# Solution required urgently

• Jun 19th 2006, 10:17 AM
lalitchugh
Solution required urgently
Hi,

• Find x from each of the following equations.
1.1)x*tan^2(45)*cos(60) = sin(60)*cot(30)

1.2)(1+x)*tan(45)*cosec^2(45) = sin(30)*sec^2(45)

1.3)X*cos(60)–sin(30) = x*tan(30)*cot(60)

1.sec(120)

2.sec(350)

3.tan(40)

Request to respond as soon as possible.

Best Regards,
Lalit Chugh
• Jun 19th 2006, 05:31 PM
malaygoel
Quote:

Originally Posted by lalitchugh
Hi,

• Find x from each of the following equations.
1.1)x*tan^2(45)*cos(60) = sin(60)*cot(30)

1.2)(1+x)*tan(45)*cosec^2(45) = sin(30)*sec^2(45)

1.3)X*cos(60)–sin(30) = x*tan(30)*cot(60)

1.sec(120)

2.sec(350)

3.tan(40)

Request to respond as soon as possible.
Best Regards,
Lalit Chugh

(1.1)$\displaystyle xtan^{2}45cos60 = sin60cot30$
$\displaystyle tan45=1$
$\displaystyle cos60=\frac{1}{2}$
$\displaystyle sin60=\frac{\sqrt{3}}{2}$
$\displaystyle cot30=\sqrt{3}$
Hence
$\displaystyle x=\frac{\frac{\sqrt{3}}{2}*\sqrt{3}}{1^2*\frac{1}{ 2}}$
$\displaystyle x=3$
• Jun 19th 2006, 07:10 PM
Soroban
Hello, lalitchugh!

You're expected to know the trig values for 30°, 60°, and 45°.
Once you plug in the values, the rest is Arithmetic and Algebra I . . .

Quote:

$\displaystyle 1.2)\;\;(1+x)\cdot\tan(45^o)\cdot\csc^2(45^o) \;= \;\sin(30^o)\cdot\sec^2(45^o)$
We have: .$\displaystyle (1 + x)(1)(\sqrt{2})^2\;=\;\left(\frac{1}{2}\right)( \sqrt{2})^2$

Then: .$\displaystyle (1 + x)\cdot 2\;=\;1\quad\Rightarrow\quad 1 + x \:=\:\frac{1}{2}\quad\Rightarrow\quad x\,=\,-\frac{1}{2}$

Quote:

$\displaystyle 1.3)\;\;x\cdot\cos(60^o) - \sin(30^o) \;= \;x\cdot\tan(30^o)\cdot\cot(60^o)$
We have: .$\displaystyle x\left(\frac{1}{2}\right) - \frac{1}{2} \;= \;x\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{ \sqrt{3}}\right)$

Then: .$\displaystyle \frac{1}{2}x - \frac{1}{2}\;=\;\frac{1}{3}x\quad\Rightarrow\quad \frac{1}{6}x\,=\, \frac{1}{2}\quad\Rightarrow\quad x = 3$
• Jun 19th 2006, 08:48 PM
lalitchugh
Solution required urgently
Hi,

Can you pls advice how we can evaluate

1. tan 40

2. sec 350

I tried a lot but could not solve these.

Best Regards,
Lalit Chugh
• Jun 20th 2006, 08:29 AM
ThePerfectHacker
Quote:

Originally Posted by lalitchugh
Hi,

Can you pls advice how we can evaluate

1. tan 40

2. sec 350

I tried a lot but could not solve these.

Best Regards,
Lalit Chugh

I do not see a simple way to do it. It leads to a cubic equation which is probably not what you are looking for.
• Jun 20th 2006, 08:50 AM
lalitchugh
Solution required urgently
Hi,

Many thanks for the update.

Cheers
Lalit Chugh
• Jun 22nd 2006, 11:13 PM
killer baby
hwy buddy
Quote:

Originally Posted by lalitchugh
Hi,

Can you pls advice how we can evaluate

1. tan 40

2. sec 350

I tried a lot but could not solve these.

Best Regards,
Lalit Chugh

HEY BUDDY THESE VALUES CAN BE LOOKED UPON FROM TRIGNOMETRIC TABLES ONLY THERE IS NO SOLUTIONF FOR THEM
• Jun 23rd 2006, 09:02 AM
ThePerfectHacker
Quote:

Originally Posted by killer baby
HEY BUDDY THESE VALUES CAN BE LOOKED UPON FROM TRIGNOMETRIC TABLES ONLY THERE IS NO SOLUTIONF FOR THEM

There is a solution for them, it is only too complicated.

Do not talk with caps on! :mad:
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