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$ta{n}^{-1}\frac{x-1}{x-2}+ta{n}^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

(E)Using $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{x+y}{1-xy}.$

$\Rightarrow ta{n}^{-1}\frac{\frac{(x-1)}{(x-2)}+\left(\frac{x+1}{x+2}\right)}{1-\frac{(x-1)}{(x-2)}\times \left(\frac{x+1}{x+2}\right)}=\frac{\pi }{4}.$

$\Rightarrow \frac{\frac{(x-1)(x+2)-1(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}}=tan\frac{\pi }{4}$

$\Rightarrow \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=1.$

$\Rightarrow \frac{x^2+2x-x-2+x^2-2x+x-2}{\left(x^2-4\right)-\left(x^2-1^2\right)}=1.$ $\left\{\begin{array}{cc}\hfill ?(a-b)(a+b)=\hfill & \hfill a^2-b^2\hfill \end{array}\right\}$

$\Rightarrow \frac{2x^2-4}{x^2-4-x^2+1}=1\Rightarrow \frac{2x^2-4}{-3}=1\Rightarrow 2x^2-4=-3$

$sin\left(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x\right)=1.$

(E) $\Rightarrow sin\left(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x\right)=sin\frac{\pi }{2}\left\{?sin\frac{\pi }{2}=1\right\}$

$\Rightarrow si{n}^{-1}\frac{1}{5}+co{s}^{-1}x=si{n}^{-1}\left(sin\frac{\pi }{2}\right)=\frac{\pi }{2}$

$\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}\frac{1}{5}$

$\Rightarrow co{s}^{-1}x=co{s}^{-1}\frac{1}{5}$ $\left\{?\frac{\pi }{2}=si{n}^{-1}x+co{s}^{-1}x\right\}$

$=\frac{\pi }{2}-si{n}^{-1}\frac{1}{5}=co{s}^{-1}\frac{1}{5}$ $\{\begin{array}{l}\Rightarrow \frac{\pi }{2}-si{n}^{-1}x=co{s}^{-1}x\\ x=\frac{1}{5}\\ \Rightarrow \frac{\pi }{2}-si{n}^{-1}\frac{1}{5}=co{s}^{-1}\frac{1}{5}\end{array}.$

= x = $\frac{1}{5}.$

 $tan\frac{1}{2}\left[si{n}^{-1}\frac{2x}{1+x^2}+co{s}^{-1}\frac{1-y^2}{1+y^2}\right],\left|x\right|<1,y>$

(M) Let x = tanØ . then tan^-1x= 0 and y = tan ω then tan^-1y = ω. we have,

tan $\frac{1}{2}$ $\{sin{}^{-1}\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{1-ta{n}^2\omega }{1+ta{n}^2\omega }\}$

$=tan\frac{1}{2}\left\{si{n}^{-1}(sin2\theta )+co{s}^{-1}(cos2\omega )\right\}$ $\left\{\therefore sin2\theta \frac{2tan\theta }{1+ta{n}^2\theta }cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }tanx+y=\frac{tanx+tany}{1-tanxtany}\right\}$

$=tan\frac{1}{2}\{2\theta +2\omega \}=tan(\theta +\omega )$

$=\frac{tan\theta +tan\omega }{1-tan\theta tan\omega }$

= $\frac{x=y}{1-xy}$

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cot $\left(ta{n}^{-1}a+co{t}^{-1}a\right)=cot\frac{\pi }{2}\left\{?ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}\right\}$

(E) $=0$

Yes, admission for the BTech programme at NIILM University for the 2025 academic year is currently open. Interested students are encouraged to apply early, as seats are allotted based on merit. It's important to regularly check the university's official website for updates on deadlines, required documents, and other admission-related announcements.

