AP EAPCET 2023 16th May Morning Shift Question Paper with Solutions

$𝑓\left(𝑥\right)=\frac{2𝑥+1}{3}$

       $𝑦=\frac{2𝑥+1}{3}$

$\frac{1}{𝛼}=\frac{2𝛼+1}{3}$ (on putting $𝑥=𝛼$ and $𝑦=\frac{1}{𝛼}$ )

$\begin{array}{cc}\Rightarrow & 3=2{𝛼}^2+𝛼\\ \Rightarrow & 2{𝛼}^2+𝛼-3=0\\ \Rightarrow & 2{𝛼}^2+3𝛼-2𝛼-3=0\\ \Rightarrow & 𝛼(2𝛼+3)-1(2𝛼+3)=0\\ \Rightarrow & (2𝛼+3)(𝛼-1)=0\\ & 𝛼=1,-\frac{3}{2}\end{array}$

∴ Sum of all possible values of such

$𝛼=1-\frac{3}{2}=-\frac{1}{2}$

$\begin{array}{cc}& \text{ Given, }𝑓\left(𝑥\right)=\left|𝑥\right|\\ & 𝑔\left(𝑥\right)=\left|𝑥\right|+𝑎,(𝑎>0)\end{array}$

$\begin{array}{cc}& \therefore \text{ For }0\le 𝑥\le 𝑏\\ & \left[\right(𝑥,𝑦)/(𝑔\left(𝑥\right)\le 𝑦\le 𝑓\left(𝑥\right)]\end{array}$

∴ This curve represents all the points in the interior of parallelogram.

Given, there exists a ${𝑘}^{\text{th }}$ order non-singular sub-matrix in a matrix $𝑃$ of order $(𝑚\times 𝑛)$

$|𝑃{|}_{𝑘\times 𝑘}\ne 0$

So, by definition of rank of matrix.

Rank of $𝑃=𝑘$ and $𝑘\le 𝜌\le min(𝑚,𝑛)$ is also possible.

From given $𝐴=\left[\begin{array}{ccc}{𝑎}_{11}& {𝑎}_{12}& {𝑎}_{13}\\ {𝑎}_{21}& {𝑎}_{22}& {𝑎}_{23}\end{array}\right]$ and

$𝐵=\left[\begin{array}{ccc}{𝑏}_{11}& {𝑏}_{12}& {𝑏}_{13}\\ {𝑏}_{21}& {𝑏}_{22}& {𝑏}_{23}\end{array}\right]$

  Now, Option (a),            ${𝐴}^{𝑇}𝐵{𝐵}^{𝑇}𝐴={𝐵}^{𝑇}𝐴{𝐴}^{𝑇}𝐵$

$\begin{array}{cc}{𝐴}^{𝑇}𝐼𝐴={𝐵}^{𝑇}𝐼𝐵& [\because 𝐴{𝐴}^{𝑇}=𝐵{𝐵}^{𝑇}=𝐼]\\ {𝐴}^{𝑇}𝐴={𝐵}^{𝑇}𝐵& [\because 𝐼𝐴=𝐴\text{ and }𝐼𝐵=𝐵]\end{array}$

But ${𝐴}^{𝑇}𝐴\ne {𝐵}^{𝑇}𝐵$ because of value of elements are not similar in case of $𝐴$ as well as $𝐵$.

So, option (a) is incorrect.

Option (b) From above

Order of ${𝐴}^{𝑇}{𝐴}_{(3\times 3)}=$ Order of ${𝐵}^{𝑇}{𝐵}_{(3\times 3)}$

∴ Option (b) is correct.

Option (c),

The order of $𝐴+𝐵$ is $(2\times 3)$ and ${𝐴}^{𝑇}𝐵$ is $(3\times 3)$ and order of $𝐵{𝐴}^{𝑇}$ is $(3\times 3)$. So, it is not equal.

Option (d)

Rank of $𝐴$ and $𝐵$ depends on determinant of $𝐴$ and $𝐵$.

