Exercise 3.7 - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 3.7
Question 1.
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ}$, prove that
(i) $\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}=4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}$
(ii) $\cos \mathrm{A}+\cos \mathrm{B}-\cos \mathrm{C}=-1+4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$
(iii) $\sin ^2 A+\sin ^2 B+\sin ^2 C=2+2 \cos A \cos B \cos C$
(iv) $\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}=2 \sin \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}$
(v) $\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$
(vii) $\sin (\mathrm{B}+\mathrm{C}-\mathrm{A})+\sin (\mathrm{C}+\mathrm{A}-\mathrm{B})+\sin (\mathrm{A}+\mathrm{B}-\mathrm{C})=4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}$.
Solution:
(i)
LHS $=(\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B})+\sin 2 \mathrm{C}$
$=2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{C} \cos \mathrm{C}$
$\left[\sin (A+B)=\sin \left(180^{\circ}-C\right)=\sin C\right]$
$=2 \sin C \cos (\mathrm{A}-\mathrm{B})+2 \sin \mathrm{C} \cos \mathrm{C}$
$=2 \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\cos \mathrm{C}]$
$\left\{\cos \mathrm{C}=\cos \left[180^{\circ}-(\mathrm{A}+\mathrm{B})\right]=-\cos (\mathrm{A}+\mathrm{B})\right\}$
$=2 \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})]$
$=2 \sin C\left\{2 \sin \frac{2 A}{2} \sin \frac{2 B}{2}\right\}$
$=4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}=\mathrm{RHS}$

(ii)
$
\begin{aligned}
\text { Hint: }\left[\cos \frac{A+B}{2}=\cos \frac{A}{2}\right)-\cos C= & 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}-\left[1-2 \sin ^2 C / 2\right] \\
& =2 \sin \frac{C}{2} \cos \frac{A-B}{2}-1+2 \sin ^2 \frac{C}{2} \\
& =-1+2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}+\sin \frac{C}{2}\right] \\
& =-1+2 \sin \frac{C}{2}\left[\cos \frac{A-B}{2}+\cos \frac{A+B}{2}\right] \\
& =-1+2 \sin \frac{C}{2}\left[2 \cos \frac{2 A}{4} \cos \frac{2 B}{4}\right] \\
& =-1+2 \sin \frac{C}{2}\left[2 \cos \frac{A}{2} \cos \frac{B}{2}\right] \\
& =-1+4 \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}=R H S
\end{aligned}
$

(iii)
$
\begin{aligned}
\text { LHS } & =\frac{1-\cos 2 \mathrm{~A}}{2}+\frac{1-\cos 2 \mathrm{~B}}{2}+\frac{1-\cos 2 \mathrm{C}}{2} \\
& =\frac{3}{2}-\frac{1}{2}[\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}] \\
\text { Hint: } & =\frac{3}{2}-\frac{1}{2}\left[2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+2 \cos ^2 \mathrm{C}-1\right] \\
{\left[\sin ^2 \mathrm{~A}=\frac{1-\cos 2 \mathrm{~A}}{2}\right] \quad } & \frac{3}{2}-\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})-\cos ^2 \mathrm{C}+\frac{1}{2} \\
& =2+\cos \mathrm{C} \cos (\mathrm{A}-\mathrm{B})-\cos 2 \mathrm{C} \\
& =2+\cos \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})(\cos (\mathrm{A}+\mathrm{B})] \\
{\left[\cos \left(180^{\circ}-\mathrm{C}\right)-\cos \mathrm{C}-\cos \mathrm{C}\right] } & \\
=2+\cos \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})+\cos (\mathrm{A} & +\mathrm{B})]
\end{aligned}
$
$
\begin{aligned}
& =2+\cos \mathrm{C}[2 \cos \mathrm{A} \cos \mathrm{B}] \\
& =2+2 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}=\mathrm{RHS}
\end{aligned}
$

(iv)
$
\begin{aligned}
\text { LHS } & =\frac{1-\cos 2 \mathrm{~A}}{2}+\frac{1-\cos 2 \mathrm{~B}}{2}-\frac{1-\cos 2 \mathrm{C}}{2} \\
& =\left[\frac{1}{2}+\frac{1}{2}-\frac{1}{2}\right]-\frac{1}{2}[\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}-\cos 2 \mathrm{C}] \\
& =\frac{1}{2}-\frac{1}{2}\left[2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})-\left(2 \cos ^2 \mathrm{C}-1\right)\right] \\
\text { Hint: } & =\frac{1}{2}-\cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+\cos ^2 \mathrm{C}-\frac{1}{2} \\
{\left[\sin ^2 \mathrm{~A}=\frac{1-\cos 2 \mathrm{~A}}{2}\right] } & =\cos \mathrm{C} \cos (\mathrm{A}-\mathrm{B})+\cos ^2 \mathrm{C} \\
& =\cos \mathrm{C}\left[\cos (\mathrm{A}-\mathrm{B})-\cos ^2(\mathrm{~A}+\mathrm{B})\right] \\
& =\cos \mathrm{C}[2 \sin \mathrm{A} \sin \mathrm{B}]=2 \sin \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}=\mathrm{RHS}
\end{aligned}
$