$ta{n}^{-1}\left[2cos\left(2si{n}^{-1}\frac{1}{2}\right)\right]=ta{n}^{-1}\left[2cos\left(2si{n}^{-1}sin\frac{\pi }{6}\right)\right]$

(E) $=ta{n}^{-1}\left[2cos\left(2\times \frac{\pi }{6}\right)\right]$

$=ta{n}^{-1}\left[2cos\frac{\pi }{3}\right]$

$=ta{n}^{-1}2\times \frac{1}{2}$

$=ta{n}^{-1}1$

$=ta{n}^{-1}\left(tan\frac{\pi }{4}\right)$

$=\frac{\pi }{4}$

Given tan ^-1 $\frac{cosx-sinx}{cosx+sinx}$ , $\frac{-\pi }{4},$ x< $\frac{3x}{4}$

(M) Dividing numerator & denominator cos x we get,

tan ^-1 $\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}}=ta{n}^{-1}\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}}=ta{n}^{-1}\frac{1-tanx}{1+tanx.}$

We know that $tan\frac{\pi }{4}=1$ = 1 so,

$=ta{n}^{-1}\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx}=ta{n}^{-1}\left[tan\left(\frac{\pi }{4}-x\right)\right]$ $\{?tan(x-y)=\frac{tanx-tany}{1+tanx·tany}\}$

$=\frac{\pi }{4}-x$ .

62. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n (H) = 100

Number of students who know English = n (E) = 50

Number of students who know both English & Hindi = 25 = n (H$\cap$E)

As each of students knows either Hindi or English,

Total number of students in the group,

n (H$\cup$E) = n (H) + n (E) - n (H$\cap$E)

= 100 + 25

= 125,

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L.H.S= 2 tan ^-1 $\frac{1}{2}$ + tan ^-1 $\frac{1}{7}$

(E) Using2 tan ^-1x= tan ^-1 $\frac{2a}{1-x^2}$ we can write.

L.H.S = tan ^-1 $\frac{2\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{1-{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{1\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1$\frac{1}{7}$

= tan ^-1 $\frac{1}{4\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1$\frac{1}{7}$ = tan ^-1 $\frac{4}{3}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1$\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}$ { Ø tan -1 x + tan -1 y = tan - 1 $\frac{x+y}{1-xy}$ }

= tan ^-1 $\frac{\frac{7\times 4+1\times 3}{3\times 7}}{\frac{3\times 7-4\times 1}{3\times 7}}$

= tan -1 $\frac{28+3}{21-4}$ = tan ^-1 $\frac{31}{17}$ = R H S

61. Let T and C be sets of students taking tea and coffee.

Then, n (T) = 150, number of students taking tea

n (C) = 225, number of students taking coffee

n (T$\cap$C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

n (T$\cup$C) = n (T) + n (C) n (T$\cap$C)

= 150 + 225 100

= 275

Number of students taking neither tea coffee

= Total number of students No of students taking either tea or coffee

= 600 275

= 325.

L.H.S = tan ^-1$\frac{2}{11}$ + tan ^-1 $\frac{7}{24}$

Using tan ^-1x+ tan ^-1y= tan ^-1 $\frac{x+y}{1-xy}$ , xy<1

L.H.S =tan ^-1 $\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\times \frac{7}{24}}$ = tan ^-1 $\frac{\frac{2\times 24+7\times 2}{11\times 24}}{\frac{11\times 24-7\times 2}{11\times 24}}$

tan ^-1 $\frac{48+14}{264-14}$ = tan ^-1 $\frac{125}{250}$ = tan ^-1 $\frac{1}{2}$ = R.H.S

60. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A$\cap$B = {x, y}$\cap$ {y, z} = {y}≠$?$

B$\cap$C = {y, z} $\cap$ {x, z} = {z}≠$?$

A$\cap$C = {x, y}$\cap$ {x, z} = {x}≠$?$

But A$\cap$B$\cap$C = (A$\cap$B)$\cap$ C

= {y} $\cap$ (x, z}

=$?$

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We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3 (cos^-1x) = 4x³-3x

3cos^-1x = cos^-1 (4x³- 3x)

Hence Proved

To apply for the BTech programme at NIILM University, students must have completed their Class 12 with Physics, Chemistry, and Mathematics (PCM) from a recognised board. For lateral entry, candidates need a Diploma in engineering. Meeting the eligibility criteria is essential to move forward in the admission process.