Exercise 3.9-Additional Question - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
In any triangle $\mathrm{ABC}$ prove that $a^2=(b+c)^2 \sin ^2 \frac{\mathrm{A}}{2}+(b-c)^2 \cos ^2 \frac{\mathrm{A}}{2}$
Solution:
$
\begin{aligned}
\text { RHS } & =(b+c)^2 \sin ^2 \frac{\mathrm{A}}{2}+(b-c)^2 \cos ^2 \frac{\mathrm{A}}{2} \\
& =\left(b^2+c^2+2 b c\right) \sin ^2 \frac{\mathrm{A}}{2}+\left(b^2+c^2-2 b c\right) \cos ^2 \frac{\mathrm{A}}{2} \\
& =\left(b^2+c^2\right)\left[\sin ^2 \frac{\mathrm{A}}{2}+\cos ^2 \frac{\mathrm{A}}{2}\right]+2 b c\left[\sin ^2 \frac{\mathrm{A}}{2}-\cos ^2 \frac{\mathrm{A}}{2}\right] \\
& =b^2+c^2-2 b c \cos \mathrm{A}=a^2=\mathrm{LHS}
\end{aligned}
$
Question 2.
$
\frac{\cos \mathrm{A}}{a}+\frac{\cos \mathrm{B}}{b}+\frac{\cos \mathrm{C}}{c}=\frac{a^2+b^2+c^2}{2 a b c}
$

Solution:
$
\begin{aligned}
\text { LHS } & =\frac{\cos \mathrm{A}}{a}+\frac{\cos \mathrm{B}}{b}+\frac{\cos \mathrm{C}}{c} \\
& =\frac{b^2+c^2-a^2}{2 a b c}+\frac{a^2+c^2-b^2}{2 a b c}+\frac{a^2+b^2-c^2}{2 a b c} \\
& =\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2 a b c} \\
& =\frac{a^2+b^2+c^2}{2 a b c}=\mathrm{RHS}
\end{aligned}
$
Question 3.
Prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{a^2-b^2}{c^2}$.
Solution:
$
\begin{aligned}
& \text { By sine formula } \frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}=2 \mathrm{R} \\
& \frac{a^2-b^2}{c^2}=\frac{(2 \mathrm{R} \sin \mathrm{A})^2-(2 \mathrm{R} \sin \mathrm{B})^2}{(2 \mathrm{R} \sin \mathrm{C})^2} \\
& =\frac{4 R^2 \sin ^2 \mathrm{~A}-4 \mathrm{R}^2 \sin ^2 \mathrm{~B}}{4 \mathrm{R}^2 \sin ^2 \mathrm{C}}=\frac{\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}}{\sin ^2 \mathrm{C}} \\
& =\frac{\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}{\sin ^2 \mathrm{C}} \quad[\because \sin \mathrm{C}=\sin (\mathrm{A}+\mathrm{B})] \\
& =\frac{\sin (A+B) \sin (A-B)}{\sin ^2(A+B)}=\frac{\sin (A-B)}{\sin (A+B)} \\
&
\end{aligned}
$
Question 4.
Prove that $\cos \frac{B-C}{2}=\frac{b+c}{a} \sin \frac{A}{2}$.

Solution:
$
\begin{aligned}
& \frac{b+c}{a} \sin \frac{\mathrm{A}}{2}=\frac{2 \mathrm{R} \sin \mathrm{B}+2 \mathrm{R} \sin \mathrm{C}}{2 \mathrm{R} \sin \mathrm{A}} \sin \frac{\mathrm{A}}{2} \\
& =\frac{\sin \mathrm{B}+\sin \mathrm{C}}{\sin \mathrm{A}} \sin \frac{\mathrm{A}}{2} \\
& =\frac{2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}}{2 \sin \frac{A}{2} \cos \frac{A}{2}} \sin \frac{A}{2}=\frac{\sin \frac{B+C}{2} \cos \frac{B-C}{2}}{\cos \frac{A}{2}} \\
& =\frac{\sin \left(\frac{180-\mathrm{A}}{2}\right) \cos \frac{\mathrm{B}-\mathrm{C}}{2}}{\cos \frac{\mathrm{A}}{2}}=\frac{\sin \left(90-\frac{\mathrm{A}}{2}\right) \cos \frac{\mathrm{B}-\mathrm{C}}{2}}{\cos \frac{\mathrm{A}}{2}} \\
& =\cos \frac{\mathrm{B}-\mathrm{C}}{2} \quad \because \sin \left(90-\frac{\mathrm{A}}{2}\right)=\cos \frac{\mathrm{A}}{2} \\
&
\end{aligned}
$

Question 5.
In any triangle $A B C$, prove that $\frac{a^2 \sin (B-C)}{\sin A}+\frac{b^2 \sin (C-A)}{\sin B}+\frac{c^2 \sin (A-B)}{\sin C}=0$
Solution:
$
\begin{aligned}
& \frac{a^2 \sin (\mathrm{B}-\mathrm{C})}{\sin \mathrm{A}}=\frac{(2 \mathrm{R} \sin \mathrm{A})^2 \sin (\mathrm{B}-\mathrm{C})}{\sin \mathrm{A}}=\frac{4 \mathrm{R}^2 \sin ^2 \mathrm{~A} \sin (\mathrm{B}-\mathrm{C})}{\sin \mathrm{A}} \\
& =4 \mathrm{R}^2 \sin \mathrm{A} \sin (\mathrm{B}-\mathrm{C})=4 \mathrm{R}^2 \sin (\mathrm{B}+\mathrm{C}) \sin (\mathrm{B}-\mathrm{C}) \\
& =4 \mathrm{R}^2\left(\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}\right)=4 \mathrm{R}^2 \sin ^2 \mathrm{~B}-4 \mathrm{R}^2 \sin ^2 \mathrm{C} \\
& =b^2-c^2 \\
& \text { Similarly, } \frac{b^2 \sin (\mathrm{C}-\mathrm{A})}{\sin \mathrm{B}}=c^2-a^2 \\
& \frac{c^2 \sin (\mathrm{A}-\mathrm{B})}{\sin \mathrm{C}}=a^2-b^2 \\
& \therefore \frac{a^2 \sin (\mathrm{B}-\mathrm{C})}{\sin \mathrm{A}}+\frac{b^2 \sin (\mathrm{C}-\mathrm{A})}{\sin \mathrm{B}}+\frac{c^2 \sin (\mathrm{A}-\mathrm{B})}{\sin \mathrm{C}} \\
& =b^2-c^2+c^2-a^2+a^2-b^2=0 \\
&
\end{aligned}
$