NCERT Solutions for Exercise 8.1 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Q2 In Fig. 8.13, find $\tan P-\cot R$ .

Answer:

We have, $\mathrm{\Delta }$ PQR is a right-angled triangle, length of PQ and PR are 12 cm and 13 cm respectively.
So, by using Pythagoras theorem,
$QR=\sqrt{{13}^2-{12}^2}$
$QR=\sqrt{169-144}$
$QR=\sqrt{25}=5cm$

Now, According to question,
$\tan P-\cot R$ = $\frac{RQ}{QP}-\frac{QR}{PQ}$
= 5/12 - 5/12 = 0

Q3 If $\sin A=\frac{3}{4},$ calculate $\cos A$ and $\tan A$ .

Answer:

Suppose $\mathrm{\Delta }$ ABC is a right-angled triangle in which $\mathrm{\angle }B={90}^0$ and we have $\sin A=\frac{3}{4},$
So,

Let the length of AB be 4 unit and the length of BC = 3 unit So, by using Pythagoras theorem,
$AB=\sqrt{16-9}=\sqrt{7}$ units
Therefore,
$\cos A=\frac{AB}{AC}=\frac{\sqrt{7}}{4}$ and $\tan A=\frac{BC}{AB}=\frac{3}{\sqrt{7}}$

Q4 Given $15\cot A=8,$ find $\sin A$ and $\sec A$ .

Answer:

We have,
$15\cot A=8,$ $\Rightarrow \cot A=8/15$
It implies that In the triangle ABC in which $\mathrm{\angle }B={90}^0$ . The length of AB be 8 units and the length of BC = 15 units

Now, by using Pythagoras theorem,
$AC=\sqrt{64+225}=\sqrt{289}$
$\Rightarrow AC=17$ units

So, $\sin A=\frac{BC}{AC}=\frac{15}{17}$
and $\sec A=\frac{AC}{AB}=\frac{17}{8}$

Q5 Given $\sec \theta =\frac{13}{12},$ calculate all other trigonometric ratios.

Answer:

We have,
$\sec \theta =\frac{13}{12},$

It means the Hypotenuse of the triangle is 13 units and the base is 12 units.
Let ABC is a right-angled triangle in which $\mathrm{\angle }$ B is 90 and AB is the base, BC is perpendicular height and AC is the hypotenuse.

By using Pythagoras theorem,
$BC=\sqrt{169-144}=\sqrt{25}$
BC = 5 unit

Therefore,
$\sin \theta =\frac{BC}{AC}=\frac{5}{13}$
$\cos \theta =\frac{BA}{AC}=\frac{12}{13}$

$\tan \theta =\frac{BC}{AB}=\frac{5}{12}$

$\cot \theta =\frac{BA}{BC}=\frac{12}{5}$

$\sec \theta =\frac{AC}{AB}=\frac{13}{12}$

$\csc \theta =\frac{AC}{BC}=\frac{13}{5}$

Q6 If $\mathrm{\angle }A$ and $\mathrm{\angle }B$ are acute angles such that $\cos A=\cos B$ , then show that $\mathrm{\angle }A=\mathrm{\angle }B$ .

Answer:

We have, A and B are two acute angles of triangle ABC and $\cos A=\cos B$

According to question, In triangle ABC,
$\cos A=\cos B$

$\frac{AC}{AB}=\frac{BC}{AB}$
$\Rightarrow AC=AB$
Therefore, $\mathrm{\angle }$ A = $\mathrm{\angle }$ B [angle opposite to equal sides are equal]

Q7 If $\cot \theta =\frac{7}{8},$ evaluate: $(i)\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$ $(ii){\cot }^2\theta$

Answer:

Given that,
$\cot \theta =\frac{7}{8}$
$\therefore$ perpendicular (AB) = 8 units and Base (AB) = 7 units
Draw a right-angled triangle ABC in which $\mathrm{\angle }B={90}^0$
Now, By using Pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
$AC=\sqrt{64+49}=\sqrt{113}$

So, $\sin \theta =\frac{BC}{AC}=\frac{8}{\sqrt{113}}$
and $\cos \theta =\frac{AB}{AC}=\frac{7}{\sqrt{113}}$

$\Rightarrow \cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{7}{8}$
$(i)\frac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}$
$\Rightarrow \frac{(1-{\sin }^2\theta )}{(1-{\cos }^2\theta )}=\frac{{\cos }^2\theta }{{\sin }^2\theta }={\cot }^2\theta$
$=(\frac{7}{8}{)}^2=\frac{49}{64}$

$(ii){\cot }^2\theta$
$=(\frac{7}{8}{)}^2=\frac{49}{64}$

Q8 If $3\cot A=4,$ check wether $\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$ or not.

