look more professional and trig inclined

if Tan(A+B)=TanA+TanB/1-TanATanB

then Tan2A=2TanA/1-(TanA)^2...........(r)

take Tan2A=5
and TanA=t

from (r) above we have
5=2t/1-t^2

5t^2+2t-5=0
as a quadratic equation..
with your calc. an answer will surface..
also, with d formula method..
we'll eventually get +8.2 or -12.2
as t which equals TanA=Tan(Tita)


1b)

recall cos2A=1-2sin^2A

therefore, from the question above..take Tita as A
cos2A+sinA=0
1-2sin^2A + sinA=0
it equals
2sin^2A-sinA-1=0....quadratic eqn
if u solve this
you'll get
(2sinA+1)(sinA-1)=0


sinA-1=0 A=arcsin(1) A=90deg.

2sinA+1=0 sinA=-1/2
A=arcsin(-0.5)=-30deg

from the quadrant

sinA=sin(180--30)
sinA=sin210
A=210degs

Angles are 90 and 210..
note -30 would have been correct but doesnt satisfy the range 0 to 360.

jerryvyne:
for no 1a

you can apply two distinctive methods


One, as queenara has solved;
and two;
the use of double Angles formula, it makes you look more professional and trig inclined

if Tan(A+B)=TanA+TanB/1-TanATanB

then Tan2A=2TanA/1-(TanA)^2...........(r)

take Tan2A=5
and TanA=t

from (r) above we have
5=2t/1-t^2

5t^2+2t-5=0
as a quadratic equation..
with your calc. an answer will surface..
also, with d formula method..
we'll eventually get +8.2 or -12.2
as t which equals TanA=Tan(Tita)


1b)

recall cos2A=1-2sin^2A

therefore, from the question above..take Tita as A
cos2A+sinA=0
1-2sin^2A + sinA=0
it equals
2sin^2A-sinA-1=0....quadratic eqn
if u solve this
you'll get
(2sinA+1)(sinA-1)=0


sinA-1=0 A=arcsin(1) A=90deg.

2sinA+1=0 sinA=-1/2
A=arcsin(-0.5)=-30deg

from the quadrant

sinA=sin(180--30)
sinA=sin210
A=210degs

Angles are 90 and 210..
note -30 would have been correct but doesnt satisfy the range 0 to 360.

Yeahh.Good one,but he might not get this too well.

Darivie04:
You made a mistake in your first step by dividing both sides by 2.
You made another mistake too...

thanks for the correction...

jerryvyne:
for no 1a

you can apply two distinctive methods


One, as queenara has solved;
and two;
the use of double Angles formula, it makes you look more professional and trig inclined

if Tan(A+B)=TanA+TanB/1-TanATanB

then Tan2A=2TanA/1-(TanA)^2...........(r)

take Tan2A=5
and TanA=t

from (r) above we have
5=2t/1-t^2

5t^2+2t-5=0
as a quadratic equation..
with your calc. an answer will surface..
also, with d formula method..
we'll eventually get +8.2 or -12.2
as t which equals TanA=Tan(Tita)


1b)

recall cos2A=1-2sin^2A

therefore, from the question above..take Tita as A
cos2A+sinA=0
1-2sin^2A + sinA=0
it equals
2sin^2A-sinA-1=0....quadratic eqn
if u solve this
you'll get
(2sinA+1)(sinA-1)=0


sinA-1=0 A=arcsin(1) A=90deg.

2sinA+1=0 sinA=-1/2
A=arcsin(-0.5)=-30deg

from the quadrant

sinA=sin(180--30)
sinA=sin210
A=210degs

Angles are 90 and 210..
note -30 would have been correct but doesnt satisfy the range 0 to 360.

Well done boss

For the partial fractions there are calculators that do that for you(with steps) online

The last question isn't clear

Queenara:


Yeahh.Good one,but he might not get this too well.

thanks..
Huum maybe he can go through well..
I would hv preferred just solving on a paper n send either

Thukzee01:

Well done boss

You r d boss

Queenara:
@burnaboss, below is the correction to 1a.

tan 2@ = 5
2@ = arc tan 5
2@= 78.69
@ =78.69/2
@=39.345° (app 39.3°)

Therefore, tan @= tan 39.3°
=0.8185

How did you convert the 39.3° to the bolded

Darivie04:
For the partial fractions there are calculators that do that for you(with steps) online

The last question isn't clear

Find all values of Z = 64^1/3

jerryvyne:
for no 1a

you can apply two distinctive methods


One, as queenara has solved;
and two;
the use of double Angles formula, it makes you look more professional and trig inclined

if Tan(A+B)=TanA+TanB/1-TanATanB

then Tan2A=2TanA/1-(TanA)^2...........(r)

take Tan2A=5
and TanA=t

from (r) above we have
5=2t/1-t^2

5t^2+2t-5=0
as a quadratic equation..
with your calc. an answer will surface..
also, with d formula method..
we'll eventually get +8.2 or -12.2
as t which equals TanA=Tan(Tita)


1b)

recall cos2A=1-2sin^2A

therefore, from the question above..take Tita as A
cos2A+sinA=0
1-2sin^2A + sinA=0
it equals
2sin^2A-sinA-1=0....quadratic eqn
if u solve this
you'll get
(2sinA+1)(sinA-1)=0


sinA-1=0 A=arcsin(1) A=90deg.

2sinA+1=0 sinA=-1/2
A=arcsin(-0.5)=-30deg

from the quadrant

sinA=sin(180--30)
sinA=sin210
A=210degs

Angles are 90 and 210..
note -30 would have been correct but doesnt satisfy the range 0 to 360.

Please we would appreciate if you could write this down on paper, for easier understanding

burnaboss:


How did you convert the 39.3° to the bolded

It wasn't converted.
We're asked to find tan tita from the question.
Tan 39.3° gave the bolded, where 39.3° is tita.

Queenara:




It wasn't converted.
We're asked to find tan tita from the question.
Tan 39.3° gave the bolded, where 39.3° is tita.

OK. Thanks a lot

burnaboss:


OK. Thanks a lot

Yeah. You welcome

burnaboss:


Please we would appreciate if you could write this down on paper, for easier understanding

I'm sorry, my camera is faulty!

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Please someone should help with the last question

jerryvyne:
for no 1a

you can apply two distinctive methods


One, as queenara has solved;
and two;
the use of double Angles formula, it makes you look more professional and trig inclined

if Tan(A+B)=TanA+TanB/1-TanATanB

then Tan2A=2TanA/1-(TanA)^2...........(r)

take Tan2A=5
and TanA=t

from (r) above we have
5=2t/1-t^2

5t^2+2t-5=0
as a quadratic equation..
with your calc. an answer will surface..
also, with d formula method..
we'll eventually get +8.2 or -12.2
as t which equals TanA=Tan(Tita)


1b)

recall cos2A=1-2sin^2A

therefore, from the question above..take Tita as A
cos2A+sinA=0
1-2sin^2A + sinA=0
it equals
2sin^2A-sinA-1=0....quadratic eqn
if u solve this
you'll get
(2sinA+1)(sinA-1)=0


sinA-1=0 A=arcsin(1) A=90deg.

2sinA+1=0 sinA=-1/2
A=arcsin(-0.5)=-30deg

from the quadrant

sinA=sin(180--30)
sinA=sin210
A=210degs

Angles are 90 and 210..
note -30 would have been correct but doesnt satisfy the range 0 to 360.

Brilliant!!!