Given, in matrix $𝐴$ if all the sub-matrices of $𝐾$ th order are singular and there is one non-singular sub-matrix of order $𝑟$.

$\begin{array}{cc}\therefore & |𝐴{|}_{𝐾\times 𝐾}=0\\ \text{ But }& |𝐴{|}_{𝑟\times 𝑟}\ne 0\end{array}$

So, by definition of rank of matrix.

Rank of $𝐴$ is $𝑟\le 𝜌\left(𝐴\right)<𝐾$

Given, $\sqrt{\surd -3-4𝑖}=𝑟{𝑒}^{𝑖𝜃}$

$\begin{array}{cc}\sqrt{\surd -4+1-4𝑖}& =𝑟{𝑒}^{𝑖𝜃}\\ \sqrt{\surd (2𝑖-1{)}^2}& =𝑟{𝑒}^{𝑖𝜃}\end{array}$

[By making perfect square and ${𝑖}^2=-1$ ]

$2𝑖-1=𝑟{𝑒}^{𝑖𝜃}...\left(𝑖\right)$

Let polar form be

$𝑟{𝑒}^{𝑖𝜃}=𝑟\cos 𝜃+𝑖𝑟\sin 𝜃$

By Eq. (i), we get

$2𝑖-1=𝑟\cos 𝜃+𝑖𝑟\sin 𝜃$

On comparing, we get

$\begin{array}{cc}& 𝑟\cos 𝜃=-1,𝑟\sin 𝜃=2\\ \therefore & \cos 𝜃=\frac{-1}{𝑟}\text{ and }\sin 𝜃=\frac{2}{𝑟}\\ \therefore & {𝑟}^2{\cos }^2𝜃=1\text{ and }{𝑟}^2{\sin }^2𝜃=4...\left(𝑖𝑖\right)\end{array}$

$\begin{array}{cc}& \text{ Now, }{𝑟}^2\tan 𝜃={𝑟}^2\frac{\sin 𝜃}{\cos 𝜃}={𝑟}^2\frac{\frac{𝑟}{-1}}{\frac{-1}{𝑟}}=-2{𝑟}^2\\ & \begin{array}{c}[{𝑟}^2={𝑟}^2{\cos }^2𝜃+{𝑟}^2{\sin }^2𝜃=5\\ \therefore {\sin }^2𝜃+{\cos }^2𝜃=1]\\ =-2\times 5=-10\end{array}\end{array}$

$\because {𝑧}_1=2-3𝑖$

$\begin{array}{cc}& \therefore \text{ Sum of roots }=\frac{-𝑏}{1}\\ & -𝑏=𝑖+2-3𝑖+2+3𝑖\\ & 𝑏=-(𝑖+4)\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \therefore 𝛼𝛽+𝛽𝛾+𝛼𝛾=\frac{𝑐}{1}\\ & 𝑖(2-3𝑖)+(2-3𝑖)(2+3𝑖)+𝑖(2+3𝑖)=𝑐\\ & \therefore 𝑐=2𝑖-3{𝑖}^2+4-9{𝑖}^2+2𝑖+3{𝑖}^2\\ & \therefore 𝑐=4𝑖+13[\because {𝑖}^2=-1]\end{array}\\ & \text{ Product of roots }=\frac{-𝑑}{1}\\ & \begin{array}{cc}& -𝑑=𝑖(2-3𝑖)(2+3𝑖)=𝑖(4-9{𝑖}^2)\\ & 𝑑=-13𝑖\\ & \begin{array}{cc}\therefore 𝑏+𝑐+𝑑& =-𝑖-4+4𝑖+13-13𝑖\\ & =9-10𝑖\end{array}\end{array}\end{array}$

