Exercise 3.1 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Trigonometry
Ex 3.1
Question 1.
Identify the quadrant in which an angle of each given measure lies
(i) $25^{\circ}$
(ii) $825^{\circ}$
(iii) $-55^{\circ}$
(iv) $328^{\circ}$
Solution:
(i) $25^{\circ}=\mathrm{I}$ quadrant
(ii) $825^{\circ}=105^{\circ}\left(90^{\circ}+15^{\circ}\right)=$ II quadrant
(iii) $-55^{\circ}=$ IV quadrant
(iv) $328^{\circ}=$ IV quadrant $\left(270^{\circ}+58^{\circ}\right)$

(v) $-230^{\circ}=360^{\circ}-230^{\circ}=130^{\circ}=\left(90^{\circ}+40^{\circ}\right)$ II quadrant
Question 2.
For each given angle, find a coterminal angle with measure of $\theta$ such that $\theta^{\circ}<\theta<360^{\circ}$
(i) $395^{\circ}$
(ii) $525^{\circ}$
(iii) $1150^{\circ}$
(iv) $-270^{\circ}$
(v) $-450^{\circ}$
Solution:
(i) $395^{\circ}=360^{\circ}+35^{\circ}$
$\therefore$ coterminal angle $=35^{\circ}$
(ii) $525^{\circ}-360^{\circ}=165^{\circ}$
coterminal angle $=165^{\circ}$
(iii) $1150^{\circ}=360 \times 3+70^{\circ}=70^{\circ}$
coterminal angle $=70^{\circ}$
(iv) $-270^{\circ}=$ coterminal angle $=+90^{\circ}\left\{270^{\circ}+90^{\circ}=360^{\circ}\right\}$
(v) $-450^{\circ}=-360^{\circ}-90^{\circ}=-90^{\circ}$
$\therefore$ coterminal angle $=360^{\circ}-90^{\circ}=270^{\circ}$

Question 3.
If $a \cos \theta-b \sin \theta=c$, show that $a \sin \theta+b \cos \theta= \pm \sqrt{a^2+b^2-c^2}$.
Solution:
$\mathrm{a} \cos \theta-\mathrm{b} \sin \theta=\mathrm{c}$
$\Rightarrow(\mathrm{a} \cos \theta-\mathrm{b} \sin \theta)^2=\mathrm{c}^2$
(i.e) $\mathrm{a}^2 \cos ^2 \theta+\mathrm{b}^2 \sin ^2 \theta-2 \mathrm{ab} \sin \theta \cos \theta=\mathrm{c}^2$
(i.e) $\mathrm{a}^2\left(1-\sin ^2 \theta\right)+\mathrm{b}^2\left(1-\cos ^2 \theta\right)-2 \mathrm{ab} \sin \theta \cos \theta=\mathrm{c}^2$
$\mathrm{a}^2-\mathrm{a}^2 \sin ^2 \theta+\mathrm{b}^2-\mathrm{b}^2 \cos ^2 \theta-2 \mathrm{ab} \sin \theta \cos \theta=\mathrm{c}^2$
$\mathrm{a}^2+\mathrm{b}^2-\mathrm{c}^2=\mathrm{a}^2 \sin ^2 \theta+\mathrm{b}^2 \cos ^2 \theta+2 \mathrm{ab} \sin \theta \cos \theta$
(i.e) $(a \sin \theta+b \cos \theta)^2=a^2+b^2-c^2$
$\Rightarrow a \sin \theta+b \cos \theta= \pm \sqrt{a^2+b^2-c^2}$
Question 4.
If $\sin \theta+\cos \theta=m$, show that $\cos ^6 \theta+\sin ^6 \theta=\frac{4-3\left(m^2-1\right)^2}{4}$, where $m^2 \leq 2$.
Solution:
$
\begin{aligned}
\text { Given, } m & =\sin \theta+\cos \theta \\
m^2-1 & =(\sin \theta+\cos \theta)^2-1=\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta-1 \\
& =1+2 \sin \theta \cos \theta-1=2 \sin \theta \cos \theta \\
\left(m^2-1\right)^2 & =(2 \sin \theta \cos \theta)^2=4 \sin ^2 \theta \cos ^2 \theta \\
\text { R.H.S. } & =\frac{4-3\left(m^2-1\right)^2}{4}=\frac{4-3\left(4 \sin ^2 \theta \cos ^2 \theta\right)}{4} \\
& =4\left[\frac{1-3 \sin ^2 \theta \cos ^2 \theta}{4}\right]=1-3 \sin ^2 \theta \cos ^2 \theta \\
\text { L.H.S } & =\cos ^6 \theta+\sin ^6 \theta=\left(\cos ^2 \theta\right)^3+\left(\sin ^2 \theta\right)^3\left[a^3+b^3=(a+b)\left(a^2-a b+b^2\right)\right] \\
& =\left(\cos ^2 \theta+\sin ^2 \theta\right)\left[\cos ^4 \theta+\sin ^4 \theta-\cos ^2 \theta \sin ^2 \theta\right] \\
& =1\left[\left(\cos ^2 \theta+\sin ^2 \theta\right)^2-2 \sin ^2 \theta \cos ^2 \theta-\sin ^2 \theta \cos ^2 \theta\right] \\
& =1-3 \sin ^2 \theta \cos ^2 \theta \\
& (1)=(2) \Rightarrow \text { LHS }=\text { RHS }
\end{aligned}
$

