Maths Solve the Following Question.(2 Marks)

The general solution of cos θ = cos ∝ is θ = 2nπ ± ∝, n ∈ Z ∴ the general solution of (1) is given by 2θ = 2nπ ± π, n ∈ Z

$\therefore \theta=n \pi \pm \frac{\pi}{2}, n \in Z$

Dividing both sides by $\sqrt{(1)^2+(-1)^2}=\sqrt{2}$, we get

$\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta=-\frac{1}{\sqrt{2}}$

$\therefore \cos \frac{\pi}{4} \cos \theta-\sin \frac{\pi}{4} \sin \theta=-\cos \frac{\pi}{4}$

$\therefore \cos \left(\theta-\frac{\pi}{4}\right)=\cos \left(\pi-\frac{\pi}{4}\right)$

$\ldots[\because \cos (\pi-\theta)=-\cos \theta]$

$\therefore \cos \left(\theta-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}$

$\ldots(1)$

The general solution of $\cos \theta=\cos \alpha$ is

$\theta=2 n \pi \pm \alpha, n \in Z$

$\therefore$ the general solution of $(1)$ is given by

$\theta-\frac{\pi}{4}=2 n \pi \pm \frac{3 \pi}{4}, n \in Z$

$\therefore \theta=2 n \pi+\pi=(2 n+1) \pi, n \in Z$

Taking negative sign, we get$\theta-\frac{\pi}{4}=2 n \pi-\frac{3 \pi}{4}, n \in Z$

$\therefore \theta=2 n \pi-\frac{\pi}{2}, n \in Z$

$\therefore$ the required general solution is

$\theta=(2 n+1) \pi, n \in Z$ or $\theta=2 n \pi-\frac{\pi}{2}, n \in Z$.

$\theta=n \pi \pm \alpha, n \in Z$.

Now, $\tan ^2 \theta=3=(\sqrt{x})^2$

$\therefore \tan ^2 \theta=\left(\tan \frac{\pi}{3}\right)^2 \ldots\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]$

$\therefore \tan ^2 \theta=\tan ^2 \frac{\pi}{3}$

$\therefore$ the required general solution is

$\theta=n \pi \pm \frac{\pi}{3}, n \in Z$.

Now, $\tan \theta=-\sqrt{x}$

$\therefore \tan \theta=\tan \frac{\pi}{3} \ldots\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right]$

$\begin{aligned} & \therefore \tan \theta=\tan \left(\pi-\frac{\pi}{3}\right) \ldots[\because \tan (\pi-\theta)=-\tan \theta] \\ & \therefore \tan \theta=\tan \frac{2 \pi}{3}\end{aligned}$

∴ the required general solution is

$\theta=n \pi+\frac{2 \pi}{3}, n \in Z$.

$\therefore \tan ^{-1}\left[\frac{\left(\frac{x+y}{1-x y}\right)+z}{1-\left(\frac{x+y}{1-x y}\right) z}\right]=\frac{\pi}{2}$

$\therefore \tan ^{-1}\left[\frac{x+y+z-x y z}{1-x y-x z-y z}\right]=\frac{\pi}{2}$

$\therefore \frac{x+y+z-x y z}{1-x y-y z-z x}=\tan \frac{\pi}{2}$, which does not exist

∴ 1 – xy – yz – zx = 0 ∴ xy + yz + zx = 1.

$\begin{array}{r}=\tan ^{-1} x-\tan ^{-1} y+\tan ^{-1} y-\tan ^{-1} z+\tan ^{-1} z-\tan ^{-1} x \\ \ldots[\because x>0, y>0, z>0]\end{array}$

$=0=$ RHS.