We know that

$(𝑎-𝑏𝜔)(𝑎-𝑏𝜔{)}^2={𝑎}^2+𝑎𝑏+{𝑏}^2$

For $𝑎=1,𝑏=1$

$(1-𝜔)(1-{𝜔}^2)=1+1+1=3$

For $𝑎=2,𝑏=1$

$(2-𝜔)(2-𝜔{)}^2=2^2+2+1$

Continuing the trend, summation becomes

$\begin{array}{cc}& \sum _{𝑘=1}^7(𝑘-𝑤)(𝑘-{𝑤}^2)\frac{𝜋}{175}\\ & =\{(1^2+2^2+\dots 7^2)+(1+2\dots +7)+1\times 7\}\frac{𝜋}{175}\end{array}$

$\begin{array}{cc}& =\{\frac{7\left(8\right)\left(15\right)}{6}+\frac{7\times 8}{2}+7\}\frac{𝜋}{175}\\ & =\{140+28+7\}\frac{𝜋}{175}=𝜋\\ & \text{ Now, }\cos \left(\sum _{𝑘=1}^7\right(𝑘-𝑤\left)(𝑘-{𝑤}^2)\frac{𝜋}{175}\right)\\ & =\cos 𝜋=-1\end{array}$

$\because 𝑧=\sqrt{\surd \frac{1-𝑖}{1+𝑖}}$

Let $𝑧=𝑥+𝑖𝑦$

$\begin{array}{cc}& 𝑥+𝑖𝑦=\sqrt{\surd \frac{1-𝑖}{1+𝑖}\times \frac{1-𝑖}{1-𝑖}}\\ & \Rightarrow 𝑥+𝑖𝑦=\pm \frac{1-𝑖}{\sqrt{\surd 2}}\\ & \Rightarrow {𝑧}_1=\frac{1}{\sqrt{\surd 2}}-\frac{1}{\sqrt{\surd 2}}𝑖\\ & \Rightarrow {𝑧}_2=\frac{-1}{\sqrt{\surd 2}}+\frac{1}{\sqrt{\surd 2}}𝑖\end{array}$

Now, we find $\arg \left({𝑧}_1\right)$ and $\arg \left({𝑧}_2\right)$

$\arg \left({𝑧}_1\right)={\tan }^{-1}\left(\frac{-\frac{1}{\sqrt{\surd 2}}}{\frac{1}{\sqrt{\surd 2}}}\right)$

$\begin{array}{cc}& \begin{array}{cc}\therefore & \arg \left({𝑧}_1\right)=-\frac{𝜋}{4}...\left(𝑖\right)\\ & \arg \left({𝑧}_2\right)={\tan }^{-1}⎛⎝\frac{\frac{1}{\sqrt{\surd 2}}}{\frac{-1}{\sqrt{\surd 2}}}⎞⎠={\tan }^{-1}(-1)\\ & \arg \left({𝑧}_2\right)=\frac{3𝜋}{4}...\left(𝑖𝑖\right)\end{array}\\ & \text{ So, the value of }\arg \left({𝑧}_1\right)+\arg \left({𝑧}_2\right)\\ & =\frac{-𝜋}{4}+\frac{3𝜋}{4}=\frac{𝜋}{2}\end{array}$

