Exercise 3.3-Additional Questions - Chapter 3 Trigonometry 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions Solved
Question 1.
Prove that: $\sin 600^{\circ} \cdot \tan \left(-690^{\circ}\right)+\sec 840^{\circ} \cdot \cot \left(-945^{\circ}\right)=\frac{3}{2}$
Solution:
$
\begin{aligned}
& \sin 600^{\circ}=\sin \left(2 \times 360^{\circ}-120^{\circ}\right)=-\sin 120^{\circ} \quad\left[\because \sin \left(360^{\circ}-\theta\right)=-\sin \theta\right] \\
& =-\sin \left(180^{\circ}-60^{\circ}\right)=-\sin 60^{\circ}=-\frac{\sqrt{3}}{2} \quad\left[\therefore \sin \left(180^{\circ}-\theta\right)=\sin \theta\right] \\
& \tan \left(-690^{\circ}\right)=-\tan 690^{\circ} \quad[\because \tan (-\theta)=-\tan \theta] \\
& =-\tan 30^{\circ}=\frac{1}{\sqrt{3}} \\
& \sec \left(840^{\circ}\right)=\sec \left(2 \times 360^{\circ}+120^{\circ}\right) \\
& =\sec 120^{\circ}=\sec \left(180^{\circ}-60^{\circ}\right) \quad\left[\sec \left(180^{\circ}-\theta\right)=\sec \theta\right] \\
& =-\sec 60^{\circ}=-2 \\
& \cot \left(-945^{\circ}\right)=-\cot \left(945^{\circ}\right) \\
& =-\cot \left(2 \times 360^{\circ}+225^{\circ}\right) \\
& =-\cot 225^{\circ}=-\cot \left(180^{\circ}+45^{\circ}\right) \\
& =-\cot 45^{\circ}=-1 \\
& \mathrm{LHS}=\sin 600^{\circ} \tan \left(-690^{\circ}\right)+\sec \left(840^{\circ}\right) \cot \left(-945^{\circ}\right) \\
& =\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{3}}\right)+(-2)(-1) \\
& =\frac{-1}{2}+2=\frac{-3}{2}=\mathrm{RHS} \\
&
\end{aligned}
$

Question 2.
Prove that $\sin \left(270^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right)-\cos \left(270^{\circ}-\theta\right) \cos \left(90^{\circ}+\theta\right)+1=0$
Solution:
LHS $=\sin \left(270^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right)-\cos \left(270^{\circ}-\theta\right) \cos \left(90^{\circ}+\theta\right)+1$
Now, $\sin \left(270^{\circ}-\theta\right)=\sin \left\{180^{\circ}+\left(90^{\circ}-\theta\right)\right\}$
$=-\cos \left(90^{\circ}-\theta\right)=-\sin \theta$
LHS $=-\cos \theta \cdot \cos \theta-(-\sin \theta)(-\sin \theta)+1$
$=-\cos ^2 \theta-\sin ^2 \theta+1$
$=-\left(\cos ^2 \theta+\sin ^2 \theta\right)+1=-1+1=0=$ RHS
Question 3.
Prove that $\cos 24^{\circ}+\cos 55^{\circ}+\cos 125^{\circ}+\cos 204^{\circ}+\cos 300^{\circ}=\frac{1}{2}$
Solution:
$
\begin{aligned}
& \cos 204^{\circ}=\cos \left(180^{\circ}+24^{\circ}\right)=-\cos 24^{\circ} \\
& \cos 125^{\circ}=\cos \left(180^{\circ}-55^{\circ}\right)=-\cos 55^{\circ} \\
& \text { LHS }=\cos 24^{\circ}+\cos 55^{\circ}+\left(-\cos 55^{\circ}\right)+\left(-\cos 24^{\circ}\right)+\cos 300^{\circ}
\end{aligned}
$
$
\begin{aligned}
& =\cos 24^{\circ}+\cos 55^{\circ}-\cos 55^{\circ}-\cos 24^{\circ}+\cos 300^{\circ} \\
& =\cos 300^{\circ}=\cos \left(360^{\circ}-60^{\circ}\right)=\cos 60^{\circ}=\frac{1}{2}=\mathrm{RHS}
\end{aligned}
$

