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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) $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$
Answer
(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.
Answer
$\tan 2A = \cot (A - 18^{\circ})$
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 $A + B = 90^{\circ}$.
Answer
$\tan A = \cot B$
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.
Answer
$\sec 4A = \csc (A - 20^{\circ})$
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
$\sin (\frac {B+C}{2}) = \cos \frac {A}{2}$
Answer
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°.
Answer
$\sin 67^{\circ} + \cos 75^{\circ}$
$= \sin (90^{\circ} - 23^{\circ}) + \cos (90^{\circ} - 15^{\circ})$
$= \cos 23^{\circ} + \sin 15^{\circ}$

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