Question 5.
If $\frac{\cos ^4 \alpha}{\cos ^2 \beta}+\frac{\sin ^4 \alpha}{\sin ^2 \beta}=1$, prove that
(i) $\sin ^4 \alpha+\sin ^4 \beta=2 \sin ^2 \alpha \sin ^2 \beta$
(ii) $\frac{\cos ^4 \beta}{\cos ^2 \alpha}+\frac{\sin ^4 \beta}{\sin ^2 \alpha}=1$
Solution:
(i) $\operatorname{Given} \frac{\cos ^4 \alpha}{\cos ^2 \beta}+\frac{\sin ^4 \alpha}{\sin ^2 \beta}=1$
$
\Rightarrow \quad \frac{\cos ^4 \alpha}{\cos ^2 \beta}=1-\frac{\sin ^4 \alpha}{\sin ^2 \beta}=\frac{\sin ^2 \beta-\sin ^4 \alpha}{\sin ^2 \beta}
$
$\Rightarrow \quad$ Given $\frac{\cos ^4 \alpha}{\cos ^2 \beta}=\frac{\sin ^2 \beta-\sin ^4 \alpha}{\sin ^2 \beta} \Rightarrow \cos ^4 \alpha \sin ^2 \beta=\left(\sin ^2 \beta-\sin ^4 \alpha\right) \cos ^2 \beta$
$\Rightarrow\left(1-\sin ^2 \alpha\right) \sin ^2 \beta=\left(\sin ^2 \beta-\sin ^4 \alpha\right)\left(1-\sin ^2 \beta\right)$
$\left(1+\sin ^4 \alpha-2 \sin ^2 \alpha\right) \sin ^2 \beta=\sin ^2 \beta-\sin ^4 \alpha-\sin ^4 \beta+\sin ^4 \alpha \sin ^2 \beta$
$\Rightarrow \sin ^2 \beta+\sin ^4 \alpha \sin ^2 \beta-2 \sin ^2 \alpha \sin ^2 \beta=\sin ^2 \beta-\sin ^4 \alpha-\sin ^4 \beta+\sin ^4 \alpha \sin ^2 \beta$
$\Rightarrow \sin ^4 \alpha+\sin ^4 \beta=2 \sin ^2 \alpha \sin ^2 \beta$
(ii) $\sin ^4 \alpha+\sin ^4 \beta=2 \sin ^2 \alpha \sin ^2 \beta$
$\Rightarrow \sin ^4 \alpha+\sin ^4 \beta-2 \sin ^2 \alpha \sin ^2 \beta=0$
$\left(\sin ^2 \alpha-\sin ^2 \beta\right)^2=0$
$\Rightarrow \quad \sin ^2 \alpha=\sin ^2 \beta$
So, $1-\sin ^2 \alpha=1-\sin ^2 \beta$
(i.e.) $\cos ^2 \alpha=\cos ^2 \beta$.
So $\frac{\cos ^4 \beta}{\cos ^2 \alpha}=\frac{\left(\cos ^2 \alpha\right)^2}{\cos ^2 \alpha}=\cos ^2 \alpha$
and $\frac{\sin ^4 \beta}{\sin ^2 \alpha}=\frac{\left(\sin ^2 \alpha\right)^2}{\sin ^2 \alpha}=\sin ^2 \alpha$
$
\therefore \quad \mathrm{LHS}=\cos ^2 \alpha+\sin ^2 \alpha=1=\mathrm{RHS}
$