Then, $x=$ tany

Now, $\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)=\tan ^{-1}\left(\frac{2 \tan y}{1-\tan ^2 y}\right)$

$\begin{aligned} & =\tan ^{-1}(\tan 2 y) \\ & =2 y=2 \tan ^{-1} x\end{aligned}$

$\ldots(1)$

$\begin{aligned} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) & =\sin ^{-1}\left(\frac{2 \tan y}{1+\tan ^2 y}\right)=\sin ^{-1}(\sin 2 y) \\ & =2 y=2 \tan ^{-1} x\end{aligned}$

$\ldots(2)$

$\begin{aligned} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right) & =\cos ^{-1}\left(\frac{1-\tan ^2 y}{1+\tan ^2 y}\right)=\cos ^{-1}(\cos 2 y) \\ & =2 y=2 \tan ^{-1} x\end{aligned}$

$\ldots(3)$

From (1), (2) and (3), we get

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

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

$\cdots\left[\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}\right]$

$=\frac{\pi}{6}+2\left(\frac{\pi}{3}\right)$

$\ldots\left[\because \sin ^{-1}(\sin x)=x, \cos ^{-1}(\cos x)=x\right]$

$\begin{aligned} & =\frac{\pi}{6}+\frac{2 \pi}{3} \\ & =\frac{5 \pi}{6}=\text { RHS. }\end{aligned}$

$\begin{aligned} & =\tan ^{-1}\left(\frac{4-2}{8+1}\right) \\ & =\tan ^{-1}\left(\frac{2}{9}\right)=\text { RHS. }\end{aligned}$

$\therefore \tan ^{-1}\left(\frac{2 x+3 x}{1-2 x \times 3 x}\right)=\tan \frac{\pi}{4}$, where $2 \mathrm{x}>0,3 \mathrm{x}>0$

$\begin{aligned} & \therefore \frac{5 x}{1-6 x^2}=\tan \frac{\pi}{4}=1 \\ & \therefore 5 x=1-6 x^2\end{aligned}$

$\begin{aligned} & \therefore 6 x^2+5 x-1=0 \\ & \therefore 6 x^2+6 x-x-1=0 \\ & \therefore 6 x(x+1)-1(x+1)=0 \\ & \therefore(x+1)(6 x-1)=0 \\ & \therefore x=-1 \text { or } x=\frac{1}{6}\end{aligned}$

But $x>0 \therefore x \neq-1$

Hence, $x=\frac{1}{6}$

$\therefore \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1}(1)$

$\therefore \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\sin ^{-1}\left(\sin \frac{\pi}{2}\right)$

$\therefore \sin ^{-1} \frac{1}{5}+\cos ^{-1} x=\frac{\pi}{2}$

$\therefore x=\frac{1}{5} \quad \cdots\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

$\therefore \tan y=\sqrt{x} \therefore x=\tan ^2 y$

Now, RHS $=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)$

$=\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 y}{1+\tan ^2 y}\right)$

$=\frac{1}{2} \cos ^{-1}(\cos 2 y)=\frac{1}{2}(2 y)=y$

$=\tan ^{-1} \sqrt{x}=\mathrm{LHS}$.

$\begin{aligned} & \therefore \frac{1}{2}(a+a \cos \mathrm{C}+c+c \cos \mathrm{A})=\frac{3 b}{2} \\ & \therefore a+c+(a \cos \mathrm{C}+c \cos \mathrm{A})=3 b\end{aligned}$

$\begin{aligned} & \therefore a+c+b=3 b \ldots[\because a \cos C+c \cos A=b] \\ & \therefore a+c=2 b\end{aligned}$

Hence, $a, b, c$ are in A.P.

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

$\begin{aligned} & \therefore \sin A=k a, \sin B=k b, \sin C=k c \\ & \therefore \sin ^2 A+\sin ^2 B=\sin ^2 C\end{aligned}$

$\begin{aligned} & \therefore k^2 a^2+k^2 b^2=k^2 c^2 \\ & \therefore a^2+b^2=c^2\end{aligned}$

∴ ∆ABC is a rightangled triangle, rightangled at C.

$

\begin{aligned}

\therefore \quad & a \cos C=b-c \cos A \text { and } a \cos B=c-b \cos A \\

& \text { L.H.S. }=a(\cos C-\cos B) \\

& =a \cos C-a \cos B \\

& =(b-c \cos A)-(c-b \cos A) \\

& =b-c \cos A-c+b \cos A \\

& =(b-c)+(b-c) \cos A \\

& =(b-c)(1+\cos A) \\

& =(b-c) \times 2 \cos ^2 \frac{A}{2} \\

& =2(b-c) \cos ^2 \frac{A}{2} \\

& =\text { R.H.S. }

\end{aligned}

$

$

\begin{aligned}

r^2 & =x^2+y^2 \\

\therefore \quad r^2 & =\left(\frac{1}{\sqrt{2}}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}=1