Answer:

Given that,
$3\cot A=4,$
$\Rightarrow \cot =\frac{4}{3}=\frac{base}{perp.}$
ABC is a right-angled triangle in which $\mathrm{\angle }B={90}^0$ and the length of the base AB is 4 units and length of perpendicular is 3 units

By using Pythagoras theorem, In triangle ABC,
$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ AC=\sqrt{16+9} \\ AC=\sqrt{25} \end{gathered}$$
AC = 5 units

So,
$\tan A=\frac{BC}{AB}=\frac{3}{4}$
$\cos A=\frac{AB}{AC}=\frac{4}{5}$
$\sin A=\frac{BC}{AC}=\frac{3}{5}$
$\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}={\cos }^{2}A-{\sin }^{2}A$
Put the values of above trigonometric ratios, we get;
$\Rightarrow \frac{1-9/4}{1+9/4}=\frac{16}{25}-\frac{9}{25}$
$\Rightarrow -\frac{5}{13}\ne \frac{7}{25}$
LHS $\ne$ RHS

Q9 In triangle $ABC$ , right-angled at $B$ , if $\tan A=\frac{1}{\sqrt{3}},$ find the value of:

$(i)\sin A\cos C+\cos A\sin C$
$(ii)\cos A\cos C+\sin A\sin C$

Answer:

Given a triangle ABC, right-angled at B and $\tan A=\frac{1}{\sqrt{3}}$ $\Rightarrow A={30}^0$

According to question, $\tan A=\frac{1}{\sqrt{3}}=\frac{BC}{AB}$
By using Pythagoras theorem,
$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ AC=\sqrt{1+3}=\sqrt{4} \end{gathered}$$
AC = 2
Now,
$$\begin{gathered} \sin A=\frac{BC}{AC}=\frac{1}{2} \\ \sin C=\frac{AB}{AC}=\frac{\sqrt{3}}{2} \\ \cos A=\frac{AB}{AC}=\frac{\sqrt{3}}{2} \\ \cos C=\frac{BC}{AC}=\frac{1}{2} \end{gathered}$$

Therefore,

$(i)\sin A\cos C+\cos A\sin C$
$$\begin{gathered} \Rightarrow \frac{1}{2}\times \frac{1}{2}+\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2} \\ \Rightarrow 1/4+3/4 \\ \Rightarrow 4/4=1 \end{gathered}$$

$(ii)\cos A\cos C+\sin A\sin C$
$$\begin{gathered} \Rightarrow \frac{\sqrt{3}}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{\sqrt{3}}{2} \\ \Rightarrow \frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\ \Rightarrow \frac{\sqrt{3}}{2} \end{gathered}$$

Q10 In $\mathrm{\Delta }PQR$ , right-angled at $Q$ , $PR+QR=25cm$ and $PQ=5cm$ . Determine the values of $\sin P,\cos Pand\tan P.$

Answer:

We have, PR + QR = 25 cm.............(i)
PQ = 5 cm
and $\mathrm{\angle }Q={90}^0$
According to question,
In triangle $\mathrm{\Delta }$ PQR,
By using Pythagoras theorem,
$$\begin{gathered} P{R}^2=P{Q}^2+Q{R}^2 \\ P{Q}^2=P{R}^2-Q{R}^2 \\ {5}^2=(PR-QR)(PR+QR) \\ 25=25(PR-QR) \end{gathered}$$
PR - QR = 1........(ii)

From equation(i) and equation(ii), we get;
PR = 13 cm and QR = 12 cm.

therefore,
$$\begin{gathered} \sin P=\frac{QR}{PR}=12/13 \\ \cos P=\frac{PQ}{RP}=5/13 \\ \therefore \tan P=\frac{\sin P}{\cos P}=12/5 \end{gathered}$$

Q11 State whether the following are true or false. Justify your answer.

(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A=\frac{12}{5}$ for some value of angle A.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.
(iv) $\cot A$ is the product of cot and A.
(v) $\sin \mathrm{\Theta }=\frac{4}{3}$ for some angle $\mathrm{\Theta }.$

Answer:

(i) False,
because $\tan 60=\sqrt{3}$ , which is greater than 1

(ii) True,
because $\sec A\ge 1$

(iii) False,
Because $\cos A$ abbreviation is used for cosine A.

(iv) False,
because the term $\cot A$ is a single term, not a product.

(v) False,
because $\sin \theta$ lies between (-1 to +1) [ $-1\le \sin \theta \le 1$ ]