(v)
Given $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \Rightarrow \frac{\mathrm{A}+\mathrm{B}+\mathrm{C}}{2}=90^{\circ}$
$
\Rightarrow \frac{A+B}{2}=90^{\circ}-\frac{C}{2}
$
So $\tan \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)=\tan \left(90^{\circ}-\frac{\mathrm{C}}{2}\right)=\cot \frac{\mathrm{C}}{2}$
(i.e)
$\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2}=\frac{1}{\tan \frac{C}{2}}$
$\Rightarrow \quad\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \tan \frac{C}{2}=1-\tan \frac{A}{2} \tan \frac{B}{2}$
(i.e.) $\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{C}}{2}+\tan \frac{\mathrm{B}}{2} \tan \frac{\mathrm{C}}{2}=1-\tan \frac{\mathrm{A}}{2} \tan \frac{\mathrm{B}}{2}$
(i.e.) $\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$

(vi)
$
\begin{aligned}
\text { LHS } & =(\sin A+\sin B)+\sin C \\
& =2 \sin \frac{A+B}{2} \cos \left(\frac{A-B}{2}\right)+2 \sin \frac{C}{2} \cos \frac{C}{2} \\
& =2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\sin \frac{C}{2}\right] \\
& =2 \cos \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)+\cos \frac{A+B}{2}\right] \\
& =2 \cos \frac{C}{2}\left[2 \cos \frac{A}{2} \cos \frac{B}{2}\right]=4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}=\text { RHS }
\end{aligned}
$

(vii)
Now A $+\mathrm{B}+\mathrm{C}=180^{\circ}$
So $\mathrm{B}+\mathrm{C}=180^{\circ}-\mathrm{A}$
$\sin (\mathrm{B}+\mathrm{C}-\mathrm{A})=\sin \left(180^{\circ}-\mathrm{A}-\mathrm{A}\right)$
$=\sin \left(180^{\circ}-2 \mathrm{~A}\right)=\sin 2 \mathrm{~A}$
Now LHS $=\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}$
$=4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}$ (from (i) ans) = RHS
Question 2 .
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=2 \mathrm{~s}$, then prove that $\sin (\mathrm{s}-\mathrm{A}) \sin (\mathrm{s}-\mathrm{B})+\sin \mathrm{s} \sin (\mathrm{s}-\mathrm{C})=\sin \mathrm{A} \sin \mathrm{B}$.

Solution:
$
\begin{aligned}
& \text { Now } \sin (s-\mathrm{A}) \sin (s-\mathrm{B})= \frac{1}{2}\{\cos [(s-\mathrm{A})-(s-\mathrm{B})]-\cos [(s-\mathrm{A})+(s-\mathrm{B})]\} \\
&= \frac{1}{2} \cos (s-\mathrm{A}-s+\mathrm{B})-\cos [2 s-(\mathrm{A}+\mathrm{B})] \\
&= \frac{1}{2}\{\cos (\mathrm{A}-\mathrm{B})-\cos \mathrm{C}\} \\
&\quad \quad \quad \therefore \cos (\mathrm{A}-\mathrm{B})=\cos (\mathrm{B}-\mathrm{A})] \\
& \text { Again } \sin s \sin s-\mathrm{C}= \frac{1}{2}[\cos \mathrm{C}-\cos (\mathrm{A}+\mathrm{B})] \\
& \text { So, } \mathrm{LHS}= \frac{1}{2}\{\cos (\mathrm{A}-\mathrm{B})-\cos \mathrm{C}+\cos \mathrm{C}-\cos (\mathrm{A}+\mathrm{B})\} \\
&=\frac{1}{2}[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})] \\
&=\frac{1}{2}[2 \sin \mathrm{A} \sin \mathrm{B}]=\sin \mathrm{A} \sin \mathrm{B}=\mathrm{RHS}
\end{aligned}
$