Question 6.
If $y=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}$, then prove that $\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=y$
Solution:
To Prove $\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}$
$
\begin{aligned}
& \text { LHS }=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha} \times \frac{(1+\sin \alpha)-\cos \alpha}{(1+\sin \alpha)-\cos \alpha} \\
& =\frac{2 \sin \alpha[(1+\sin \alpha-\cos \alpha)}{(1+\sin \alpha)^2-\cos ^2 \alpha} \\
& \text { Dr }=1+\sin ^2 \alpha+2 \sin \alpha-\cos ^2 \alpha \\
& =\sin ^2 \alpha+2 \sin \alpha+\left(1-\cos ^2 \alpha\right) \\
& =\sin ^2 \alpha+2 \sin \alpha+\sin ^2 \alpha \\
& =2 \sin ^2 \alpha+2 \sin \alpha \\
& =2 \sin \alpha(1+\sin \alpha) \\
& \therefore(1) \Rightarrow \frac{2 \sin \alpha[1+\sin \alpha-\cos \alpha]}{2 \sin \alpha(1+\sin \alpha)}=\frac{1+\sin \alpha-\cos \alpha}{1+\sin \alpha}=\text { RHS } \\
&
\end{aligned}
$
Question 7.
If $x=\sum_{n=0}^{\infty} \cos ^{2 n} \theta, y=\sum_{n=0}^{\infty} \sin ^{2 n} \theta$ and $z=\sum_{n=0}^{\infty} \cos ^{2 n} \theta \sin ^{2 n} \theta, 0<\theta<\frac{\pi}{2}$, then show that $x y z=x+y+z$.
[Hint: Use the formula $1+x+x^2+x^3+\ldots=\frac{1}{1-x}$, where $\left.|x|<1\right]$.
Solution:

$
\begin{aligned}
& x=1+\cos ^2 \theta+\cos ^4 \theta+\cos ^6 \theta+\ldots .=\frac{1}{1-\cos ^2 \theta}=\frac{1}{\sin ^2 \theta} \\
& y=.1+\sin ^2 \theta+\sin ^4 \theta+\sin ^6 \theta+\ldots \ldots . .=\frac{1}{1-\sin ^2 \theta}=\frac{1}{\cos ^2 \theta} \\
& z=1+\cos ^2 \theta \sin ^2 \theta+\cos ^4 \theta \sin ^4 \theta+\ldots \ldots=\frac{1}{1-\cos ^2 \theta \sin ^2 \theta} \\
& \text { LHS }=x y z=\frac{1}{\sin ^2 \theta} \cdot \frac{1}{\cos ^2 \theta} \cdot \frac{1}{1-\sin ^2 \theta \cos ^2 \theta} \\
& =\frac{1}{\sin ^2 \theta \cos ^2 \theta\left(1-\sin ^2 \theta \cos ^2 \theta\right)} \\
& \text { R'HS }=x+y+z=\frac{1}{\sin ^2 \theta}+\frac{1}{\cos ^2 \theta}+\frac{1}{1-\sin ^2 \theta \cos ^2 \theta} \\
& =\left(\frac{1}{\sin ^2 \theta}+\frac{1}{\cos ^2 \theta}\right)+\left(\frac{1}{1-\sin ^2 \theta \cos ^2 \theta}\right) \\
& =\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin ^2 \theta \cos ^2 \theta}+\frac{1}{1-\sin ^2 \theta \cos ^2 \theta} \\
& =\frac{1}{\sin ^2 \theta \cos ^2 \theta}+\frac{1}{1-\sin ^2 \theta \cos ^2 \theta} \\
& =\frac{1-\sin ^2 \theta \cos ^2 \theta+\sin ^2 \theta \cos ^2 \theta}{\left(\sin ^2 \theta \cos ^2 \theta\right)\left(1-\sin ^2 \theta \cos ^2 \theta\right)} \\
& =\frac{1}{\left(\sin ^2 \theta \cos ^2 \theta\right)\left(1-\sin ^2 \theta \cos ^2 \theta\right)} \\
& \text { (1) }=(2) \Rightarrow \text { LHS }=\text { RHS (i.e) } x y z=x+y+z \\
&
\end{aligned}
$
Question 8.
If $\tan ^2 \theta=1-k^2$, show that $\sec \theta+\tan ^3 \theta \operatorname{cosec} \theta=\left(2-k^2\right)^{3 / 2}$. Also, find the values of $\boldsymbol{k}$ for which this result holds.