\end{aligned}

$$
\begin{array}{ll}
\therefore \quad & r=1 \\
& x= r \cos \theta, y= r \sin \theta \\
\therefore \quad & \frac{1}{\sqrt{2}}=1 \times \cos \theta \text { and }-\frac{1}{\sqrt{2}}=1 \times \sin \theta \\
\therefore \quad & \cos \theta=\frac{1}{\sqrt{2}} \text { and } \sin \theta=-\frac{1}{\sqrt{2}} \\
\therefore & \theta=\frac{7 \pi}{4}
\end{array}
$
$\therefore \quad$ The required polar co-ordinates are $\left(1, \frac{7 \pi}{4}\right)$.

$

\therefore \quad \cos 2 \theta=\cos \frac{3 \pi}{4}\left(\text { As cos } \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos (\pi-A)=\cos A\right)

$

The general solution of $\cos \theta=\cos \alpha$ is $\theta=2 n \pi \pm \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos 2 \theta=\cos \frac{3 \pi}{4}$ is $2 \theta=2 n \pi \pm \frac{3 \pi}{4}$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos 2 \theta=-\frac{1}{\sqrt{2}}$ is $\theta=n \pi \pm \frac{3 \pi}{8}$, where $n \in Z$.

$

\therefore \quad \cos \theta=\cos \frac{3 \pi}{4}\left(\text { As } \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \text { and } \cos (\pi-A)=-\cos A\right)

$

The general solution of $\cos \theta=\cos \alpha$ is $\theta=2 n \pi \pm \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos \theta=\cos \frac{3 \pi}{4}$ is $\theta=2 n \pi \pm \frac{3 \pi}{4}$, where $n \in Z$.

$\therefore \quad$ The general solution of $\sec \theta=\sqrt{2}$ is $\theta=2 n \pi \pm \frac{3 \pi}{4}$, where $n \in Z$.

$

\therefore \quad \cos \theta=\cos \frac{2 \pi}{3}\left(\text { As cos } \frac{\pi}{3}=\frac{1}{2} \text { and } \cos (\pi-A)=-\cos A\right)

$

The general solution of $\cos \theta=\cos \alpha$ is $\theta=2 n \pi \pm \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos \theta=\cos \frac{2 \pi}{3}$ is $\theta=2 n \pi \pm \frac{2 \pi}{3}$ where $n \in Z$.

$\therefore \quad$ The general solution of $\cos \theta=-\frac{1}{2}$ is $\theta=2 n \pi \pm \frac{2 \pi}{3}$, where $n \in Z$.

$\therefore \quad \sin \theta=\sin \frac{4 \pi}{3}\left(\right.$ As $\sin \frac{4 \pi}{3}=\frac{\sqrt{3}}{2}$ and $\left.\sin (\pi+ A )=-\sin A \right)$

The general solution of $\sin \theta=\sin \alpha$ is $\alpha= n \pi+(-1)^{ n } \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\sin \theta=\sin \frac{4 \pi}{3}$ is $\theta=n \pi+(-1)^{ n } \frac{4 \pi}{3}$, where $n \in Z$.

$\therefore \quad$ The general solution of $\sin \theta=-\frac{\sqrt{3}}{2}$ is $\theta=n \pi+(-1)^{ n } \frac{4 \pi}{3}$, where $n \in Z$.

$

\therefore \quad \cos \theta=\cos \frac{\pi}{4}

$

The general solution of $\cos \theta=\cos \alpha$ is $\theta=2 n \pi \pm \alpha$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos \theta=\cos \frac{\pi}{4}$ is $\theta=2 n \pi \pm \frac{\pi}{4}$, where $n \in Z$.

$\therefore \quad$ The general solution of $\cos \theta=\frac{1}{\sqrt{2}}$ is $\theta=2 n \pi \pm \frac{\pi}{4}$, where $n \in Z$.

