Mathematics Test - 22

Consider the third angle as A

Given, tan^-1 2, tan^-1 3 are two angles of a triangle.

We know that,

\(\tan ^{-1} {a}+\tan ^{-1} {b}=\tan ^{-1}\left(\frac{{a}+{b}}{1-{ab}}\right)\)

Consider

\(\tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}\left(\frac{2+3}{1-2.3}\right)\)

\(\tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}\left(\frac{5}{-5}\right)\)

\(\tan ^{-1} 2+\tan ^{-1} 3=\tan ^{-1}(-1)\)

\(\tan ^{-1} 2+\tan ^{-1} 3=\frac{3 \pi}{4}\)

Now, the sum of angles of the triangle is 180º i..e π.

\(\tan ^{-1} 2+\tan ^{-1} 3+{A}=\pi\)

\(\frac{3 \pi}{4}+A=\pi\)

\({A}=\frac{\pi}{4}\)

If tan^-1 2, tan^-1 3 are two angles of a triangle, then the third angle is \(\frac{\pi}{4}\).

Given,

\(2(\sec A-\cos A)^{2}+2(\operatorname{cosec} A-\sin A)^{2}-2(\cot A-\tan A)^{2}-2\)

\(=2\left[(\sec A-\cos A)^{2}+(\operatorname{cosec} A-\sin A)^{2}-(\cot A-\tan A)^{2}-1\right]\)

\(=2\left[\sec ^{2} A+\cos ^{2} A-2 \sec A \cos A+\operatorname{cosec}^{2} A+\sin ^{2} A-2 \operatorname{cosec} A \sin A-\cot ^{2} A-\tan ^{2} A+2 \cot A \tan A-1\right]\)

\(=2[\sec ^{2} A+\cos ^{2} A-2+\operatorname{cosec}^{2} A+\sin ^{2} A-2-\cot ^{2} A-\tan ^{2} =2[(\cos ^{2} A+\sin ^{2} A)+(\sec ^{2} A-\tan ^{2} A)+(\operatorname{cosec}^{2} A-\cot ^{2} A)-3]\)

\({\left[\because \cos ^{2} A+\sin ^{2} A=1, \sec ^{2} A-\tan ^{2} A=1 \text { and } \operatorname{cosec}^{2} A-\cot ^{2} A=1\right]}\)

\(=2[1+1+1-3]\)

\(=0\)

The direction cosinesof the vector are the cosines of angles that the vector forms with the coordinate axes.

The X-axis makes angles 0°, 90°, and 90° with the X, Y, and Z-axis respectively.

So, direction cosine of X-axis:

cos 0 = 1, cos 90 = 0, and cos 90 = 0

So, the sum of the squares of direction cosines of X-axis = 1²+ 0 + 0 = 1

∴ The sum of the squares of the direction cosines is equal to one.

i.e., \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\)

Given,

\(7,13,15,11,4\)

Formula used:

\(\mathrm{S} . \mathrm{D}=\sqrt{\frac{\sum|x-m|^{2}}{n}}\)

Mean \((m)=\frac{\text { Total of observations }}{\text { number of observations }}\)

\(\mathrm{S.D}=\) standard deviation

\(\sum=\) summation

\(x=\) observation

\(m=\) mean of the observations

\(n=\) number of observation

Mean of \(7,13,15,11,4\)

\(=\frac{50}{5}=10\)

\(\text { S.D }=\sqrt{\frac{(7-10)^{2}+(13-10)^{2}+(15-10)^{2}+(11-10)^{2}+(4-10)^{2}}{5}}\)

\(=\sqrt{\frac{9+9+25+1+36}{5}}\)

\(=\sqrt{\frac{80}{5}} \)

\(=\sqrt{16}=4\)

\(\therefore\) The standard deviation is \(4\).

Given,

A pack of \(52\) cards.

As we know there are \(26\) cards in total which are black. Let \(A\) and \(B\) denote respectively the eventsthat the first and second drawn cards are black.

Now, \(P (A) = P\) (black card in first draw) \(= \frac{26}{52} =\frac{1}{2}\)

Because the second card is drawn without replacement so, now the total number of black cards will be \(25\) and the total cards will be \(51\) that is the conditional probability of \(B\) given that \(A\) has already occurred.