Solution:
$
\begin{aligned}
& \tan ^2 \theta=1-k^2 \\
&-k^2=\tan ^2 \theta-1 \\
& 2-k^2=\tan ^2 \theta-1+2=1+\tan ^2 \theta=\sec ^2 \theta \\
& 2-k^2=\sec ^2 \theta \\
& \text { RHS }=\left(2-k^2\right)^{3 / 2}=\left(\sec ^2 \theta\right)^{3 / 2}=\sec ^3 \theta \\
& \text { Now LHS }=\sec \theta+\tan ^3 \theta \operatorname{cosec} \theta \\
&=\frac{1}{\cos \theta}+\frac{\sin ^3 \theta}{\cos ^3 \theta} \cdot \frac{1}{\sin \theta}=\frac{1}{\cos \theta}+\frac{\sin ^2 \theta}{\cos ^3 \theta} \\
&=\frac{\cos { }^2 \theta+\sin ^2 \theta}{\cos ^3 \theta}=\frac{1}{\cos ^3 \theta}=\sec ^3 \theta
\end{aligned}
$
$
\text { (1) = (2) } \Rightarrow \mathrm{LHS}=\mathrm{RHS}
$
$
\begin{aligned}
& 2-k^2 \geq 0 \Rightarrow-k^2 \geq-2 \Rightarrow \therefore k^2 \leq 2 \Rightarrow k=\sqrt{2} \\
& \therefore k \in(-1,1)
\end{aligned}
$
Question 9.
If $\sec \theta+\tan \theta=p$, obtain the values of $\sec \theta, \tan \theta$ and $\sin \theta$ in terms of $p$. Solution:
Given, $\sec \theta+\tan \theta=p$
we know $\sec ^2 \theta-\tan ^2 \theta=1$

$
\begin{aligned}
& \text { (i.e) }(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1 \\
& \Rightarrow \sec \theta-\tan \theta=\frac{1}{\sec \theta+\tan \theta}=\frac{1}{p}
\end{aligned}
$
Now $\sec \theta+\tan \theta=p$
$
\begin{aligned}
& \sec \theta-\tan \theta=\frac{1}{p} \\
&(1)+(2) \Rightarrow 2 \sec \theta=p+\frac{1}{p}=\frac{p^2+1}{p} \\
& \Rightarrow \sec \theta=\frac{p^2+1}{2 p} \\
&(1)-(2) \Rightarrow 2 \tan \theta=p-\frac{1}{p}=\frac{p^2-1}{p} \\
& \Rightarrow \tan \theta=\frac{p^2-1}{2 p} \\
& \text { Now } \frac{\tan \theta}{\sec \theta}= \frac{\sin \theta}{\cos \theta} \times \cos \theta=\sin \theta \\
& \therefore \sin \theta=\frac{\tan \theta}{\sec \theta}=\frac{p^2-1}{2 p} / \frac{p^2+1}{2 p}=\frac{p^2-1}{p^2+1} \\
& \text { So, } \sec \theta=\frac{p^2+1}{2 p} ; \tan \theta=\frac{p^2-1}{2 p} ; \sin \theta=\frac{p^2-1}{p^2+1}
\end{aligned}
$
Question 10.
If $\cot \theta(1+\sin \theta)=4 \mathrm{~m}$ and $\cot \theta(1-\sin \theta)=4 \mathrm{n}$, then prove that $\left(\mathrm{m}^2-\mathrm{n}^2\right)^2=\mathrm{mn}$.
Solution:
$
\begin{aligned}
& \cot \theta(1+\sin \theta)=4 \mathrm{~m} \\
& \Rightarrow \quad \frac{\cos \theta}{\sin \theta}(1+\sin \theta)=4 m \\
& \Rightarrow \quad \cot \theta+\cos \theta=4 m \\
& \Rightarrow \quad m=\frac{\cot \theta+\cos \theta}{4}=4 n \\
& \Rightarrow \quad \cot \theta(1-\sin \theta)=4 n
\end{aligned}
$