We know that $\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}$

Using identities, $\tan (\pi-\theta)=-\tan \theta$ and $\tan (2 \pi-\theta)=-\tan \theta$, we get

$\tan \frac{5 \pi}{6}=-\frac{1}{\sqrt{3}}$ and $\tan \frac{11 \pi}{6}=-\frac{1}{\sqrt{3}}$

An $0 \leq \frac{5 \pi}{6}<2 \pi$ and $0 \leq \frac{11 \pi}{6}<2 \pi$

$\therefore \frac{5 \pi}{6}$ and $\frac{11 \pi}{6}$ are required principal solutions.

$

\begin{aligned}

& \tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right) \text { if } x y>1 \\

& \text { Here } x y=1 \times 2=2>1 \\

\therefore \quad & \tan ^{-1} 1+\tan ^{-1} 2=\pi+\tan ^{-1}\left(\frac{1+2}{1-(1)(2)}\right) \\

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

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

= & \pi-\tan ^{-1} 3\left(\text { As }, \tan ^{-1}(-x)=-\tan ^{-1} x\right) \\

\therefore & \tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi

\end{aligned}

$

Let the cartesian coordinates be $(x, y)$

Then, $x=r \cos \theta=\frac{1}{2} \cos \frac{7 \pi}{3}=\frac{1}{2} \cos \left(2 \pi+\frac{\pi}{3}\right)$

$=\frac{1}{2} \cos \frac{\pi}{3}=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$

$y=r \sin \theta=\frac{1}{2} \sin \frac{7 \pi}{3}=\frac{1}{2} \sin \left(2 \pi+\frac{\pi}{3}\right)$

$=\frac{1}{2} \sin \frac{\pi}{3}=\frac{1}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$

$\therefore$ the cartesian coordinates of the given point are $\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$

Let the cartesian coordinates be $(x, y)$

Then, $x=r \cos \theta=\frac{3}{4} \cos \frac{3 \pi}{4}=\frac{3}{4} \cos \left(\pi-\frac{\pi}{4}\right)$

$=-\frac{3}{4} \cos \frac{\pi}{4}=-\frac{3}{4} \times \frac{1}{\sqrt{2}}=-\frac{3}{4 \sqrt{2}}$

$y=r \sin \theta=\frac{3}{4} \sin \frac{3 \pi}{4}=\frac{3}{4} \sin \left(\pi-\frac{\pi}{4}\right)$

$=\frac{3}{4} \sin \frac{\pi}{4}=\frac{3}{4} \times \frac{1}{\sqrt{2}}=\frac{3}{4 \sqrt{2}}$

$\therefore$ the cartesian coordinates of the given point are

$\left(-\frac{3}{4 \sqrt{2}}, \frac{3}{4 \sqrt{2}}\right)$

Let the cartesian coordinates be $(x, y)$

Then, $x=r \cos \theta=\sqrt{2} \cos \frac{\pi}{4}=\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=1$

$y=r \sin \theta=\sqrt{2} \sin \frac{\pi}{4}=\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)=1$

$\therefore$ the cartesian coordinates of the given point are $(1,1)$.

$=2 b c\left(\frac{b^2+c^2-a^2}{2 b c}\right)+2 a c\left(\frac{c^2+a^2-b^2}{2 c a}\right)+2 a b\left(\frac{a^2+b^2-c^2}{2 a b}\right) \ldots($ By cosine rule $]$

$=b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2=a^2+b^2+c^2=$ RHS.

∴ the point lies in the fourth quadrant.

Let the polar coordinates be (r, θ).

Then, $r^2=x^2+y^2=(1)^2+(-\sqrt{3})^2=1+3=4$

$\therefore r=2 \ldots[\because r>0]$

$\cos \theta=\frac{x}{r}=\frac{1}{2}$

and $\sin \theta=\frac{y}{r}=-\frac{\sqrt{3}}{2}$

$\therefore \tan \theta=-\sqrt{3}$

Since, the point lies in the fourth quadrant and

$0 \leqslant \theta<2 \pi$.

