NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Q1 Evaluate :

$(ii) \frac{\tan 26^{o}}{\cot 64^{o}}$

Answer:

$\frac{\tan 26^{o}}{\cot 64^{o}}$
The above equation can be written as ;

$\tan (90^o-64^o)/\cot 64^o$ .........(i)
It is known that, $\tan (90^o-\theta) = \cot \theta$
Therefore, equation (i) becomes,
$\cot64^o/\cot 64^o = 1$

So, the answer is 1.

Q1 Evaluate :

$(iii) \cos 48^{o}-\sin 42^{o}$

Answer:

$\cos 48^{o}-\sin 42^{o}$
The above equation can be written as ;
$\cos (90^o-42^{o})-\sin 42^{o}$ ....................(i)
It is known that $\cos (90^o-\theta) = \sin \theta$
Therefore, equation (i) becomes,
$\sin42^{o}-\sin 42^{o} = 0$

So, the answer is 0.

Q1 Evaluate :

$(iv) cosec \: 31^{o}-\sec 59^{o}$

Answer:

$cosec \: 31^{o}-\sec 59^{o}$

This equation can be written as;
$cosec 31^o - \sec(90^o-31^o)$ .................(i)
We know that $\sec(90^o-\theta) = cosec \theta$

Therefore, equation (i) becomes;
$cosec 31^o - cosec\ 31^o$ = 0

So, the answer is 0.

Q2 Show that :

$(i) \tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$

Answer:

$\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}= 1$
Taking Left Hand Side (LHS)
= $\tan 48^{o}\tan 23^{o}\tan 42^{o}\tan 67^{o}$
$\Rightarrow \tan 48^{o}\tan 23^{o}\tan (90^o-48^{o})\tan (90^o-23^{o})$
$\Rightarrow \tan 48^{o}\tan 23^{o}\cot 48^{o}\cot23^{o}$ [it is known that $\tan (90^0-\theta = \cot\theta)$ and $\cot\theta\times \tan \theta =1$
$=1$

Hence proved.

Q2 Show that :

$(ii) \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0$

Answer:

$\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0$

Taking Left Hand Side (LHS)
= $\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}$
= $\cos 38^{o}\cos (90^o-38^{o})-\sin 38^{o}\sin (90^o-38^{o})$
= $\cos 38^{o}\sin38^{o}-\sin 38^{o}\cos 38^{o}$ [it is known that $\sin(90^0-\theta) =\cos \theta$ and $\cos(90^0-\theta) =\sin \theta$ ]
= 0

Q3 If $\tan 2A= \cot (A-18^{o})$ , where $2A$ is an acute angle, find the value of $A$ .

Answer:

We have,
$\tan$ 2A = $\cot$ (A - $\18^{0}$ )
we know that, $\\\cot (90^0-\theta) = \tan\theta$
$\\\Rightarrow \cot (90^0-2A) = \cot (A-18^0)\\\Rightarrow 90^0-2A = A - 18^0\\\Rightarrow 3A = 108\\\Rightarrow A = 108/3 = 36^0$

Q4 If $\tan A= \cot B$ , prove that $A+B= 90^{o}$ .

Answer:

We have,
$\tan A= \cot B$
and we know that $\tan (90^0 - \theta)= \cot \theta$
therefore,
$\tan A= \tan(90^0- B)$
A = 90 - B
A + B = 90
Hence proved.

Q5 If $\sec 4A= cosec (A-20^{o})$ , where $4A$ is an acute angle, find the value of $A$ .

Answer:

We have,
$\sec 4A= cosec (A-20^{o})$ , Here 4A is an acute angle
According to question,
We know that $cosec(90^0-\theta)= \sec \theta$
$cosec(90^0-4A)= cosec (A-20^{o})$

$\\\Rightarrow 90 - 4A = A - 20\\\\ \Rightarrow 5A=110\\\\ \Rightarrow A=\frac{110}{5}\\\:\\ \Rightarrow A=22^o$

Q6 If $A,B$ and $C$ are interior angles of a triangle $ABC$ , then show that

$\sin (\frac{B+C}{2})= \cos \frac{A}{2}$

Answer:

Given that,
A, B and C are interior angles of $\Delta ABC$
To prove - $\sin (\frac{B+C}{2})= \cos \frac{A}{2}$

Now,
In triangle $\Delta ABC$ ,
A + B + C = $180^0$
$\Rightarrow B + C = 180 - A$
$\Rightarrow B + C/2 = 90^0 - A/2$
$\sin \frac{B+C}{2}=\sin (90^0-A/2)$
$\sin \frac{B+C}{2}=\cos A/2$
Hence proved.

Q7 Express $sin 67^{o}+\cos 75^{o}$ in terms of trigonometric ratios of angles between $0^{o}$ and $45^{o}$ .

Answer:

By using the identity of $\sin\theta$ and $\cos\theta$
$sin 67^{o}+\cos 75^{o}$
We know that,
$\sin(90-\theta) =\cos \theta$ and $\cos(90-\theta) =\sin \theta$
the above equation can be written as;
$=\sin (90^0-23^0)+\cos(90^0-15^0)$
$=\sin (15^0)+\cos(23^0)$