${𝑥}^2+𝑥+1=0$

$𝑥=\frac{-1\pm \sqrt{\surd 1-4}}{2}=\frac{-1\pm 𝑖\sqrt{\surd 3}}{2}$

$𝑥=\frac{-1+𝑖\sqrt{\surd 3}}{2},\frac{-1-𝑖\sqrt{\surd 3}}{2}$

$𝑥={𝜔}^2,𝜔$

$𝛼=𝜔;𝛽={𝜔}^2$

$\therefore {𝛼}^{2023}={𝜔}^{2023};{𝛽}^{1012}={\left({𝜔}^2\right)}^{1012}$

${𝛼}^{2023}=𝜔;{𝛽}^{1012}={𝜔}^2$   $[\because {𝜔}^3=1]$

Therefore, the required equation is

${𝑥}^2+𝑥+1=0$

$\begin{array}{cc}& \because 2{𝑥}^2-4𝑥+3=0\\ & \begin{array}{cc}\Rightarrow 𝛼+𝛽& =2\text{ and }𝛼\cdot 𝛽=\frac{3}{2}\\ {𝛼}^2+{𝛽}^2& =(𝛼+𝛽{)}^2-2𝛼𝛽=4-3=1\\ {𝛼}^4+{𝛽}^4& ={({𝛼}^2+{𝛽}^2)}^2-2{𝛼}^2{𝛽}^2\\ & =1-2\times \frac{9}{4}=\frac{-7}{2}\end{array}\\ & \begin{array}{c}\therefore \frac{2({𝛼}^4+{𝛽}^4)+3({𝛼}^2+{𝛽}^2)}{𝛼+𝛽}=\frac{2\left(\frac{-7}{2}\right)+3\left(1\right)}{2}\\ =\frac{-7+3}{2}=\frac{-4}{2}=-2\end{array}\end{array}$

Given, $𝑎{𝑥}^2+𝑏𝑥+𝑐=0$

$\because 𝛼$ and $𝛽$ are the root of the given equation,

$\begin{array}{cc}& \Rightarrow 𝛼+𝛽=\frac{-𝑏}{𝑎}\text{ and }𝛼𝛽=\frac{𝑐}{𝑎}\\ & \therefore \frac{1}{𝛼}+\frac{1}{𝛽}=\frac{𝛼+𝛽}{𝛼𝛽}=\frac{-𝑏}{𝑎}\times \frac{𝑎}{𝑐}=-\frac{𝑏}{𝑐}\end{array}$

∴ Required equation is

$\begin{array}{cc}& {𝑥}^2-(𝛼+𝛽+\frac{1}{𝛼}+\frac{1}{𝛽})𝑥+(𝛼+𝛽)(\frac{1}{𝛼}+\frac{1}{𝛽})=0\\ & \Rightarrow {𝑥}^2-(\frac{-𝑏}{𝑎}-\frac{𝑏}{𝑐})𝑥+\frac{-𝑏}{𝑎}\cdot \frac{-𝑏}{𝑐}=0\\ & \Rightarrow {𝑥}^2+\left(\frac{𝑏𝑐+𝑎𝑏}{𝑎𝑐}\right)𝑥+\frac{{𝑏}^2}{𝑎𝑐}=0\\ & \Rightarrow 𝑎𝑐{𝑥}^2+(𝑏𝑐+𝑎𝑏)𝑥+{𝑏}^2=0\end{array}$

Given equation,

$16{𝑥}^3-44{𝑥}^2+36𝑥-9=0$

Let $𝛼,𝛽,𝛾$ are the roots of the given equation,

$\begin{array}{c}𝛼+𝛽+𝛾=\frac{11}{4}...\left(𝑖\right)\\ 𝛼𝛽+𝛽𝛾+𝛾𝛼=\frac{9}{4}...\left(𝑖𝑖\right)\\ 𝛼𝛽𝛾=\frac{9}{16}...\left(𝑖𝑖𝑖\right)\end{array}$

$\because 𝛼,𝛽,𝛾$ are in HP

$\begin{array}{cc}\therefore & \frac{2}{𝛽}=\frac{1}{𝛼}+\frac{1}{𝛾}=\frac{𝛼+𝛾}{𝛼𝛾}\\ \therefore & 2𝛼𝛾=𝛼𝛽+𝛽𝛾\end{array}$

On putting this in Eq. (ii), we get

$3𝛼𝛾=\frac{9}{4}\Rightarrow 𝛼𝛾=\frac{3}{4}...\left(𝑖𝑣\right)$

On putting this in Eq. (iii), we get $𝛽=\frac{3}{4}$

Now, on putting $𝛽=\frac{3}{4}$ in Eq. (i), we get

$𝛼+𝛾=\frac{11}{4}-\frac{3}{4}=2...\left(𝑣\right)$

From Eqs. (iv) and (v), we get

$\begin{array}{cc}4{𝛼}^2-8𝛼+3& =0\\ (2𝛼-1)(2𝛼-3)& =0\Rightarrow 𝛼=\frac{1}{2},\frac{3}{2}\end{array}$

Now, put $𝛼=\frac{1}{2},\frac{3}{2}$ in Eq. (iv), we get

$𝛾=\frac{3}{2},\frac{1}{2}$

∴ Its greatest root is $\frac{3}{2}$.

