Question

In triangle ABC,angle C=120 degree, then the value of cos^2 A + cos^2 B -cosA.cosB=
A 3/4 B 3/2 C 1/2 D 1/4

Answer 1

\[A+B+120= 180 \implies A+B = 60 \]

Answer 2

\[\cos^2A + \cos^2B - \cos A\cos B = \left(\cos A -\cos B\right)^2 + \cos A \cos B\]

Answer 3

use cosC - cosD, and 2cosAcosB identities ^

Answer 4

Yup I got it.... I had done that initially too but was not getting the answer... Thanxx a lot.i have a two more questions... Could help me out with them too?

Answer 5

I'l try, ask..

Answer 6

The number of real solutions of the equation sin(e^x)=2^x+ 2^(-x) is-
A 1 B 0 C 2 D none

Answer 7

hint : \(-1 \le \sin \theta \le 1\)

Answer 8

\(\large \sin(e^x) = 2^x + 2^{-x}\)

\(\large \sin(e^x) = 2^x +\dfrac{1}{2^x}\)

\(\large 2^x\sin(e^x) = 2^{2x} + 1\)


\(\large 2^x[4-\sin(e^x) ] = -1\)

Answer 9

It should be zero right? Because all the values of the result are greater than zero?

Answer 10

Clearly the left side is always positive,
but the right side is negative - so no solutions.

Answer 11

The left hs can be positive or negative right coz the power is on e whereas the rhs can only be positive because anything to the power anything cannot be lesser than zero

Answer 12

wait there is a mistake, \(2^{2x} \ne 4*2^x\)

Answer 13

There is another way to deal with your question.

Answer 14

Yup. Basically, sina cannot be greater than or lesser than 1 but the hrs sums up to be always greater than 1 right?

Answer 15

So number of solutions are zero

Answer 16

Yep sorry I have applied opposite inequalities sign in my statements. Here is the correct one.

Answer 17

\(2^x + \cfrac{1}{2^x} \ge 2\)
or
\(2^x + \cfrac{1}{2^x} \le -2 \)

But, since

\(-1 \le \sin (e^x) \le 1\)

So there are no solutions for given equation.

Answer 18

Did you get this one ?

Answer 19

Thanks a lot @vishweshshrimali5 and @ganeshie8

Answer 20

No problem.

Answer 21

By the way can you help me with another question?

Answer 22

ya sure

Answer 23

If x1 and x2 are two values lying in [0,360] degrees range,for which Tanx=p, tan((x1)/2).tan((x2)/2)= what?

Answer 24

Sorry is that tan(x) = pi

Answer 25

No it's just a constant. In the question it was given lambda but I did not know how to represent it so p

Answer 26

No problem.

Answer 27

Also, in the given domain :

\[\tan x _{1} = \tan (180 + x_{1})\]

So
\[x_{2} = 180 + x_{1}\]

Answer 28

How did you get that?

Answer 29

There was no need of that previous formula.

You can use only the above result.

Answer 30

Which result are u talking about ?

Answer 31

I know the previous formula . I do not know how you arrived to the second conclusion

Answer 32

Ohh:

See we know that :

\[\tan (\theta) = \tan(\pi + \theta) = \tan (2\pi + \theta) = ...\]

Answer 33

But because domain is only \([0,2\pi]\) so only possibility that two angles \(x_{1}\) and \(x_{2}\) to exist such that
\[\tan (x_{1}) = \tan (x_{2}) = \lambda\]
is
\[x_{2} = x_{1} + \pi\]

Answer 34

Ok

Answer 35

Now you have to find out the value of :

\[\tan(\cfrac{x_{1}}{2}) \tan(\cfrac{x_{2}}{2})\]

Now put \(x_{2} = \pi + x_{1}\) and use this result :

\[\tan(x+\cfrac{\pi}{2}) = -\cot(x) = -\cfrac{1}{\tan(x)}\]

Answer 36

But the answer is to be given in numbers

Answer 37

Answer is -1

Answer 38

Yes and your answer is absolutely correct.

Answer 39

How do I get that?

Answer 40

See:

\[\tan(\cfrac{x_{2}}{2}) = \tan(\cfrac{x_{1} + \pi}{2})\]
\[ = \tan(\cfrac{x_{1}}{2} + \cfrac{\pi}{2})\]
\[= -\cfrac{1}{\tan(\cfrac{x_1}{2})}\]

So, the required product will become -1.

Answer 41

Did you get it ?

Answer 42

Yes thank you

Answer 43

No problem

Answer 44

what is the value of tan pi/16 +2tan pi/8 +4

Answer 45

can you'll help @vishweshshrimali5

Answer 46

you*

Answer 47

@amistre64 and @ganeshie8

Answer 48

can youll help

Answer 49

what is the value of tan pi/16 +2tan pi/8 +4

Answer 50

calculator not allowed ?

Answer 51

no, the value is supposed to be in trgonometric ratios

Answer 52

Ok let me try

Answer 53

Ok try putting pi/16 as x

Answer 54

Also, 4 = 4 tan(pi/4)

Answer 55

So, expression will become,

tan x + 2 tan 2x + 4 tan 4x

Answer 56

On simplifying this you will get,

\[\cfrac{\sin(2x)}{\cos(2x) + 1} + \cfrac{2\sin(4x)}{\cos(4x) + 1} + \cfrac{4\sin(8x)}{\cos(8x) + 1}\]

Put \(x = \cfrac{\pi}{16}\)
You will get,

\[\cfrac{\sin(2\cfrac{\pi}{16})}{\cos(2\cfrac{\pi}{16}) + 1} + \cfrac{2\sin(4\cfrac{\pi}{16})}{\cos(4\cfrac{\pi}{16}) + 1} + \cfrac{4\sin(8\cfrac{\pi}{16})}{\cos(8\cfrac{\pi}{16}) + 1}\]

Solve and simplify it.

Answer 57

You will have to find out \(\sin(\cfrac{\pi}{8})\) and \(\cos(\cfrac{\pi}{8})\) which you can easily find out using the following formulae:

(1) \(\sin{x} = \sqrt{\cfrac{1- \cos{2x}}{2}}\)

(2) \(\cos{x} = \sqrt{\cfrac{1 + \cos{2x}}{2}}\)

Answer 58

I know its very tedious but that's all I can think at present. Sorry !

Answer 59

olkay thankyou

Answer 60

btw the answer has to come in trgo ratios... the answer is cot pi/16.... how ddo i get it though?