Now, \(P (\frac{B}{A}) = P\) (black card in second draw) \(= \frac{25}{51}\)

Thus the probability that both the cards are black.

\(=P(A \cap B)=\frac{1}{2} \times \frac{25}{51}=\frac{25}{102}\)

Therefore, the probability that both the cards are black is \(\frac{25}{102}\).

Given \(y=\tan ^{-1}\left[\frac{3 a^{2} x-x^{3}}{a\left(a^{2}-3 x^{2}\right)}\right]\)

Let \(x=a \tan \theta\)

\(y=\tan ^{-1}\left[\frac{3 a^{2} a \tan \theta-(a \tan \theta)^{3}}{a\left[a^{2}-3(\operatorname{atan} \theta)^{2}\right]}\right]\)

Taking a³ common from both denominator and numerator

\(y=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^{3} \theta}{\left(1-3 \tan ^{2} \theta\right)}\right]\)

\(y=\tan ^{-1}[\tan 3 \theta]\)

\(y=3 \theta=3 \tan ^{-1}\left(\frac{x}{a}\right)\)

Now \(\frac{d y}{d x}=\frac{d}{d x}\left[3 \tan ^{-1}\left(\frac{x}{a}\right)\right]\)

\(\frac{d y}{d x}=\frac{3}{1+\frac{x^{2}}{a^{2}}} \times \frac{1}{a}\)

\(\frac{d y}{d x}=\frac{3 a}{a^{2}+x^{2}}\)

Given,

\(|z-1|=|z+i|\)

Let \(x=x+\) iy

\(|x+i y-1|=|x+i y+1|\)

\(|(x-1)+i y|=|x+i(y+1)|\)

Squaring both sides

\(2(x-1)({iy})+({x}-1)^{2}+{y}^{2} {i}^{2}={x}^{2}+{i}^{2}({y}+1)^{2}+2 {xi}({y}+1)\)

\(\Rightarrow({x}-1)^{2}-{y}^{2}+2 {xyi}-2 {yi}={x}^{2}-{y}^{2}-1-2 {y}+2 {xyi}+2 {xi}\)

\(\Rightarrow x^{2}+1-2 x-y^{2}-2 y i=x^{2}-y^{2}-2 y+2 x i-1\)

\(\Rightarrow 2 y-2 x+2=2 x i+2 y i\)

\(\Rightarrow y-x+1=i(x+y)=0\)

\(\Rightarrow {y}-{iy}-{x}-{xi}+1=0\)

\(\Rightarrow y(1-i)-x(1+i)+1=0\)

Therefore it can be written as:

by \(-a x+1=0 \quad((1-i)=b\) and \((1+i)=a\) as they are constant)

\(\therefore\) It represent the straight line passing through origin.

Given:

\(\frac{(\sin x+\cos x+1)\left(\sin x+\cos {x}-1\right)}{(\tan x+\cot x)\left(\sin ^{2} x \cos ^{2} x\right)}\)

\(=\frac{(\sin x+\cos x)^{2}-1}{\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\left(\sin ^{2} x \cos ^{2} x\right)}\)

\(=\frac{\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x-1}{\frac{\sin ^{2} x+\cos x}{\sin x \cdot \\cos ^{2} x} \times \sin ^{2} x \cos ^{2} x}\)

\(=\frac{2 \sin x \cos x}{\sin x \cos x}\)

\(=2\)

Given,

A dice is tossed thrice

Then the sample space S = {1, 2, 3, 4, 5, 6}

Let P (A) = probability of getting an odd number in the first throw.

P (A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

Let P (B) = probability of getting an even number.

⇒ P (B) =  \(\frac{3}{6}\) = \(\frac{1}{2}\)

Now, probability of getting an even number in three times = \(\frac{1}{2} × \frac{1}{2} × \frac{1}{2} = \frac{1}{8}\)

So, probability of getting an odd number at least once

= \(1\) – probability of getting an odd number in no throw

= \(1\) – probability of getting an even number in three times

= \(1\) – \(\frac{1}{8}\)

= \(\frac{7}{8}\)

∴ Probability of getting an odd number at least once is \(\frac{7}{8}\).