⇒Trigonometric Functions

Correct answer option is (B)

$\begin{array}{rl}& (a-b{)}^2{\cos }^2\frac{C}{2}+(a+b{)}^2{\sin }^2\frac{C}{2}\\ & =(a^2-2ab+b^2){\cos }^2\frac{C}{2}+(a^2+2ab+b^2){\sin }^2\frac{C}{2}\\ & =(a^2+b^2)({\cos }^2\frac{C}{2}+{\sin }^2\frac{C}{2})\\ & -2ab{\cos }^2\frac{C}{2}+2ab{\sin }^2\frac{C}{2}\\ & =a^2+b^2-2ab({\cos }^2\frac{C}{2}-{\sin }^2\frac{C}{2})\\ & =a^2+b^2-2ab\cdot {\cos }^2\\ & =a^2+b^2-(a^2+b^2-c^2)\dots [By\text{ Cosine rule }]\\ & =c^2=4^2=16\end{array}$

Correct answer option is (B)

In $\mathrm{△}\mathrm{ABC}$, by sine rule,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

According to the given condition,

In $\mathrm{△}\mathrm{ABC},\mathrm{a}=2\mathrm{b}$ and

$\begin{array}{rl}& \mathrm{A}-\mathrm{B}={60}^{\circ }\Rightarrow \mathrm{A}={60}^{\circ }+\mathrm{B}\\ & \Rightarrow \frac{\sin ({60}^{\circ }+\mathrm{B})}{2\mathrm{b}}=\frac{\sin \mathrm{B}}{\mathrm{b}}\\ & \Rightarrow \frac{\sin \mathrm{B}}{\sin (\mathrm{B}+{60}^{\circ })}=\frac{1}{2}\\ & \Rightarrow 2\sin \mathrm{B}=\sin \mathrm{B}\cos {60}^{\circ }+\cos \mathrm{B}\sin {60}^{\circ }\\ & \Rightarrow 2\sin \mathrm{B}=\sin \mathrm{B}\left(\frac{1}{2}\right)+\cos \mathrm{B}\left(\frac{\sqrt{3}}{2}\right)\\ & \Rightarrow \frac{3}{2}\sin \mathrm{B}=\frac{\sqrt{3}}{2}\cos \mathrm{B}\end{array}$

$\begin{array}{rl}& \Rightarrow \tan B=\frac{1}{\sqrt{3}}\Rightarrow B={30}^{\circ }\\ & \therefore A={30}^{\circ }+{60}^{\circ }={90}^{\circ }\end{array}$

$\therefore \mathrm{△}\mathrm{ABC}$ is right angled.

Correct answer option is (B)

Let ${\mathrm{p}}_1,{\mathrm{p}}_2,{\mathrm{p}}_3$ be the altitudes of $\mathrm{△}\mathrm{PQR}$ Area of $\mathrm{△}\mathrm{PQR}$

$\begin{array}{rl}& =\frac{1}{2}\times \text{ base }\times \text{ height }\\ & =\frac{1}{2}\times {\mathrm{p}}_1\times \mathrm{a}\end{array}$

Area $=\frac{1}{2}{p}_{1}a$

$\therefore {\mathrm{p}}_1=2\times \frac{\text{ Area }}{\mathrm{a}}$ ..... (i)

$\therefore$ similarly,

${\mathrm{p}}_2=\frac{2\times \text{ Area }}{\mathrm{b}}$ ..... (ii)

${\mathrm{p}}_3=\frac{2\times \text{ Area }}{\mathrm{c}}$ ..... (iii)

$\therefore$ By sine Rule,

$\frac{\mathrm{a}}{\sin \mathrm{P}}=\frac{\mathrm{b}}{\sin \mathrm{Q}}=\frac{\mathrm{c}}{\sin \mathrm{R}}$

Let $\frac{a}{\sin P}=\frac{b}{\sin Q}=\frac{c}{\sin R}=k$

$\therefore \sin \mathrm{P}=\frac{\mathrm{a}}{\mathrm{k}},\sin \mathrm{Q}=\frac{\mathrm{b}}{\mathrm{k}},\sin \mathrm{R}=\frac{\mathrm{c}}{\mathrm{k}}$

$\sin P,\sin Q$ and $\sin R$ are in A.P.

