Additional Questions - Chapter 6 Trigonometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Exercise 6.1
Question $1 .$
Find the six trigonometric ratios of the angle 6 using the diagram

Solution:
$\begin{aligned}
\text { Hypotenuse } &=\sqrt{12^{2}+5^{2}} \\
&=\sqrt{144+25}=\sqrt{169} \\
&=13
\end{aligned}$

Question $2 .$
If $3 \cot \theta=1$, then find the value of $\frac{3 \cos \theta-4 \sin \theta}{5 \sin \theta+4 \cos \theta}$
Solution:
$3 \cot \theta=1$
$\begin{aligned}
\cot \theta &=\frac{1}{3} \\
\frac{\text { adjacent }}{\text { opposite }} &=\frac{1}{3} \\
\text { Hypotenuse } &=\sqrt{3^{2}+1}=\sqrt{10} \\
\frac{3 \cos \theta-4 \sin \theta}{5 \sin \theta+4 \cos \theta} &=\frac{3 \times \frac{1}{\sqrt{10}}-4 \times \frac{3}{\sqrt{10}}}{5 \times \frac{3}{\sqrt{10}}+4 \times \frac{1}{\sqrt{10}}}=\frac{3-12}{15+4}=\frac{-9}{19}
\end{aligned}$

Question $3 .$
If $\sin \theta=\frac{a}{\sqrt{\left(a^{2}+b^{2}+c^{2}+2 b c\right)}}$, then show that $(b+c) \sin \theta=(a) \cos \theta$.

Solution:

$\begin{aligned} \text { Adjacent side }=& \sqrt{a^{2}+b^{2}+c^{2}+2 b c-a^{2}}=\sqrt{(b+c)^{2}}=b+c \\(b+c) \sin \theta=a \cos \theta \end{aligned}$ $(b+c) \times \frac{a}{\sqrt{\left(a^{2}+b^{2}+c^{2}+2 b c\right)}}=a \times \frac{(b+c)}{\sqrt{a^{2}+b^{2}+c^{2}+2 b c}}$ Hence proved Question 4. $\begin{aligned} \text { If } 3(\tan \theta)+4(\sec \theta \times \sin \theta)=24 \text {. Then find all the trigonometric ratios of the angle } \theta . \\ \text { Solution: } \\ 3 \tan \theta+4(\sec \theta \times \sin \theta) &=24 \\ 3 \tan \theta+4\left(\frac{1}{\cos \theta} \times \sin \theta\right) &=24 \\ 7 \tan \theta &=24 \\ \tan \theta &=\frac{24}{7} \\ \text { hypetenuse } &=\sqrt{24^{2}+49}=\sqrt{576+49}=\sqrt{625}=25 . \\ \sin \theta=\frac{24}{25} ; \cos \theta=\frac{7}{25} ; \operatorname{cosec} \theta &=\frac{25}{24} ; \sec \theta=\frac{25}{7} ; \cot \theta=\frac{7}{24} \end{aligned}$

Question $4 .$
If $3(\tan \theta)+4(\sec \theta \times \sin \theta)=24$. Then find all the trigonometric ratios of the angle $\theta$.
Solution:

$\begin{aligned}
3 \tan \theta+4(\sec \theta \times \sin \theta) &=24 \\
3 \tan \theta+4\left(\frac{1}{\cos \theta} \times \sin \theta\right) &=24 \\
7 \tan \theta &=24 \\
\tan \theta &=\frac{24}{7} \\
\text { hypetenuse } &=\sqrt{24^{2}+49}=\sqrt{576+49}=\sqrt{625}=25 \\
\sin \theta=\frac{24}{25} ; \cos \theta=\frac{7}{25} ; \operatorname{cosec} \theta &=\frac{25}{24} ; \sec \theta=\frac{25}{7} ; \cot \theta=\frac{7}{24}
\end{aligned}$

Question $5 .$
From the given figure, find all the trigonometric ratios of angle $\theta$. Solution:
$\begin{aligned}
x &=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}=\sqrt{25}=5 \\
\sin \theta &=\frac{5}{13} ; \cos \theta=\frac{12}{13} ; \\
\tan \theta &=\frac{5}{12} ; \operatorname{cosec} \theta=\frac{13}{5} ; \\
\sec \theta &=\frac{13}{12} ; \cot \theta=\frac{12}{5}
\end{aligned}$
 

Exercise $6.2$
Question 1.
Find the value of $\sin 3 x \cdot \sin 6 x \cdot \sin 9 x$ when $x=10^{\circ}$
Solution:
$\sin 3\left(10^{\circ}\right) \sin 6\left(10^{\circ}\right) \sin 9\left(10^{\circ}\right)=\frac{1}{2} \times \frac{\sqrt{3}}{2} \times 1=\frac{\sqrt{3}}{4}$

Question $2 .$
Find the value of $\cot 15^{\circ} \cdot \cot 30^{\circ} \cdot \cot 45^{\circ} \cdot \cot 60^{\circ} \cdot \cot 75^{\circ}$
Solution:
$\cot \left(90^{\circ}-75\right) \cot \left(90^{\circ}-60^{\circ}\right) \cot 45^{\circ} \cot 60^{\circ} \cot 75^{\circ}$
$=\tan 75^{\circ} \tan 60^{\circ}(1) \cot 60^{\circ} \cot 75^{\circ}=1$

Exercise $6.2$
Question $1 .$
Find the value of $\sin 3 x . \sin 6 x . \sin 9 x$ when $x=10^{\circ}$
Solution:
$\sin 3\left(10^{\circ}\right) \sin 6\left(10^{\circ}\right) \sin 9\left(10^{\circ}\right)=\frac{1}{2} \times \frac{\sqrt{3}}{2} \times 1=\frac{\sqrt{3}}{4}$
 

Question $2 .$
Find the value of $\cot 15^{\circ} \cdot \cot 30^{\circ} \cdot \cot 45^{\circ} \cdot \cot 60^{\circ} \cdot \cot 75^{\circ}$
Solution:
$\cot \left(90^{\circ}-75\right) \cot \left(90^{\circ}-60^{\circ}\right) \cot 45^{\circ} \cot 60^{\circ} \cot 75^{\circ}$
$=\tan 75^{\circ} \tan 60^{\circ}(1) \cot 60^{\circ} \cot 75^{\circ}=1$
 

Exercise $6.3$
Question $3 .$
Find the value of $\cos 19^{\circ} 59^{\prime}+\tan 12^{\circ} 12^{\prime}+\sin 49^{\circ} 20^{\prime}$.
Solution:
$\cos 19^{\circ} 59^{\prime}+\tan 12^{\circ} 12^{\prime}+\sin 49^{\circ} 20^{\prime}=0$
 

Question $4 .$
Given that $\sin \alpha=\frac{1}{\sqrt{2}}$ and $\tan \beta=\sqrt{3}$. Find the value of $\alpha+\beta$.
Solution:
$\begin{aligned}
\alpha &=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ} \\
\beta &=60^{\circ} \\
\alpha+\beta &=105^{\circ}
\end{aligned}$

Question $5 .$
Find the value of $\frac{\cos 63^{\circ} 20^{r}}{\sin 26^{2} 40^{\prime}}$
Solution:
$\frac{\cos 63^{\circ} 20^{\prime}}{\sin 26^{\circ} 40^{\prime}}=\frac{\cos 63^{\circ} 20^{\prime}}{\cos 63^{\circ} 20^{\prime}}=1$