Exercise 6.3 - Chapter 6 Trigonometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.3$
Question $1 .$
Find the value of the following:
(i) $\left(\frac{\cos 47^{\circ}}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin 72^{\circ}}{\cos 18^{\circ}}\right)^{2}-2 \cos ^{2} 45^{\circ}$
(ii) $\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}-8 \cos ^{2} 60^{\circ}$
(iii) $\tan 15^{\circ} \tan 30^{\circ} \tan 45^{\circ} \tan 60^{\circ} \tan 75^{\circ}$
(iv) $\frac{\cot \theta}{\tan \left(90^{\circ}-\theta\right)}+\frac{\cos \left(90^{\circ}-\theta\right) \tan \theta \sec \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}-\theta\right)}$

Solution:

(i)
$\begin{aligned}
\left(\frac{\cos 47^{\circ}}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin 72^{\circ}}{\cos 18^{\circ}}\right)^{2}-2 \cos ^{2} 45^{\circ} &=\left(\frac{\cos \left(90^{\circ}-43^{\circ}\right)}{\sin 43^{\circ}}\right)^{2}+\left(\frac{\sin \left(90^{\circ}-18^{\circ}\right)}{\cos 18}\right)^{2}-2 \cos ^{2} 45^{\circ} \\
&=\left(\frac{\sin 43}{\sin 43}\right)^{2}+\left(\frac{\cos 18}{\cos 18}\right)^{2}-2\left(\frac{1}{\sqrt{2}}\right)^{2} \\
&=1^{2}+1^{2}-2^{\circ} \times \frac{1}{z}=2-1=1
\end{aligned}$
(ii)$\begin{aligned}
&\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}+\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}-8 \cos ^{2} 60^{\circ} \\
&=\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}+\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}+\frac{\cos \theta}{\cos \theta}-8\left(\frac{1}{2}\right)^{2} \\
&=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}+\frac{\cos \theta}{\cos \theta}-8 \times \frac{2}{4}=1+1+1-2=1
\end{aligned}$
(iii) $\tan 15^{\circ} \tan 30^{\circ} \tan 45^{\circ} \tan 60^{\circ} \tan 75^{\circ}$
$\begin{aligned}
&=\tan 15^{\circ} \times \frac{1}{\sqrt{3}} \cdot 1 \cdot \sqrt{3} \cdot \tan 75^{\circ} \\
&=\tan 15^{\circ} \times \tan \left(90-15^{\circ}\right)=\tan 15^{\circ} \times \cot 15^{\circ}=\tan 15^{\circ} \times \frac{1}{\tan 15^{\circ}}=1
\end{aligned}$
(iv) $\frac{\cot \theta}{\tan \left(90^{\circ}-\theta\right)}+\frac{\cos \left(90^{\circ}-\theta\right) \tan \theta \sec \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}-\theta\right) \cot \left(90^{\circ}-\theta\right) \operatorname{cosec}\left(90^{\circ}-\theta\right)}$
$=\frac{\cot \theta}{\cot \theta}+\frac{\cos (90-\theta) \cdot \tan \theta \cdot \operatorname{cosec} \theta}{\cos \theta \cdot \tan \theta \cdot \sec \theta}=1+\frac{\sin \theta}{\cos \theta} \cdot\left(\frac{1}{\sin \theta} / \frac{1}{\cos \theta}\right)=1+\frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta} \times \frac{\cos \theta}{1}=2$