First of all we're not trying to find a solution of the equation here, what you are suggesting is to prove that $\mathrm{lhs} =\sqrt2$ To do so we square the lhs (first read it fully) and we get $2$...

does it make sense now? just consider the angles more arbitrary than the diagram you made

There is a general identity that the locus of the foot of perpendiculars to all tangents of the parabola is the tangent at the vertex. Here is a link to the proof. So it is easier to find the line ...

Find the equation of the line $(L_1)$ joining the two given points, and find the point of intersection of the line with the circle. To calculate the intersection point, find the line $L_d$ ...

Generally, whenever we encounter greatest integer functions, we create a bound for it and try to apply sandwich theorem for the limit. Here, $na-1\le[na]\le na$
$u_{xx} - 7u_{tx} + 12u_{tt} = 0$ ...$(i)$ $u(t,x) = sinx$ on $t+3x=0$ ...$(ii)$ $u(t,x)=x$ on $t+4x=0$ ...$(iii)$ First of all, we observe from equations $(ii)$ and $(iii)$ that $u(... View answer Accepted answer 1 votes It is impossible to find an upper bound because it does not exist. We can conclude this by observing the behaviour of the polynomial as$x→∞$, which again goes to$∞$which is greater than zero. If ... View answer 0 votes So it turned out I was doing some calculation mistakes before. As suggested in the question, I got the answer this way: $$P(0<X<2(\lambda +1))=1-P(X=0)+P(X\ge 2\lambda +2)$$ $$\implies P(X=0)+P(... View answer Accepted answer 0 votes Your approach is correct if the three parts are linear or their order mattered like giving flowers to 3 people. But here because it is a circle, the total order is not important here. out of the 12... View answer 0 votes I think the relations (i) and (ii) should be sufficient to solve for \theta and \phi and hence AC. (i) would give : 2\cos\theta = 3\cos\phi and (ii) : 2\sin\theta + 3\sin\phi = 6\sin\theta \... View answer Accepted answer 0 votes They are linearly dependent as you see, we have here three equations in four variables so we can always find solutions where all four are not zero. Hence the system is linearly dependent. View answer 0 votes \lim_{n\to\infty}(3^n+7^n)^\frac1n=\lim_{n\to\infty}7(1+(\frac 37)^n)^\frac1n=7\cdot 1^0=7 View answer 0 votes Doing it by the method of vectors would be easiest I think. Consider a triangle ABC. Take point B to be origin, A to be pointed by a and C by c, now O is centroid(point of intersection of ... View answer 0 votes If you want to use trigonometric substitution, it is bettar to do$$x-3=\sqrt3\tan\theta $$so that the denominator reduces to a single term which helps make the solution easier which was not possible ... View answer 0 votes Hint: substitute y=c\sec \theta View answer Accepted answer 0 votes Yes, you are correct that they would result in the same equation. In the case of parabolas that have directrices that are parallel to one of the axes, you would think that the second method is better ... View answer 0 votes The most basic, easy to understand and straightforward method according to me is this:- Consider any point on the plane that has to be reflected, here P_1:2x+3y+4z−3=0. Let (0,0,\frac 34) (or any ... View answer 0 votes For power series,$$\sum_{n=1}^{\infty}a_nX^n$$a_n=\frac {(-1)^n}{n.5^n} , and X=-(x+3) By ratio test,$$\lim_{n\to∞}|{\frac{a_{n+1}}{a_n}}|=\frac 15$$Radius of convergence =5 So the series ... View answer 0 votes In case of equal biasing in all coins. Let, for the biased coin, the probability of landing heads is p and tails is 1-p. Then if you understood the formula given in question, The change we need ... View answer 0 votes after taking out e^{-2b^2a^2}, substitute \sqrt 2bx=t. this leaves us with \int_{-\infty}^{\infty}e^{-t^2}\text dt with some factor multiplied. Now we substitute t = \sqrt { y} leaves us with ... View answer 0 votes I believe that both have the same meaning and the first and second both are short forms for what should have been$$ \sum_{i=1}^n(\sum_{j=1}^nX_{kj}) X_{ik}$$If the first was like [i=1,j=1],[i=2,... View answer 0 votes Let r be the inradius of triangle ABC. The sides mentioned are - a is the side opposite to angle A and so on. Now we want the value of CM-AN=AC-AM-AN =AC-2AN = b - 2r (because of the square,... View answer Accepted answer -1 votes sorry for bad quality but the answer is correct. View answer Accepted answer -1 votes$$A^2=I_n$$pre multiplying A^{-1} on both sides, gives$$A=A^{-1}$\$