Question 4.
Prove that $\cos ^2 \frac{\pi}{8}+\cos ^2 \frac{3 \pi}{8}+\cos ^2 \frac{5 \pi}{8}+\cos ^2 \frac{7 \pi}{8}=2$
Solution:
$
\begin{aligned}
\cos \frac{5 \pi}{8} & =\sin \left(\frac{\pi}{2}-\frac{5 \pi}{8}\right) \\
& =\sin \left(\frac{4 \pi-5 \pi}{8}\right)=\sin \left(-\frac{\pi}{8}\right)=-\sin \frac{\pi}{8} \quad[\because \sin (-\theta)=-\sin \theta] \\
\cos \frac{7 \pi}{8} & =\sin \left(\frac{\pi}{2}-\frac{7 \pi}{8}\right)=-\sin \left(\frac{4 \pi-7 \pi}{8}\right) \\
& \left.\left.=-\sin \left(-\frac{3 \pi}{8}\right)=-\sin \frac{3 \pi}{8}\right) \sin \left(\frac{\pi}{2}-\theta\right)\right] \\
\mathrm{LHS} & =\cos ^2 \frac{\pi}{8}+\cos ^2 \frac{3 \pi}{8}+\cos ^2 \frac{5 \pi}{8}+\cos ^2 \frac{7 \pi}{8} \\
& =\cos ^2 \frac{\pi}{8}+\cos ^2 \frac{3 \pi}{8}+\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8} \\
& =\left(\cos ^2 \frac{\pi}{8}+\sin ^2 \frac{\pi}{8}\right)+\left(\cos ^2 \frac{3 \pi}{8}+\sin ^2 \frac{3 \pi}{8}\right)=1+1=2=\text { RHS }
\end{aligned}
$
Question 5.
Prove that $\left[1+\cot \alpha-\sec \left(\alpha+\frac{\pi}{2}\right)\right]\left[1+\cot \alpha+\sec \left(\alpha+\frac{\pi}{2}\right)\right]=2 \cot \alpha$
Solution:
$
\begin{aligned}
& \sec \left(\alpha+\frac{\pi}{2}\right)=-\operatorname{cosec} \alpha \\
& \text { LHS }=[(1+\cot \alpha)+\operatorname{cosec} \alpha][(1+\cot \alpha)-\operatorname{cosec} \alpha] \\
& =(1+\cot \alpha)^2-\operatorname{cosec}^2 \alpha \\
& =1+\cot ^2 \alpha+2 \cot \alpha-\operatorname{cosec}^2 \alpha \\
& {\left[\because 1+\cot ^2 \alpha=\operatorname{cosec}^2 \alpha\right]} \\
& =\operatorname{cosec}^2 \alpha+2 \cot \alpha-\operatorname{cosec}^2 \alpha \\
& =2 \cot ^\alpha \alpha=\text { RHS }
\end{aligned}
$