$\because 16!=16\times 15\times 14\times 13\times 12\times 11$

$\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$

$\Rightarrow 16!=2^{4+1+2+1+3+1+2+1}\times 15\times 7\times 13$

$\times 3\times 11\times 5\times 9\times 7\times 3\times 5\times 3\times 1$

$\therefore 16!=2^{15}𝑘\Rightarrow 𝑘=\frac{16!}{2^{15}}$

$\because 2^{𝑛}$ divides 16 ! and $2^{𝑛+1}$ does not divide 16! (given)

$\therefore 2^{𝑛}=2^{15}$

Hence, $𝑛=15$

Coefficient of ${𝑥}^{𝑟}$ in the expansion of

$\frac{1}{\text{exception 25:}}=\frac{1}{(1-2𝑥{)}^{\frac{2}{3}}}=(1-2𝑥{)}^{\frac{-2}{3}}$

By using binomial expansion

$\begin{array}{cc}& \frac{[\frac{2}{3}(\frac{2}{3}+1)(\frac{2}{3}+2)\dots (\frac{2}{3}+𝑟-1)(2𝑥{)}^{𝑟}]}{(𝑟!)}\\ & =\frac{\frac{2}{3}\cdot \frac{5}{3}\cdot \frac{8}{3}\dots \frac{3𝑟-1}{3}{2}^{𝑟}{𝑥}^{𝑟}}{(𝑟!)}\\ & \Rightarrow \frac{2\cdot 5\cdot 8\cdot \dots (3𝑟-1)}{𝑟!}{\left(\frac{2}{3}\right)}^{𝑟}{𝑥}^{𝑟}\\ & \therefore \text{ Coefficient of }{𝑥}^{𝑟}=\frac{2\cdot 5\cdot 8\dots (3𝑟-1)}{𝑟!}{\left(\frac{2}{3}\right)}^{𝑟}\end{array}$

$\begin{array}{cc}& \because 3\times {}^5{𝐶}_0+8\times {}^5{𝐶}_1+13\times {}^5{𝐶}_2+18\times {}^5{𝐶}_3+23\times {}^5{𝐶}_4+28\times {}^5{𝐶}_5\\ & =3\times {}^5{𝐶}_0+8\times {}^5{𝐶}_1+13\times {}^5{𝐶}_2+18\times {}^5{𝐶}_2+23\times {}^5{𝐶}_1+28\times {}^5{𝐶}_0\end{array}$

$[\because {}^{𝑛}{𝐶}_{𝑟}={}^{𝑛}{𝐶}_{𝑛-𝑟}]$

$\begin{array}{cc}& =31\times {}^5{𝐶}_0+31\times {}^5{𝐶}_1+{31}^5\times {𝐶}_2\\ & =31({}^5{𝐶}_0+{}^5{𝐶}_1+{}^5{𝐶}_2)\\ & =31[\frac{5!}{5!0!}+\frac{5!}{4!1!}+\frac{5!}{3!2!}]\\ & =31[1+5+10]=31\times 2^4\end{array}$

(On comparing with $𝑘\times 2^4$ from questions)

$\therefore 𝑘=31$

${\because }^{(𝑛-1)}{𝐶}_6+{}^{(𝑛-1)}{𝐶}_7<{}^{𝑛}{𝐶}_8$

$\begin{array}{cc}& {}^{𝑛}{𝐶}_7<{}^{𝑛}{𝐶}_8[\because {}^{𝑛}{𝐶}_{𝑟}+{}^{𝑛}{𝐶}_{𝑟+1}={}^{𝑛+1}{𝐶}_{𝑟+1}]\\ & \Rightarrow \frac{𝑛!}{(𝑛-7)(𝑛-8)!7!}<\frac{𝑛!}{(𝑛-8)!8\times 7!}\\ & \Rightarrow 8<𝑛-7\\ & \Rightarrow 𝑛>15\end{array}$

So, the least value of $𝑛$ is 16 .

