Exercise 3.4-Additional Questions - Chapter 3 - Trigonometry - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions
Question 1.
Prove that $\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})=\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}$
Solution:
$
\begin{aligned}
& \text { LHS }=\sin (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B}) \\
& =(\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B})(\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}) \\
& =\sin ^2 \mathrm{~A} \cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A} \sin ^2 \mathrm{~B} \\
& =\left(1-\cos ^2 \mathrm{~A}\right) \cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}\left(1-\cos ^2 \mathrm{~B}\right) \\
& =\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A} \cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~A} \cos ^2 \mathrm{~B} \\
& =\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}=\text { RHS }
\end{aligned}
$
Question 2.
Prove that
(i) $\sin \mathrm{A}+\sin \left(120^{\circ}+\mathrm{A}\right)+\sin \left(240^{\circ}+\mathrm{A}\right)=0$
(ii) $\cos \mathrm{A}+\cos \left(120^{\circ}+\mathrm{A}\right)+\cos \left(120^{\circ}-\mathrm{A}\right)=0$
Solution:
(i) $\sin \mathrm{A}+\sin \left(120^{\circ}+\mathrm{A}\right)+\sin \left(240^{\circ}+\mathrm{A}\right)$
$=\sin \mathrm{A}+\sin 120^{\circ} \cos \mathrm{A}+\cos 120^{\circ} \sin \mathrm{A}+\sin 240^{\circ} \cos \mathrm{A}+\cos 240^{\circ} \sin \mathrm{A}$

Now,
$
\begin{aligned}
& \sin 120^{\circ}=\sin \left(180^{\circ}-60^{\circ}\right)=\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\
& \sin 240^{\circ}=\sin \left(180^{\circ}+60^{\circ}\right)=-\sin 60^{\circ}=\frac{-\sqrt{3}}{2} \\
& \cos 120^{\circ}=\cos \left(180^{\circ}-60^{\circ}\right)=-\cos 60^{\circ}=\frac{-1}{2} \\
& \cos 240^{\circ}=\cos \left(180^{\circ}+60^{\circ}\right)=-\cos 60^{\circ}=\frac{-1}{2}
\end{aligned}
$
By substituting these values in (1), we get,
$
\begin{aligned}
& \sin \mathrm{A}+\frac{\sqrt{3}}{2} \cos \mathrm{A}-\frac{1}{2} \sin \mathrm{A}-\frac{\sqrt{3}}{2} \cos \mathrm{A}-\frac{1}{2} \sin \mathrm{A}=0=\mathrm{RHS} \\
& \text { (ii) } \cos 120^{\circ}=\cos \left(180^{\circ}-60^{\circ}\right)=-\cos 60^{\circ}=\frac{-1}{2} \\
& \text { LHS }=\cos \mathrm{A}+\cos \left(120^{\circ}+\mathrm{A}\right)+\cos \left(120^{\circ}-\mathrm{A}\right) \\
& =\cos \mathrm{A}+\cos 120^{\circ} \cos \mathrm{A}-\sin 120^{\circ} \sin \mathrm{A}+\cos 120^{\circ} \cos \mathrm{A}+\sin 120^{\circ} \sin \mathrm{A} \\
& =\cos \mathrm{A}+2 \cos 120^{\circ} \cos \mathrm{A} \\
& =\cos \mathrm{A}+2\left(\frac{-1}{2}\right) \cos \mathrm{A}=\cos \mathrm{A}-\cos \mathrm{A}=0=\text { RHS }
\end{aligned}
$
Question 3.
If $\tan A=\frac{5}{6}, \tan B=\frac{1}{11}$, show that $A+B=45^{\circ}$.
Solution:
$
\begin{aligned}
\tan (A+B) & =\frac{\tan A+\tan B}{1-\tan A \tan B} \\
& =\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6} \cdot \frac{1}{11}}=\frac{\frac{55+6}{66}}{\frac{66-5}{66}}=\frac{\left(\frac{61}{66}\right)}{\left(\frac{61}{66}\right)} \\
& =\tan 45^{\circ} \\
\tan (A+B)=1 & \Rightarrow A+B=45^{\circ}
\end{aligned}
$

Question 4.
If $A+B=45^{\circ}$, show that $(1+\tan A)(1+\tan B)=2$ and hence deduce the value of $\tan 22 \frac{1^{\circ}}{2}$.
Solution:
$
\begin{aligned}
& \text { Given } A+B=45^{\circ} \Rightarrow \tan (A+B)=\tan 45^{\circ} \\
& \frac{\tan A+\tan B}{1-\tan A \cdot \tan B}=1 \\
& \text { (i.e.) } \tan A+\tan B=1-\tan A \cdot \tan B
\end{aligned}
$
(i.e.) $1+\tan \mathrm{A}+\tan \mathrm{B}=2-\tan \mathrm{A} \tan \mathrm{B}$ (add 1 on both sides)
$1+\tan \mathrm{A}+\tan \mathrm{B}+\tan \mathrm{A} \tan \mathrm{B}=2$
(i.e.) $(1+\tan A)(1+\tan B)=2$
Take $A=B$ then $2 A=45^{\circ} \Rightarrow A=22 \frac{1}{2}=B$
$
\begin{gathered}
\therefore\left(1+\tan 22 \frac{1}{2}\right)^2=2 \Rightarrow 1+\tan 22 \frac{1}{2}= \pm \sqrt{2} \\
\therefore \tan 22 \frac{1}{2}= \pm \sqrt{2}-1
\end{gathered}
$
Since $22 \frac{1}{2}$ is acute, $\tan 22 \frac{1}{2}$ is positive and therefore $\tan 22 \frac{1}{2}=\sqrt{2}-1$