Question 6.
Prove that $\sec \left(\frac{3 \pi}{2}-\theta\right) \sec \left(\theta-\frac{5 \pi}{2}\right)+\tan \left(\frac{5 \pi}{2}+\theta\right) \tan \left(\theta-\frac{3 \pi}{2}\right)=-1$
Solution:
$\begin{aligned} & \sec \left(\frac{3 \pi}{2}-\theta\right)=\sec \left(270^{\circ}-\theta\right)=-\operatorname{cosec} \theta \\ & \sec \left(\theta-\frac{5 \pi}{2}\right)=\sec \left(\theta-450^{\circ}\right) \\ &=\sec \left(450^{\circ}-\theta\right) \\ &=\sec \left(450^{\circ}-\theta\right) \quad[\because \sec (-\theta)=\sec \theta]=\sec \left[-450^{\circ}-\theta\right] \\ &=\sec \left[360^{\circ}+\left(90^{\circ}-\theta\right)\right] \\ & \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec} \theta \\ & \tan \left[\frac{5 \pi}{2}+\theta\right]=\tan \left(450^{\circ}+\theta\right) \\ &=\tan \left[360^{\circ}+\left(90^{\circ}+\theta\right)\right] \\ &=\tan \left(90^{\circ}+\theta\right)=-\cot \theta \\ &=\tan \left(\theta-270^{\circ}\right) \\ &=\tan \left[-\left(270^{\circ}-\theta\right)\right]=-\tan \left(270^{\circ}-\theta\right) \\ &=-\cot \theta \\ & \tan \left[\theta-\frac{3 \pi}{2}\right] \\ & \text { LHS }=(-\operatorname{cosec} \theta)\left(\operatorname{cosec}^{\circ} \theta\right)+(-\cot \theta)(-\cot \theta) \\ &=-\operatorname{cosec} \theta+\cot ^2 \theta \\ &=-\left(\operatorname{cosec} \theta-\cot ^2 \theta\right)=-1=\text { RHS }\end{aligned}$

Question 7.
Simplify: $\frac{\cos \left(90^{\circ}+\theta\right) \sec (-\theta) \tan \left(180^{\circ}-\theta\right)}{\sec \left(360^{\circ}-\theta\right) \sin \left(180^{\circ}+\theta\right) \cot \left(90^{\circ}+\theta\right)}$
Solution:
$
\begin{aligned}
& \cos \left(90^{\circ}+\theta\right)=-\sin \theta \\
& \sec (-\theta)=\sec \theta \\
& \tan \left(180^{\circ}-\theta\right)=-\tan \theta \\
& \sec \left(360^{\circ}-\theta\right)=\sec \theta \\
& \sin \left(180^{\circ}+\theta\right)=-\sin \theta \\
& \cot \left(90^{\circ}+\theta\right)=-\tan \theta \\
& \therefore \frac{\cos \left(90^{\circ}+\theta\right) \sec (-\theta) \tan \left(180^{\circ}-\theta\right)}{\sec \left(360^{\circ}-\theta\right) \sin \left(180^{\circ}+\theta\right) \cot \left(90^{\circ}+\theta\right)}=\frac{(-\sin \theta)(\sec \theta)(-\tan \theta)}{(\sec \theta)(-\sin \theta)(-\tan \theta)}=1
\end{aligned}
$
Question 8.
Find $\mathrm{x}$ from the equation $\operatorname{cosec}\left(90^{\circ}+\mathrm{A}\right)+\mathrm{x} \cos \mathrm{A} \cot \left(90^{\circ}+\mathrm{A}\right)=\sin \left(90^{\circ}+\mathrm{A}\right)$.
Solution:
$
\begin{aligned}
\operatorname{cosec}\left(90^{\circ}+\mathrm{A}\right) & =\sec \mathrm{A}, \cot \left(90^{\circ}+\mathrm{A}\right)=-\tan \mathrm{A} \\
\mathrm{LHS} & =\sec \mathrm{A}+x \cos \mathrm{A}(-\tan \mathrm{A}) \\
& =\frac{1}{\cos \mathrm{A}}-x \cos \mathrm{A} \times \frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\frac{1}{\cos \mathrm{A}}-x \sin \mathrm{A} \\
\mathrm{RHS} & =\sin \left(90^{\circ}+\mathrm{A}\right)=\cos \mathrm{A} \\
\therefore \frac{1}{\cos \mathrm{A}}-x \sin \mathrm{A} & =\cos \mathrm{A} \\
\Rightarrow \frac{1}{\cos \mathrm{A}}-\cos \mathrm{A} & =x \sin \mathrm{A} \Rightarrow \frac{1-\cos ^2 \mathrm{~A}}{\cos \mathrm{A}}=x \sin \mathrm{A} \\
\Rightarrow \frac{\sin ^2 \mathrm{~A}}{\sin \mathrm{A} \cos \mathrm{A}} & =x
\end{aligned}
$

$x=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\tan \mathrm{A}$