We know that a number is divisible by 3 if sum of its digits is divisible by 3 .

Hence, we must have

$\begin{array}{cc}& (4+6+9+5+3)+(𝑥+𝑦)=3𝑘\\ & 27+(𝑥+𝑦)=3𝑘...\left(𝑖\right)\\ & \therefore 𝑥+𝑦\text{ is also multiple of }3...\left(𝑖𝑖\right)\end{array}$

Again since all the seven digit numbers is to have distinct digits therefore $𝑥$ and $𝑦$ must be chosen from $0,1,2,7,8$ (excluding 4, 6, 9, 5, 3) such that $𝑥+𝑦$ is a multiple of 3 by Eq. (ii).

If $𝑥=0$, then $𝑦$ is not possible.

If $𝑥=1$, then $𝑦=2,8(1,2),(1,8)$

If $𝑥=2$, then $𝑦=1,7(2,1),(2,7)$

If $𝑥=7$, then $𝑦=2,8(7,2),(7,8)$

If $𝑥=8$, then $𝑦=1,7(8,1),(8,7)$

Hence, in all 8 ordered pairs $(𝑥,𝑦)$ are possible.

$\begin{array}{cc}& \\ & \begin{array}{cc}\text{ LHS }={\sin }^4\frac{𝜋}{8}+{\sin }^4\frac{3𝜋}{8}& +{\sin }^4\frac{5𝜋}{8}+{\sin }^4\frac{7𝜋}{8}\end{array}\\ & \begin{array}{cc}& =\frac{1}{4}{(1-\cos \frac{𝜋}{4})}^2+\frac{1}{4}{(1-\cos \frac{3𝜋}{4})}^2+\frac{1}{4}{(1-\cos \frac{5𝜋}{4})}^2+\frac{1}{4}{(1-\cos \frac{7𝜋}{4})}^2\\ & =\frac{(\sqrt{\surd 2}-1{)}^2}{8}+\frac{(\sqrt{\surd 2}+1{)}^2}{8}+\frac{(\sqrt{\surd 2}+1{)}^2}{8}+\frac{(\sqrt{\surd 2}-1{)}^2}{8}\\ & =\frac{1}{4}\left(\right(\sqrt{\surd 2}+1{)}^2+(\sqrt{\surd 2}-1{)}^2)=\frac{3}{2}\end{array}\end{array}$

$\begin{array}{cc}& \text{ To find : }\tan 𝐴\tan 𝐵+\tan 𝐵\tan 𝐶+\tan 𝐶\tan 𝐴\\ & =\frac{\sin 𝐴\sin 𝐵}{\cos 𝐴\cdot \cos 𝐵}+\frac{\sin 𝐵\cdot \sin 𝐶}{\cos 𝐵\cos 𝐶}+\frac{\sin 𝐶\sin 𝐴}{\cos 𝐶\cos 𝐴}\end{array}$

$=\frac{\sin 𝐴\cdot \sin 𝐵\cdot \cos 𝐶+\sin 𝐵\cdot \sin 𝐶\cdot \cos 𝐴+\sin 𝐶\cdot \sin 𝐴\cdot \cos 𝐵}{\cos 𝐴\cdot \cos 𝐵\cdot \cos 𝐶}$