Question 3.
If $x+y+z=x y z$, then prove that $\frac{2 x}{1-x^2}+\frac{2 y}{1-y^2}+\frac{2 z}{1-z^2}=\frac{2 x}{1-x^2} \frac{2 y}{1-y^2} \frac{2 z}{1-z^2}$.
Solution:
Taking $x=\tan \mathrm{A}, y=\tan \mathrm{B}$ and $z=\tan \mathrm{C}$
$
\frac{2 x}{1-x^2}=\frac{2 \tan \mathrm{A}}{1-\tan ^2 \mathrm{~A}}=\tan 2 \mathrm{~A}
$
Similarly, $\quad \frac{2 y}{1-y^2}=\tan 2 \mathrm{~B}$ and $\frac{2 z}{1-z^2}=\tan 2 \mathrm{C}$
Given $x+y+z=x y z$
(i.e) we are given $\tan \mathrm{A}+\tan B+\tan C=\tan \mathrm{A} \tan B \tan C$
$
\begin{aligned}
& \Rightarrow \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \\
& \Rightarrow \mathrm{A}+\mathrm{B}=180^{\circ}-\mathrm{C} \\
& \text { multiply } 2 \text { on both sides } \Rightarrow 2 \mathrm{~A}+2 \mathrm{~B}=360^{\circ}-2 \mathrm{C} \\
& 2(\mathrm{~A}+\mathrm{B})=360^{\circ}-2 \mathrm{C} \\
& \Rightarrow \tan (2 \mathrm{~A}+2 \mathrm{~B})=\tan \left(360^{\circ}-2 \mathrm{C}\right)=-\tan 2 \mathrm{C} \\
& \text { (i.e) } \quad \frac{\tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}}{1-\tan 2 \mathrm{~A} \tan 2 \mathrm{~B}}=-\tan 2 \mathrm{C}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}=-\tan 2 \mathrm{C}[1-\tan 2 \mathrm{~A} \tan 2 \mathrm{~B}] \\
& \Rightarrow \tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}=-\tan 2 \mathrm{C}+\tan 2 \mathrm{~A} \tan 2 \mathrm{~B} \tan 2 \mathrm{C} \\
& \Rightarrow \tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}+\tan 2 \mathrm{C}=\tan 2 \mathrm{~A} \tan 2 \mathrm{~B} \tan 2 \mathrm{C} \\
& \text { (i.e.) } \frac{2 x}{1-x^2}+\frac{2 y}{1-y^2}+\frac{2 z}{1-z^2}=\frac{2 x}{1-x^2} \times \frac{2 y}{1-y^2} \times \frac{2 z}{1-z^2}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}=-\tan 2 \mathrm{C}[1-\tan 2 \mathrm{~A} \tan 2 \mathrm{~B}] \\
& \Rightarrow \tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}=-\tan 2 \mathrm{C}+\tan 2 \mathrm{~A} \tan 2 \mathrm{~B} \tan 2 \mathrm{C} \\
& \Rightarrow \tan 2 \mathrm{~A}+\tan 2 \mathrm{~B}+\tan 2 \mathrm{C}=\tan 2 \mathrm{~A} \tan 2 \mathrm{~B} \tan 2 \mathrm{C} \\
& \text { (i.e.) } \frac{2 x}{1-x^2}+\frac{2 y}{1-y^2}+\frac{2 z}{1-z^2}=\frac{2 x}{1-x^2} \times \frac{2 y}{1-y^2} \times \frac{2 z}{1-z^2}
\end{aligned}
$
Question 4.
If $\mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{\pi}{2}$, prove the following
(i) $\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}+\sin 2 \mathrm{C}=4 \cos \mathrm{A} \cos \mathrm{B} \cos \mathrm{C}$
(ii) $\operatorname{COS} 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}=1+4 \sin \mathrm{A} \sin \mathrm{B} \sin C$.
Solution:
$
\begin{aligned}
& \text { (i) } \text { LHS }=(\sin 2 A+\sin 2 B)+\sin 2 C \\
& =2 \sin (A+B) \cos (A-B)+2 \sin C \cos C=2 \sin \left(90^{\circ}-C\right) \cos (A-B)+2 \sin C \cos C \\
& =2 \cos C[\cos (A-B)+\sin C]+\cos (A+B)(\therefore A+B=\pi / 2-C) \\
& =2 \cos C[\cos (A-B)+\cos (A+B)] \\
& =2 \cos C[2 \cos A \cos B] \\
& =4 \cos A \cos B \cos C=\text { RHS }
\end{aligned}
$
$
\begin{aligned}
& \text { (ii) } \mathrm{LHS}=(\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B})+\cos 2 \mathrm{C} \\
& =2 \cos (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})+1-2 \sin ^2 \mathrm{C} \\
& =1+2 \sin \mathrm{C}\left(\cos (\mathrm{A}-\mathrm{B})-2 \sin ^2 \mathrm{C}\right) \\
& \left\{\therefore \cos (\mathrm{A}+\mathrm{B})=\cos \left(90^{\circ}-\mathrm{C}\right)=\sin \mathrm{C}\right\} \\
& =1+2 \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})-\sin \mathrm{C}] \\
& =1+2 \sin \mathrm{C}[\cos (\mathrm{A}-\mathrm{B})-\cos (\mathrm{A}+\mathrm{B})] \\
& =1+2 \sin \mathrm{C}[2 \sin \mathrm{A} \sin \mathrm{B}] \\
& =1+4 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}=\mathrm{RHS}
\end{aligned}
$
Question 5.
If $\triangle A B C$ is a right triangle and if $\angle A=\frac{\pi}{2}$, then prove that
(i) $\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C}=1$
(ii) $\sin ^2 \mathrm{~B}+\sin ^2 \mathrm{C}=1$
(iii) $\cos B-\cos C=-1+2 \sqrt{2} \cos \frac{B}{2} \sin \frac{C}{2}$
Solution:

(i)

$
\begin{aligned}
& \angle \mathrm{A}=90^{\circ} ; \cos \mathrm{B}=\frac{\mathrm{AB}}{\mathrm{BC}} ; \cos \mathrm{C}=\frac{\mathrm{AC}}{\mathrm{BC}} \\
& \therefore \mathrm{LHS}=\cos ^2 \mathrm{~B}+\cos ^2 \mathrm{C} \\
& =\frac{\mathrm{AB}^2}{\mathrm{BC}^2}+\frac{\mathrm{AC}^2}{\mathrm{BC}^2}=\frac{\mathrm{AB}^2+\mathrm{AC}^2}{\mathrm{BC}^2} \\
& =\frac{\mathrm{BC}^2}{\mathrm{BC}^2}=1=\mathrm{RHS} \\
&
\end{aligned}
$

(ii) From diagram, $\sin \mathrm{B}=\frac{\mathrm{AC}}{\mathrm{BC}}, \sin \mathrm{C}=\frac{\mathrm{AB}}{\mathrm{BC}}$
$
\begin{aligned}
\text { L.H.S } & =\sin ^2 B+\sin ^2 C \\
& =\frac{A C^2}{B C^2}+\frac{A B^2}{B C^2}=\frac{A C^2+\mathrm{AB}^2}{\mathrm{BC}^2} \\
& =\frac{B C^2}{B C^2}=1=\text { RHS }
\end{aligned}
$

$
\begin{aligned}
& \text { (iii) } \mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \text {, Given } \mathrm{A}=90^{\circ} \\
& \therefore \mathrm{B}+\mathrm{C}=90^{\circ} \Rightarrow \frac{\mathrm{B}+\mathrm{C}}{2}=45^{\circ} \\
& \frac{\mathrm{B}}{2}+\frac{\mathrm{C}}{2}=45^{\circ} \\
& \text { RHS }=-1+2 \sqrt{2} \cos \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2} \\
&=1+\sqrt{2}\left(2 \cos \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}\right)
\end{aligned}
$
We know that $2 \cos A \sin B=\sin (A+B)-\sin (A-B)$
$
\begin{aligned}
& =-1+\sqrt{2}\left(\sin \frac{(B+C)}{2}-\sin \frac{(B-C)}{2}\right) \\
& =-1+\sqrt{2}\left(\sin 45^{\circ}-\sin \frac{(B-C)}{2}\right) \\
& =-1+\sqrt{2}\left(\frac{1}{\sqrt{2}}-\sin \frac{(B-C)^2}{2}\right)
\end{aligned}
$

$\begin{aligned}
& =-1+1-\sqrt{2} \sin \frac{(\mathrm{B}-\mathrm{C})}{2} \\
& =-\sqrt{2} \sin \frac{(\mathrm{B}-\mathrm{C})}{2} \ldots(1) \\
\text { LHS } & =\cos \mathrm{B}-\cos \mathrm{C} \\
& =2 \sin \frac{(\mathrm{B}+\mathrm{C})}{2} \sin \frac{(\mathrm{C}-\mathrm{B})}{2} \\
& =2 \sin 45^{\circ} \sin \frac{(\mathrm{C}-\mathrm{B})}{2} \\
& =2\left(\frac{1}{\sqrt{2}}\right) \sin \left(\frac{-(\mathrm{B}-\mathrm{C})}{2}\right) \\
& =-\sqrt{2} \sin \frac{(\mathrm{B}-\mathrm{C})}{2} \ldots(2) \\
\text { From } & \text { (1) } \&(2) \Rightarrow \text { LHS }=\text { RHS }
\end{aligned}$