Exercise 6.2 - Chapter 6 Trigonometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex6.2
Question $1 .$
Verify the following equalities:
(i) $\sin ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}=1$
(ii) $1+\tan ^{2} 30^{\circ}=\sec ^{2} 30^{\circ}$
(iii) $\cos 90^{\circ}=1-2 \sin ^{2} 45^{\circ}=2 \cos ^{2} 45^{\circ}-1$
(iv) $\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}=\sin 90$

Solution:

(i) $\sin ^{2} 60^{\circ}+\cos ^{2} 60^{\circ}=\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$
$1+\tan ^{2} 30^{\circ}=\sec ^{2} 30^{\circ}$

(ii)
$\begin{aligned}
1+\tan ^{2} 30^{\circ} &=\sec ^{2} 30^{\circ} \\
1+\tan ^{2} 30^{\circ}=1+\left(\frac{1}{\sqrt{3}}\right)^{2} &=1+\frac{1}{3}=\frac{3+1}{3}+\frac{4}{3} \\
\sec 30^{\circ}=\frac{1}{\cos 30} &=\frac{1}{\frac{\sqrt{3}}{2}}=1 \times \frac{2}{\sqrt{3}}=\frac{2}{\sqrt{3}} \\
\sec ^{2} 30 &=\left(\frac{2}{\sqrt{3}}\right)^{2} \\
&=\frac{4}{3}
\end{aligned}$
$(1)=(2) \Rightarrow \quad \text { LHS }=\text { RHS }$
$\therefore$ Hence it is verified.

(iii)
$\begin{aligned}
\cos 90^{\circ} &=1-2 \sin ^{2} 45^{\circ}=2 \cos ^{2} 45^{\circ}-1 \\
\cos 90^{\circ} &=0 \\
1-2 \sin ^{2} 45^{\circ} &=1-2 \times\left(\frac{1}{\sqrt{2}}\right)^{2} \\
&=1-2 \times \frac{1}{2}=1-1=0 \\
2 \cos ^{2} 45^{\circ}=2 \times\left(\frac{1}{\sqrt{2}}\right)^{2} &=2 \times \frac{1}{2} \\
2 \cos ^{2} 45^{\circ}-1 &=1-1=0
\end{aligned}$
$(1)=(2)=(3)$. Hence it is verified.
(iv) $\sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ}=\sin 90^{\circ}$
$\begin{aligned}
\sin 30^{\circ} &=\frac{1}{2}, \sin 60^{\circ}=\frac{\sqrt{3}}{2} \\
\cos 30^{\circ} &=\frac{\sqrt{3}}{2}, \cos 60^{\circ}=\frac{1}{2} \\
\sin 90^{\circ} &=1 \\
\therefore \sin 30^{\circ} \cos 60^{\circ} &=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4} \\
\cos 30^{\circ} \sin 60^{\circ} &=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}=\frac{3}{4} \\
\therefore \sin 30^{\circ} \cos 60^{\circ}+\cos 30^{\circ} \sin 60^{\circ} &=\frac{1}{4}+\frac{3}{4}=\frac{4}{A}=1=\sin 90^{\circ} .
\end{aligned}$
Hence it is verified.

Question $2 .$
Find the value of the following:
(i) $\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}}$
(ii) $\left(\sin 90^{\circ}+\cos 60^{\circ}+\cos 45^{\circ}\right) \times\left(\sin 30^{\circ}+\cos 0^{\circ}-\cos 45^{\circ}\right)$
(iii) $\sin ^{2} 30^{\circ}-2 \cos ^{3} 60^{\circ}+3 \tan ^{4} 45^{\circ}$

Solution:
(i) $\frac{\tan 45^{\circ}}{\operatorname{cosec} 30^{\circ}}+\frac{\sec 60^{\circ}}{\cot 45^{\circ}}-\frac{5 \sin 90^{\circ}}{2 \cos 0^{\circ}}$
$=\frac{1}{2}+\frac{2}{1}-\frac{5 \times 1}{2 \times 1}$
$=\frac{1}{2}+\frac{2}{1}-\frac{5}{2}=\frac{1+4-5}{2}=\frac{5-5}{2}=\frac{0}{2}=0$
(ii) $\left(\sin 90^{\circ}+\cos 60^{\circ}+\cos 45^{\circ}\right) \times\left(\sin 30^{\circ}+\cos 0^{\circ}-\cos 45^{\circ}\right)$
$\begin{aligned}
&=\left(1+\frac{1}{2}+\frac{1}{\sqrt{2}}\right) \times\left(\frac{1}{2}+1-\frac{1}{\sqrt{2}}\right) \\
&=\left(1+\frac{1}{2}\right)^{2}-\frac{1}{2}=\frac{9}{4}-\frac{1}{2}=\frac{7}{4}
\end{aligned}$
(iii) $\sin ^{2} 30^{\circ}-2 \cos ^{3} 60^{\circ}+3 \tan ^{4} 45^{\circ}$
$\begin{aligned}
&=\left(\frac{1}{2}\right)^{3}-2\left(\frac{1}{2}\right)^{3}+3(1)^{4} \\
&=\frac{1}{4}-2 \underset{4}{8}\left(\frac{1}{8}\right)+3(1)=\frac{1}{4}-\frac{1}{4}+3=3
\end{aligned}$