$=\frac{-\cos 𝐴\cos 𝐵\cos 𝐶+\cos (𝐴+𝐵+𝐶)}{\cos 𝐴\cos 𝐵\cos 𝐶}$

As $\cos (𝐴+𝐵+𝐶)=\sin 𝐴\sin 𝐵\cos 𝐶$

$+\sin 𝐴\cos 𝐵\sin 𝐶+\cos 𝐴\sin 𝐵\sin 𝐶$

$+\cos 𝐴\cos 𝐵\cos 𝐶$

And $𝐴+𝐵+𝐶={180}^{\circ }$

$\Rightarrow \cos (𝐴+𝐵+𝐶)=-1$

$\therefore \frac{-\cos 𝐴\cos 𝐵\cos 𝐶+\cos (𝐴+𝐵+𝐶)}{\cos 𝐴\cos 𝐵\cos 𝐶}$

$=\frac{-\frac{1}{5}-1}{-\frac{1}{5}}=\frac{6}{5}\times \frac{5}{1}=6$

Here, $\cos (𝜃-\infty ),\cos 𝜃$ and $\cos (𝜃+\infty )$ are in HP.

$\begin{array}{cc}& \therefore \frac{2}{\cos 𝜃}=\frac{1}{\cos (𝜃-𝛼)}+\frac{1}{\cos (𝜃+𝛼)}\\ & \Rightarrow \frac{2}{\cos 𝜃}=\frac{\cos (𝜃+𝛼)+\cos (𝜃-𝛼)}{\cos (𝜃-𝛼)\cos (𝜃+𝛼)}\\ & \Rightarrow \frac{2}{\cos 𝜃}=\frac{\left[\begin{array}{c}\cos 𝜃\cos 𝛼-\sin 𝜃\sin 𝛼\\ +\cos 𝜃\cos 𝛼+\sin 𝜃\sin 𝛼\end{array}\right]}{\cos (𝜃-𝛼)\cos (𝜃+𝛼)}\\ & \Rightarrow \frac{2\cos (𝜃-𝛼)\cos (𝜃+𝛼)}{\cos 𝜃}=2\cos 𝜃\cos 𝛼\\ & \Rightarrow \frac{{\cos }^2𝜃{\cos }^2𝛼-{\sin }^2𝜃{\sin }^2𝛼}{\cos 𝜃\cos 𝛼}=\cos 𝜃\\ & \Rightarrow \cos 𝜃=\frac{{\cos }^2𝜃{\cos }^2𝛼-(1-{\cos }^2𝜃){\sin }^2𝛼}{\cos 𝜃\cos 𝛼}\\ & \Rightarrow \cos 𝜃=\frac{{\cos }^2𝜃({\cos }^2𝛼+{\sin }^2𝛼)-{\sin }^2𝛼}{\cos 𝜃\cos 𝛼}\\ & \Rightarrow {\cos }^2𝜃{\cos 𝛼={\cos }^2𝜃-{\sin }^2𝛼}_{2{\cos }^2𝛼}\\ & \Rightarrow {\sin }^2𝛼={\cos }^2𝜃(1-\cos 𝛼)\end{array}$

$\begin{array}{cc}& \Rightarrow {\cos }^2𝜃=\frac{{\sin }^2𝛼}{1-\cos 𝛼}\\ & \Rightarrow {\sin }^2𝜃=\frac{1-\cos 𝛼-{\sin }^2𝛼}{1-\cos 𝛼}\\ & \therefore 2{\tan }^2𝜃=\frac{2{\sin }^2𝜃}{{\cos }^2𝜃}\\ & =2(\frac{1-\cos 𝛼-{\sin }^2𝛼}{1-\cos 𝛼}\times \frac{1-\cos 𝛼}{{\sin }^2𝛼})\\ & =2\left(\frac{2{\sin }^2\frac{𝛼}{2}-{\sin }^2𝛼}{{\sin }^2𝛼}\right)=\frac{4{\sin }^2\frac{𝛼}{2}}{{\sin }^2𝛼}-2\end{array}$

$\begin{array}{cc}& =\frac{4{\sin }^2\frac{𝛼}{2}}{{(2\sin \frac{𝛼}{2}\cos \frac{𝛼}{2})}^2}-2={\sec }^2\frac{𝛼}{2}-2\\ & ={\sec }^2\frac{𝛼}{2}-1-1={\tan }^2\frac{𝛼}{2}-1\end{array}$