Additional Questions - Chapter 6 - Trigonometry - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
Given $\tan \mathrm{A}=\frac{4}{3}$, find the other trigonometric ratios of the angle A.
Solution:
Let us first draw a right $\triangle \mathrm{ABC}$.
Now, we know that $\tan \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{4}{3}$
Therefore, if $\mathrm{BC}=4 \mathrm{k}$, then $\mathrm{AB}=3 \mathrm{k}$, where $\mathrm{k}$ is a positive number.

Now, by using the pythagoras theorem, we have
$
\begin{aligned}
& \mathrm{AC}^2=\mathrm{AB}^2+\mathrm{BC}^2 \\
& =(4 \mathrm{k})^2+(3 \mathrm{k})^2=25 \mathrm{k}^2 \\
& \mathrm{AC}=5 \mathrm{k}
\end{aligned}
$
So,

Now, we can write all the trigonometric ratios using their definitions.
$
\begin{aligned}
\sin \mathrm{A} & =\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{4 k}{5 k}=\frac{4}{5} \\
\cos \mathrm{A} & =\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{3 k}{5 k}=\frac{3}{5}
\end{aligned}
$
Therefore, $\cot \mathrm{A}=\frac{1}{\tan \mathrm{A}}=\frac{3}{4}$
$\operatorname{cosec} A=\frac{1}{\sin A}=\frac{5}{4}$, and
$
\sec A=\frac{1}{\cos A}=\frac{5}{3}
$
Question 2.
Prove that $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$,
using the identity $\sec ^2 \theta=1+\tan ^2 \theta$.
Solution:
Since we will apply the identity involving $\sec \theta$ and $\tan \theta$, let us first convert the LHS (of the identity we need to prove) in terms of $\sec \theta$ and $\tan \theta$ by dividing numerator and denominator by $\cos \theta$.

$
\begin{aligned}
\text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\
& =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1} \\
& =\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\
& =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \\
& =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \\
& =\frac{-1}{\tan \theta-\sec \theta} \\
& =\frac{1}{\sec \theta-\tan \theta}
\end{aligned}
$
Which is the RHS of the identity, we are required to prove.
Question 3.
Prove that $\sec \mathrm{A}(1-\sin \mathrm{A})(\sec \mathrm{A}+\tan \mathrm{A})=1$.
Solution:

$
\begin{aligned}
\text { LHS } & =\sec \mathrm{A}(1-\sin \mathrm{A})(\sec \mathrm{A}+\tan \mathrm{A}) \\
& =\left[\frac{1}{\cos \mathrm{A}}\right](1-\sin \mathrm{A})\left[\frac{1}{\cos \mathrm{A}}+\frac{\sin \mathrm{A}}{\cos \mathrm{A}}\right] \\
& =\frac{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}{\cos ^2 \mathrm{~A}} \\
& =\frac{1-\sin ^2 \mathrm{~A}}{\cos ^2 \mathrm{~A}} \\
& =\frac{\cos ^2 \mathrm{~A}}{\cos ^2 \mathrm{~A}}=1=\mathrm{RHS}
\end{aligned}
$
Question 4.
In a right triangle $\mathrm{ABC}$, right-angled at $\mathrm{B}$, if $\tan \mathrm{A}=1$, then verify that $2 \sin \mathrm{A} \cos \mathrm{A}=1$. Solution:
In $\mathrm{ABC}, \tan \mathrm{A}=\frac{B C}{A B}$

Let $\mathrm{AB}=\mathrm{BC}=\mathrm{k}$, where $\mathrm{k}$ is a positive number.

$
\begin{aligned}
& \text { Now, } \quad \mathrm{AC}=\sqrt{\mathrm{AB}^2+\mathrm{BC}^2} \\
& =\sqrt{(k)^2+(k)^2}=k \sqrt{2}
\end{aligned}
$
Therefore,
$
\begin{aligned}
& \sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{1}{\sqrt{2}} \text { and } \\
& \cos \mathrm{A}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{1}{\sqrt{2}}
\end{aligned}
$
So, $2 \sin A \cos A=2\left[\frac{1}{\sqrt{2}}\right]\left[\frac{1}{\sqrt{2}}\right]=1$, which is the required value.
Question 5.
If $\sin (A-B)=\frac{1}{2}, \cos (A+B)=\frac{1}{2}, 0^{\circ}B$, find $A$ and $B C$

Solution:

Since, $\sin (\mathrm{A}-\mathrm{B})=\frac{1}{2}, \therefore \mathrm{A}-\mathrm{B}=30^{\circ}$
Also, since $\cos (\mathrm{A}+\mathrm{B})=\frac{1}{2}$,
$\therefore \quad \mathrm{A}+\mathrm{B}=60^{\circ}$
Solving (1) and (2)
$
\begin{aligned}
\mathrm{A}-\mathrm{B}+\mathrm{A}+\mathrm{B} & =30^{\circ}+60^{\circ} \\
2 \mathrm{~A} & =90^{\circ} \\
\mathrm{A} & =45^{\circ}
\end{aligned}
$
We get,
$\mathrm{A}=45^{\circ}$ and $\mathrm{B}=15^{\circ}$
Question 6.
Express the ratios $\cos \mathrm{A}, \tan \mathrm{A}$ and $\sec \mathrm{A}$ in terms of $\sin \mathrm{A}$.
Solution:
Since
$\cos ^2 \mathrm{~A}+\sin ^2 \mathrm{~A}=1$, therefore,
$\cos ^2 \mathrm{~A}=1-\sin ^2 \mathrm{~A}$
i.e., $\cos \mathrm{A}=\pm \sqrt{1-\sin ^2 A}$
This gives $\cos \mathrm{A}=\sqrt{1-\sin ^2 A}$
Hence, $\quad \tan \mathrm{A}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}=\frac{\sin \mathrm{A}}{\sqrt{1-\sin ^2 \mathrm{~A}}}$
and $\quad \sec \mathrm{A}=\frac{1}{\cos \mathrm{A}}=\frac{1}{\sqrt{1-\sin ^2 \mathrm{~A}}}$
Question 7.
Evaluate $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
Solution:
We know:
$\cot \mathrm{A}=\tan \left(90^{\circ}-\mathrm{A}\right)$

So,
$
\begin{aligned}
\cot 25^{\circ} & =\tan \left(90^{\circ}-25^{\circ}\right)=\tan 65^{\circ} \\
\frac{\tan 65^{\circ}}{\cot 25^{\circ}} & =\frac{\tan 65^{\circ}}{\tan 65^{\circ}}=1
\end{aligned}
$
Question 8.
Since $\sin 3 \mathrm{~A}=\cos \left(\mathrm{A}-26^{\circ}\right)$, where $3 \mathrm{~A}$ is an acute angle, find the value at $\mathrm{A}$.
Solution:
We are given that $\sin 3 \mathrm{~A}=\cos \left(\mathrm{A}-26^{\circ}\right) \ldots(1)$
Since $\sin 3 \mathrm{~A}=\cos \left(90^{\circ}-3 \mathrm{~A}\right)$ we can write $(1)$ as $\cos \left(90^{\circ}-3 \mathrm{~A}\right)=\cos \left(\mathrm{A}-26^{\circ}\right)$
Since $90^{\circ}-3 \mathrm{~A}$ and $\mathrm{A}-26^{\circ}$ are both acute angles.
$
90^{\circ}-3 \mathrm{~A}=\mathrm{A}-26^{\circ}
$
which gives $\mathrm{A}=29^{\circ}$
Question 9.
Express $\cot 85^{\circ}+\cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$.
Solution:
$
\begin{aligned}
& \cot 85^{\circ}+\cos 75^{\circ} \\
& =\cot \left(90^{\circ}-5^{\circ}\right)+\cos \left(90^{\circ}-15^{\circ}\right) \\
& =\tan 5^{\circ}+\sin 15^{\circ}
\end{aligned}
$
Question 10.
From a point on a bridge across a river, the angles of depression of the banks on opposite sides at the river are $30^{\circ}$ and $45^{\circ}$, respectively. If the bridge is at a height at $3 \mathrm{~m}$ from the banks, find the width at the river.
Solution:
$\mathrm{A}$ and $\mathrm{B}$ represent points on the bank on opposite sides at the river, so that $\mathrm{AB}$ is the width of the river. $\mathrm{P}$ is a point on the bridge at a height of $3 \mathrm{~m}$ i.e., $\mathrm{DP}=3 \mathrm{~m}$. We are interested to determine the width at the river which is the length at the side $\mathrm{AB}$ of the $\triangle \mathrm{APB}$.

Now, $\mathrm{AB}=\mathrm{AD}+\mathrm{DB}$
In right $\triangle \mathrm{APD}$,
$
\angle \mathrm{A}=30^{\circ}
$
So, $\quad \tan 30^{\circ}=\frac{\mathrm{PD}}{\mathrm{AD}}$
i.e., $\quad \frac{1}{\sqrt{3}}=\frac{3}{\mathrm{AD}}$ (or) $\mathrm{AD}=3 \sqrt{3} \mathrm{~m}$
Also, in right $\triangle \mathrm{PBD}$,
$\mathrm{B}=45^{\circ}$
So, $\mathrm{BD}=\mathrm{PD}=3 \mathrm{~m}$
Now, $\mathrm{AB}=\mathrm{BD}+\mathrm{AD}$
$=3+3 \sqrt{3}=3(1+\sqrt{3}) \mathrm{m}$
Therefore, the width at the river is $3(\sqrt{3}+1)$