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \Rightarrow \quad \frac{\cos \theta}{\sin \theta}(1-\sin \theta)=4 n \\
& \Rightarrow \quad \cot \theta-\cos \theta \quad=4 n \\
&
\end{aligned}\\
&\begin{aligned}
& \Rightarrow \quad n=\frac{\cot \theta-\cos \theta}{4} \\
& \text { To prove }\left(m^2-n^2\right)^2=m n \\
& \text { Now RHS }=m n=\left(\frac{\cot \theta+\cos \theta}{4}\right)\left(\frac{\cot \theta-\cos \theta}{4}\right) \\
& =\frac{\cot ^2 \theta-\cos ^2 \theta}{16} \\
& \text { Now }\left(m^2-n^2\right)=[(m+n)(m-n)] \\
& m+n \quad=\frac{\cot \theta+\cos \theta+\cot \theta-\cos \theta}{4}=\frac{2 \cot \theta}{4}=\frac{\cot \theta}{2} \\
& m-n \quad=\frac{\cot \theta+\cos \theta-\cot \theta+\cos \theta}{4}=\frac{2 \cos \theta}{4} \frac{\cos \theta}{2} \\
& \therefore(m+n)(m-n)=\frac{\cot \theta}{2} \frac{\cos \theta}{2}=\frac{\cot \theta \cos \theta}{4} \\
& \text { So LHS }=\left(m^2-n^2\right)^2=[(m-n)(m+n)]^2 \\
& =\left(\frac{\cot \theta \cos \theta}{4}\right)^2=\frac{\cot ^2 \theta \cos ^2 \theta}{16} \\
& \mathrm{RHS}=m n=\frac{\cot ^2 \theta-\cos ^2 \theta}{16} \\
&
\end{aligned}
\end{aligned}
\end{equation}$

$
\begin{aligned}
& =\frac{1}{16}\left[\frac{\cos ^2 \theta}{\sin ^2 \theta}-\cos ^2 \theta\right]=\frac{\cos ^2 \theta}{16 \sin ^2 \theta}\left[1-\sin ^2 \theta\right] \\
& =\frac{1}{16} \cot ^2 \theta \cos ^2 \theta=\mathrm{LHS} \\
& \therefore \mathrm{LHS}=\mathrm{RHS} \Rightarrow\left(m^2-n^2\right)^2=m n
\end{aligned}
$
Question 11.
If $\operatorname{cosec} \theta-\sin \theta=a^3$ and $\sec \theta-\cos \theta=b^3$, then prove that $a^2 b^2\left(a^2+b^2\right)=1$.

Solution:

