NCERT Solutions for Exercise 8.2 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Q 1: Evaluate the following :

$(ii)2{\tan }^2{45}^o+2{\cos }^2{30}^o-2{\sin }^2{60}^o$

Answer:

We know the value of

$\tan {45}^0=1$ and

$\cos {30}^0=\sin {60}^0=\frac{\sqrt{3}}{2}$
According to question,

$=2{\tan }^2{45}^o+2{\cos }^2{30}^o-2{\sin }^2{60}^o$
$$\begin{gathered} =2(1{)}^2+(\frac{\sqrt{3}}{2}{)}^2-(\frac{\sqrt{3}}{2}) \\ =2 \end{gathered}$$

Q 1: Evaluate the following :

$(iii)\frac{\cos {45}^o}{\sec {30}^o+\csc {30}^o}$

Answer:

$\frac{\cos {45}^o}{\sec {30}^o+\csc {30}^o}$
we know the value of

$\cos 45=1/\sqrt{2}$ , $\sec {30}^0=2/\sqrt{3}$ and $cosec30=2$ ,

After putting these values
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+2}$
$=\frac{1/\sqrt{3}}{(2+2\sqrt{3})/\sqrt{3}}$
$=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}\times \frac{2\sqrt{2}-2\sqrt{6}}{2\sqrt{2}-2\sqrt{6}}$
$$\begin{gathered} =\frac{2\sqrt{6}-2\sqrt{18}}{-16} \\ =2\frac{\sqrt{6}-3\sqrt{3}}{-16}=\frac{3\sqrt{3}-\sqrt{6}}{8} \end{gathered}$$

Q 1: Evaluate the following :

$(iv)\frac{\sin {30}^o+\tan {45}^o-cosec{60}^o}{\sec {30}^o+\cos {60}^o+\cot {45}^o}$

Answer:

$\frac{\sin {30}^o+\tan {45}^o-cosec{60}^o}{\sec {30}^o+\cos {60}^o+\cot {45}^o}$ ..................(i)
It is known that the values of the given trigonometric functions,
$$\begin{gathered} \sin {30}^0=1/2=cos{60}^0 \\ \tan {45}^0=1=\cot {45}^0 \\ \sec {30}^0=2/\sqrt{3}=cosec{60}^0 \end{gathered}$$
Put all these values in equation (i), we get;
$$\begin{gathered} \Rightarrow \frac{1/2+1-2/\sqrt{3}}{2/\sqrt{3}+1/2+1} \\ \Rightarrow \frac{3/2-2/\sqrt{3}}{3/2+2/\sqrt{3}} \\ \Rightarrow \frac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \frac{4-3\sqrt{3}}{4-3\sqrt{3}} \\ \Rightarrow \frac{12\sqrt{3}-27-16+12\sqrt{3}}{-11} \\ \Rightarrow \frac{43-24\sqrt{3}}{11} \end{gathered}$$

Q 1: Evaluate the following :

$(v)\frac{5{\cos }^2{60}^o+4{\sec }^2{30}^o-{\tan }^2{45}^o}{{\sin }^2{30}^o+{\cos }^2{30}^o}$

Answer:

$\frac{5{\cos }^2{60}^o+4{\sec }^2{30}^o-{\tan }^2{45}^o}{{\sin }^2{30}^o+{\cos }^2{30}^o}$ .....................(i)
We know the values of-
$$\begin{gathered} \cos {60}^0=1/2=\sin {30}^0 \\ \sec {30}^0=2/\sqrt{3} \\ \tan {45}^0=1 \\ \cos {30}^0=\sqrt{3}/2 \end{gathered}$$
By substituting all these values in equation(i), we get;

$$\begin{gathered} \Rightarrow \frac{5.(\frac{1}{2}{)}^2+4.(\frac{2}{\sqrt{3}}{)}^2-1}{(\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2} \\ \Rightarrow \frac{5/4-1+16/3}{1} \\ \Rightarrow \frac{1/4+16/3}{1} \\ \Rightarrow \frac{67}{12} \end{gathered}$$

Q 2: Choose the correct option and justify your choice :

$(i)\frac{2\tan {30}^o}{1+{\tan }^2{30}^o}=$

$(A)\sin {60}^o$ $(B)\cos {60}^o$ $(C)\tan {60}^o$ $(D)\sin {30}^o$

Answer:

Put the value of tan 30 in the given question-
$\frac{2\tan {30}^o}{1+{\tan }^2{30}^o}=\frac{2\times 1/\sqrt{3}}{1+(1/\sqrt{3}{)}^2}=\frac{2}{\sqrt{3}}\times \frac{3}{4}=\sqrt{3}/2=\sin {60}^0$

The correct option is (A)

Q 2: Choose the correct option and justify your choice :

$(ii)\frac{1-{\tan }^2{45}^o}{1+{\tan }^2{45}^o}=$

$(A)\tan {90}^o$ $(B)1$ $(C)\sin {45}^o$ $(D)0$

Answer:

The correct option is (D)
$\frac{1-{\tan }^2{45}^o}{1+{\tan }^2{45}^o}=$
We know that $\tan 45=1$
So, $\frac{1-1}{1+1}=0$

Q 2: Choose the correct option and justify your choice :

$(iii)\sin 2A=2\sin A$ is true when $A$ =

$(A)0^o$ $(B){30}^o$ $(C){45}^o$ $(D){60}^o$

Answer:

The correct option is (A)
$\sin 2A=2\sin A$
We know that $\sin 2A=2\sin A\cos A$
So, $2\sin A\cos A=2\sin A$
$$\begin{gathered} \cos A=1 \\ A=0^0 \end{gathered}$$

Q 2: Choose the correct option and justify your choice :

$(iv)\frac{2\tan {30}^o}{1-ta{n}^2{30}^o}=$

$(A)\cos {60}^o$ $(B)\sin {60}^o$ $(C)\tan {60}^o$ $(D)\sin {30}^o$

Answer:

Put the value of $tan{30}^o=\text{1}/\sqrt{3}$

$$\begin{gathered} =\frac{2\tan {30}^o}{1-ta{n}^2{30}^o}=\frac{2\times 1/\sqrt{3}}{1-(\frac{1}{\sqrt{3}}{)}^2} \\ =\frac{2/\sqrt{3}}{1-1/3} \\ =\frac{2}{\sqrt{3}}\times \frac{3}{2} \\ =\sqrt{3}=\tan {60}^0 \end{gathered}$$

The correct option is (C)

Q 3: If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}};$ $0^o<A+B\le {90}^o;A>B,$ find $AandB.$

Answer:

Given that,
$\tan (A+B)=\sqrt{3}=\tan {60}^0$
So, $A+B={60}^o$ ..........(i)
$\tan (A-B)=1/\sqrt{3}=\tan {30}^0$
therefore, $A-B={30}^o$ .......(ii)
By solving the equation (i) and (ii) we get;

$A={45}^o$ and $B={15}^o$

Q 4: State whether the following are true or false. Justify your answer.

$(i)\sin (A+B)=\sin A+\sin B$
$(ii)$ The value of $\sin \theta$ increases as $\theta$ increases.
$(iii)$ The value of $\cos \theta$ increases as $\theta$ increases.
$(iv)\sin \theta =\cos \theta$ for all values of $\theta$ .
$(v)\cot A$ is not defined for $A=0^o$

Answer:

(i) False,
Let A = B = ${45}^0$
Then, $$\begin{gathered} \sin ({45}^0+{45}^0)=\sin {45}^0+\sin {45}^0 \\ \sin {90}^{=}1/\sqrt{2}+q/\sqrt{2} \\ 1\ne \sqrt{2} \end{gathered}$$

(ii) True,
Take $\theta =0^0,{30}^0,{45}^0$
whent
$\theta$ = 0 then zero(0),
$\theta$ = 30 then value of $\sin \theta$ is 1/2 = 0.5
$\theta$ = 45 then value of $\sin \theta$ is 0.707

(iii) False,
$\cos 0^0=1,\cos {30}^0=\sqrt{3}/2=0.87,\cos {45}^0=1\sqrt{2}=0.707$

(iv) False,
Let $\theta$ = 0
$$\begin{gathered} \sin 0^0=\cos 0^0 \\ 0\ne 1 \end{gathered}$$

(v) True,
$\cot 0^0=\frac{\cos 0^0}{\sin 0^0}=\frac{1}{0}$ (not defined)