$\tan \theta=-\sqrt{3}=-\tan \frac{\pi}{3}$

$=\tan \left(2 \pi-\frac{\pi}{3}\right) \quad \ldots[\because \tan (2 \pi-\theta)=-\tan \theta]$

$=\tan \frac{5 \pi}{3}$

$\therefore \theta=\frac{5 \pi}{3}$

$\therefore$ the polar coordinates of the given point are $\left(2, \frac{5 \pi}{3}\right)$.

the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)

Then, $r^2=x^2+y^2=(0)^2+\left(\frac{1}{2}\right)^2=0+\frac{1}{4}=\frac{1}{4}$

$\therefore r=\frac{1}{2} \ldots[\because r>0]$

$\cos \theta=\frac{x}{r}=\frac{0}{(1 / 2)}=0$

and $\sin \theta=\frac{y}{r}=\frac{(1 / 2)}{(1 / 2)}=1$

Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π

$\cos \theta=0=\cos \frac{\pi}{2}$ and $\sin \theta=1=\sin \frac{\pi}{2}$

$\therefore \theta=\frac{\pi}{2}$

$\therefore$ the polar coordinates of the given point are $\left(\frac{1}{2}, \frac{\pi}{2}\right)$.

∴ the point lies in the first quadrant.

Let the polar coordinates be (r, θ)

Then, $r^2=x^2+y^2=(\sqrt{2})^2+(\sqrt{2})^2=2+2=4$

$\therefore r=2 \ldots[\because r>0]$

$\cos \theta=\frac{x}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

and $\sin \theta=\frac{y}{r}=\frac{\sqrt{ } 2}{2}=\frac{1}{\sqrt{2}}$

$\therefore \tan \theta=1$

Since the point lies in the first quadrant and

$0 \leq \theta \leq 2 \pi, \tan \theta=1=\tan \frac{\pi}{4}$

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

$\therefore$ the polar coordinates of the given point are $\left(2, \frac{\pi}{4}\right)$.

We know that,

$\begin{aligned} & \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \text { and } \tan (\pi+\theta)=\tan \theta \\ & \therefore \tan \frac{\pi}{6}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \frac{7 \pi}{6}\end{aligned}$

$\therefore \tan \frac{\pi}{6}=\tan \frac{7 \pi}{6}=\frac{1}{\sqrt{3}}$, where

$\begin{aligned} & 0<\frac{\pi}{6}<2 \pi \text { and } 0<\frac{7 \pi}{6}<2 \pi \\ & \therefore \cot \theta=\sqrt{3}, \text { i.e. } \tan \theta=\frac{1}{\sqrt{3}} \text { gives }\end{aligned}$

$\begin{aligned} & \tan \theta=\tan \frac{\pi}{6}=\tan \frac{7 \pi}{6} \\ & \therefore \theta=\frac{\pi}{6} \text { and } \theta=\frac{7 \pi}{6}\end{aligned}$

Hence, the required principal solution are

$\theta=\frac{\pi}{6}$ and $\theta=\frac{7 \pi}{6}$.

$\therefore \cos \frac{\pi}{3}=\cos \left(2 \pi-\frac{\pi}{3}\right)=\cos \frac{5 \pi}{3}$

$\therefore \cos \frac{\pi}{3}=\cos \frac{5 \pi}{3}=\frac{1}{2}$, where

$\begin{aligned} & 0<\frac{\pi}{3}<2 \pi \text { and } 0<\frac{5 \pi}{3}<2 \pi \\ & \therefore \cos \theta=\frac{1}{2} \text { gives } \cos \theta=\cos \frac{\pi}{3}=\cos \frac{5 \pi}{3}\end{aligned}$

$\therefore \theta=\frac{\pi}{3}$ and $\theta=\frac{5 \pi}{3}$

Hence, the required principal solutions are

$\theta=\frac{\pi}{3}$ and $\theta=\frac{5 \pi}{3}$

$\theta=n \pi+\alpha, n \in Z$

Now, $\cot \theta=0 \therefore \tan \theta$ does not exist

$\therefore \tan \theta=\tan \frac{\pi}{2}\left[\because \tan \frac{\pi}{2}\right.$ does not exist $]$

∴ the required general solution is

$\theta=n \pi+\frac{\pi}{2}, n \in Z$.

$\theta=n \pi+\alpha, n \in Z$

Now, $\tan \theta=\frac{1}{\sqrt{3}}=\tan \frac{\pi}{6} \ldots\left[\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$

∴ the required general solution is

$\theta=n \pi+\frac{\pi}{6}, n \in Z$.