$\begin{aligned}
& \operatorname{cosec} \theta-\sin \theta \quad=a^3 \\
& \Rightarrow \frac{1}{\sin \theta}-\sin \theta=a^3 \\
& \text { (i.e) } \frac{1-\sin ^2 \theta}{\sin \theta}=a^3 \\
& \Rightarrow a^3=\frac{\cos ^2 \theta}{\sin \theta} \Rightarrow a=\frac{(\cos \theta)^{2 / 3}}{(\sin \theta)^{1 / 3}} \\
& \text { Similary, } \sec \theta-\cos \theta=b^3 \\
& \Rightarrow \frac{1}{\cos \theta}-\cos \theta=b^3 \\
& \frac{1-\cos ^2 \theta}{\cos \theta}=b^3 \\
& \Rightarrow \frac{\sin ^2 \theta}{\cos \theta}=b^3 \Rightarrow b=\frac{(\sin \theta)^{2 / 3}}{(\cos \theta)^{1 / 3}} \\
& \text { Now } a^3=\frac{\cos ^2 \theta}{\sin \theta} ; b^3=\frac{\sin ^2 \theta}{\cos \theta} \\
& \therefore a^3 b^3=\frac{\cos ^2 \theta}{\sin \theta} \cdot \frac{\sin ^2 \theta}{\cos \theta}=\sin \theta \cos \theta \Rightarrow a b=(\sin \theta \cos \theta)^{1 / 3} \\
& \Rightarrow \quad a^2 b^2=\left\{(\sin \theta \cos \theta)^{1 / 3}\right\}^2=(\sin \theta \cos \theta)^{2 / 3} \\
& =\sin ^{2 / 3} \theta \cos ^{2 / 3} \theta \\
& \text { To prove } a^2 b^2\left(a^2+b^2\right)=1 \\
& a^2+b^2=\left\{\left[\frac{(\cos \theta)^{2 / 3}}{(\sin \theta)^{1 / 3}}\right]^2+\left[\frac{(\sin \theta)^{2 / 3}}{(\cos \theta)^{1 / 3}}\right]^2\right\} \\
& =\frac{(\cos \theta)^{4 / 3}}{(\sin \theta)^{2 / 3}}+\frac{(\sin \theta)^{4 / 3}}{(\cos \theta)^{2 / 3}} \\
& =\frac{\cos ^2 \theta+\sin ^2 \theta}{(\sin \theta \cos \theta)^{2 / 3}}=\frac{1}{(\sin \theta \cos \theta)^{2 / 3}} \\
& \text { LHS } \quad=a^2 b^2\left(a^2+b^2\right)=(\sin \theta \cos \theta)^{2 / 3} \times \frac{1}{(\sin \theta \cos \theta)^{2 / 3}} \\
&
\end{aligned}$

$
=1=\mathrm{RHS}
$
Question 12.
Eliminate $\theta$ from the equations a $\sec \theta-\mathrm{c} \tan \theta=\mathrm{b}$ and $\mathrm{b} \sec \theta+\mathrm{d} \tan \theta=\mathrm{c}$.
Solution:
Taking $\sec \theta=\mathrm{X}$ and $\tan \theta=\mathrm{Y}$ we get the equations as
$
\begin{aligned}
& a \mathrm{X}-c \mathrm{Y}=b \\
& b \mathrm{X}+d \mathrm{Y}=c
\end{aligned}
$
$(1) \times d \Rightarrow \quad a d \mathrm{X}-c d \mathrm{Y}=b d$
$(2) \times c \Rightarrow \quad b c \mathrm{X}+c d \mathrm{Y}=c^2$
$(3)+(4) \Rightarrow \quad \mathrm{X}(a d+b c)=b d+c^2$
$\therefore \quad \mathrm{X}=\frac{b d+c^2}{a d+b c}$
(i.e) $\sec \theta=\frac{b d \quad c}{a d \quad b c}$
$(2) \times a \Rightarrow a b \mathrm{X}+a d \mathrm{Y}=a c$
Now $(1) \times b \Rightarrow a b \mathrm{X}-b c \mathrm{Y}=b^2$
$
\begin{aligned}
& (5)-(6) \Rightarrow \quad Y(a d+b c)=a c-b^2 \\
& \Rightarrow \quad Y=\frac{a c-b^2}{a d+b c} \\
& \text { (i.e) } \quad \tan \theta=\frac{a c-b^2}{a d+b c} \\
& \therefore \sec \theta=\frac{b d+c^2}{a d+b c} \text { and } \tan \theta=\frac{a c-b^2}{a d+b c}
\end{aligned}
$

$
\begin{aligned}
& \text { We know } \sec ^2 \theta-\tan ^2 \theta=1 \\
& \Rightarrow \quad \frac{\left(b d+c^2\right)^2}{(a d+b c)^2}-\frac{\left(a c-b^2\right)^2}{(a d+b c)^2}=1 \\
& \Rightarrow \quad \frac{\left(b d+c^2\right)^2-\left(a c-b^2\right)^2}{(a d+b c)^2}=1 \\
& \Rightarrow \quad\left(b d+c^2\right)^2-\left(a c-b^2\right)^2=(a d+b c)^2 \\
& \Rightarrow \quad\left(b d+c^2\right)^2=(a d+b c)^2+\left(a c-b^2\right)^2
\end{aligned}
$