$\therefore$ a, b, c are in A.P.

$\therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in H.P. ..... (iv)

From equations (i), (ii), (iii) and (iv), we get ${\mathrm{p}}_1,{\mathrm{p}}_2$ and ${\mathrm{p}}_3$ are in H.P.

Correct answer option is (A)

$\begin{array}{rl}& \text{ In }\mathrm{△}ABC\\ & \frac{a+b}{7}=\frac{b+c}{8}=\frac{c+a}{9}=k\text{... [Given]}\\ & \therefore a+b=7k\text{.... (i)}\\ & b+c=8k\text{.... (ii)}\\ & c+a=9k\text{.... (iii)}\end{array}$

Adding above equations,

$\begin{array}{rl}& 2\mathrm{a}+2\mathrm{b}+2\mathrm{c}=24\mathrm{k}\\ & \mathrm{a}+\mathrm{b}+\mathrm{c}=12\mathrm{k}\text{.... (iv)}\end{array}$

Solving equations (i), (ii), (iii), (iv) We get,

$\begin{array}{rl}& \mathrm{c}=5\mathrm{k},\mathrm{a}=4\mathrm{k},\mathrm{b}=3\mathrm{k}\\ & \therefore {\mathrm{c}}^2={\mathrm{a}}^2+{\mathrm{b}}^2\end{array}$

$\therefore \mathrm{△}\mathrm{ABC}$ is right angled triangle

$\therefore \mathrm{\angle }\mathrm{C}={90}^{\circ }$

$\begin{array}{r}\begin{array}{c}\text{ Area of }\mathrm{△}\mathrm{ABC}=\frac{1}{2}\mathrm{ab}\sin \mathrm{C}\\ =\frac{1}{2}\mathrm{ab}\sin 90\\ =\frac{1}{2}\times 4\mathrm{k}\times 3\mathrm{k}=6{\mathrm{k}}^2\end{array}\\ \therefore \text{ Now, }\frac{[\mathrm{A}(\mathrm{△}\mathrm{ABC}){]}^2}{{\mathrm{k}}^4}=\frac{{\left(6{\mathrm{k}}^2\right)}^2}{{\mathrm{k}}^4}=\frac{36{\mathrm{k}}^4}{{\mathrm{k}}^4}=36\end{array}$

Correct answer option is (A)

$\begin{array}{ll}& \text{ In }\mathrm{△}\mathrm{ABC},\\ & \mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{B}+\mathrm{\angle }\mathrm{C}={180}^{\circ }\\ \therefore & \mathrm{\angle }\mathrm{A}+\frac{\pi }{2}+\mathrm{\angle }\mathrm{B}={180}^{\circ }\\ \therefore & \mathrm{\angle }\mathrm{A}+\mathrm{\angle }\mathrm{B}=\frac{\pi }{2}\\ \therefore & \frac{\mathrm{\angle }\mathrm{A}}{2}+\frac{\mathrm{\angle }\mathrm{B}}{2}=\frac{\pi }{4}\\ & \tan \left(\frac{\mathrm{A}}{2}\right)\text{ and }\tan \left(\frac{\mathrm{B}}{2}\right)\text{ are roots of equation }\\ & {\mathrm{a}}_1{x}^2+{\mathrm{b}}_{1}x+{\mathrm{c}}_1=0\text{... [Given]}\\ \therefore & \mathrm{Sum}\text{ of roots }=\frac{-{\mathrm{b}}_1}{{\mathrm{a}}_1}\\ & \tan \left(\frac{\mathrm{A}}{2}\right)+\tan \left(\frac{\mathrm{B}}{2}\right)=\frac{-{\mathrm{b}}_1}{{\mathrm{a}}_1}\end{array}$

Also, $\tan \left(\frac{\mathrm{A}}{2}\right)\cdot \tan \left(\frac{\mathrm{B}}{2}\right)=\frac{{\mathrm{c}}_1}{{\mathrm{a}}_1}$

Using $\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2}\tan \frac{B}{2}}$, we get

$\tan \left(\frac{\pi }{4}\right)=\frac{\frac{-b_1}{a_1}}{1-\frac{c_1}{a_1}}$

$1=\frac{-b_1}{a_1-c_1}$

$a_1-c_1=-b_1$

$a_1+b_1=c_1$