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

## NCERT Solutions for Class 10 Maths Introduction to Trigonometry Exercise 8.3

In this page we have NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry for EXERCISE 8.3 . Hope you like them and do not forget to like , social_share and comment at the end of the page.
Formula to Remember
$\sin (90^{\circ} -A) = \cos A$
$\cos (90^{\circ} -A) = \sin A$
$\sec (90^{\circ} -A) = \csc A$
$\csc (90^{\circ} -A) = \sec A$
$\tan (90^{\circ} -A) = \cot A$
$\cot (90^{\circ} -A) = \tan A$

Question 1
Evaluate :
(i) $\frac {\sin 18^{\circ}}{\cos 72^{\circ}}$ (ii) $\frac {\tan 26^{\circ}}{\cot 64^{\circ}}$ (iii)$\cos 48^{\circ} - \sin 42^{\circ}$ (iv)$\csc 31^{\circ} - \sec 59^{\circ}$
Solution
(i) $\frac {\sin 18^{\circ}}{\cos 72^{\circ}}$ $= \frac {\sin (90^{\circ} - 18^{\circ})}{\cos 72^{\circ}}$
Now As $\sin (90 -A) = \cos A$ $=\frac {\cos 72^{\circ}}{\cos 72^{\circ}} = 1$

(ii) $\frac {\tan 26^{\circ}}{\cot 64^{\circ}}$
$= \frac {\tan (90^{\circ} - 36^{\circ})}{\cot 64^{\circ}}$
Now As $\tan (90 -A) = \cot A$ $= \frac {\cot 64^{\circ}}{\cot 64^{\circ}} = 1$

(iii)$\cos 48^{\circ} - \sin 42^{\circ}$
$= \cos (90^{\circ} - 42^{\circ}) - \sin 42^{\circ}$
Now As $\cos (90 -A) = \sin A$ $= \sin 42^{\circ} - \sin 42^{\circ}= 0$

(iv) $\csc 31^{\circ} - \sec 59^{\circ}$
$= \csc (90^{\circ} - 59^{\circ}) - \sec 59^{\circ}$
$= \sec 59^{\circ} - \sec 59^{\circ}= 0$

Question 2
Show that :
(i) $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
(ii) $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
(i) LHS
$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$= \tan (90^{\circ} - 42^{\circ}) \tan (90^{\circ} - 67^{\circ}) \tan 42^{\circ} \tan 67^{\circ}$
As $\tan (90 -A) = \cot A$
$= \cot 42^{\circ} \cot 67^{\circ} \tan 42^{\circ} \tan 67^{\circ}$
$= (\cot 42^{\circ} \tan 42^{\circ}) (\cot 67^{\circ} \tan 67^{\circ})$ Now $\cot A \tan A = \frac {1}{\tan A} \tan A=1$
Therefore
$= 1=RHS$

(ii) LHS
$\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ}$
$= \cos (90^{\circ} - 52^{\circ}) \cos (90^{\circ}- 38^{\circ}) - \sin 38^{\circ} sin 52^{\circ}$
Now As $\cos (90 -A) = \sin A$
$= \sin 52^{\circ} \sin 38^{\circ} - \sin 38^{\circ} \sin 52^{\circ}= 0=RHS$

Question 3
If $\tan 2A = \cot (A - 18^{\circ})$, where 2A is an acute angle, find the value of A.
$\tan 2A = \cot (A - 18^{\circ})$
As $\tan A = \cot (90 -A)$
$\cot (90^{\circ} - 2A) = cot (A -18^{\circ})$
Equating angles,
$90^{\circ} - 2A = A- 18^{\circ}$
or $A = 36^{\circ}$

Question 4
If $\tan A = \cot B$, prove that $A + B = 90^{\circ}$.
$\tan A = \cot B$
As $\cot A = \tan (90 -A)$
$\tan A = \tan (90 - B)$
Equating Angles
$A = 90^{\circ} - B$
$A + B = 90^{\circ}$

Question 5
If $\sec 4A = \csc (A - 20^{\circ})$, where 4A is an acute angle, find the value of A.
$\sec 4A = \csc (A - 20^{\circ})$
Now As 4A is an acute angle, $\sec 4A= \csc (90 -4A)$ Therefore $\csc (90 - 4A) = \csc (A - 20)$
Equating angles,
$90 - 4A= A- 20$
or $A = 22^{\circ}$

Question 6
If A, B and C are interior angles of a triangle ABC, then show that
$\sin (\frac {B+C}{2}) = \cos \frac {A}{2}$
In a triangle, sum of all the interior angles
$A + B+ C = 180$
$B+ C = 180 - A$
Dividing by 2 on both the sides $\frac {B+C}{2} = \frac {(180-A)}{2}$
$\frac {B+C}{2} = (90-\frac {A}{2})$
So, LHS $\sin (\frac {B+C}{2}) = \sin (90 -\frac {A}{2})$
or $\sin (\frac {B+C}{2}) = \cos \frac {A}{2}$

Question 7
Express $\sin 67^{\circ} + \cos 75^{\circ}$ in terms of trigonometric ratios of angles between 0° and 45°.
$\sin 67^{\circ} + \cos 75^{\circ}$
$= \sin (90^{\circ} - 23^{\circ}) + \cos (90^{\circ} - 15^{\circ})$
$= \cos 23^{\circ} + \sin 15^{\circ}$

• Notes
• Assignments
• Revison Notes
• NCERT solution

Go back to Class 10 Main Page using below links

### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20

Latest Articles
Synthetic Fibres and Plastics Class 8 Practice questions

Class 8 science chapter 5 extra questions and Answers

Mass Calculator

3 Fraction calculator

Garbage in Garbage out Extra Questions7