1. 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

2. 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) $xtan^{2}45cos60 = sin60cot30$
$tan45=1$
$cos60=\frac{1}{2}$
$sin60=\frac{\sqrt{3}}{2}$
$cot30=\sqrt{3}$
Hence
$x=\frac{\frac{\sqrt{3}}{2}*\sqrt{3}}{1^2*\frac{1}{ 2}}$
$x=3$

3. 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 . . .

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

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

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

Then: . $\frac{1}{2}x - \frac{1}{2}\;=\;\frac{1}{3}x\quad\Rightarrow\quad \frac{1}{6}x\,=\, \frac{1}{2}\quad\Rightarrow\quad x = 3$

4. 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

5. 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.

6. Solution required urgently

Hi,

Many thanks for the update.

Cheers
Lalit Chugh

7. hwy buddy

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

8. 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!
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