Question 1. Write the values cos 0°, cos 45°, cos 60° and cos 90°. What happens to the values of cos as angle increases from 0° to 90°? Solution
$\cos 0^{\circ}= 1$
$\cos 45^{\circ}= \frac {1}{\sqrt {2}}$
$\cos 60^{\circ}= \frac {1}{2}$
$\cos 90^{\circ}= 0$
We can see that values of cos decreases as angle increases from 0° to 90°
Question 2. Write the values of sin 0°, sin 30°, sin 45°, sin 60° and sin 90°. What happens to the values of sin as angle increases from 0° to 90°? Solution
$\sin 0^{\circ}= 0$
$\sin 30^{\circ}= \frac {1}{2}$
$\sin 45^{\circ}= \frac {1}{\sqrt {2}}$
$\sin 60^{\circ}= \frac {\sqrt {3}}{2}$
$\sin 90^{\circ}= 1$
We can see that values of sin increases as angle increases from 0° to 90°
Question 3. Write the values of tan 0°,tan 30°, tan 45°, tan 60° and tan 90°. What happens to the values of tan as angle increases from 0° to 90°? Solution
$\tan 0^{\circ}= 0$
$\tan 30^{\circ}= \frac {1}{\sqrt {3}}$
$\tan 45^{\circ}= 1$
$\tan 60^{\circ}= \sqrt {3}$
$\tan 90^{\circ}= undefined$
We can see that values of tan increases as angle increases from 0° to 90°
Question 4. If $\sin A =\frac {3}{5}$ .find cos A and tan A. Solution
Given $\sin A= \frac{3}{5}$
Or $\frac {P}{H}=\frac{3}{5}$
Let P=3k and H=5k
Now By Pythagoras theorem
$P^2 + B^2 =H^2$
$9k^2 + B^2 =25k^2$
Or $ B =4k$
Now $\Cos A = {B \over H} = \frac {4}{5}$
$\tan A = {{\sin A} \over {\cos A}} = \frac {3}{4}$
Question 5. If $cosec A = \sqrt {10}$. find other five trigonometric ratios. Solution
Given $cosec A = \sqrt {10}$
Or $\frac {H}{P}=\sqrt {10}$
Let $H=\sqrt {10} k$ and P=k
Now By Pythagoras theorem
$P^2 + B^2 =H^2$
$k^2 + B^2 = 10k^2$
B= 3k
Now $sinA = {P \over H} = \frac {1}{\sqrt {10}}$
$cosA = {B \over H} = \frac {3}{\sqrt {10}}$
$tan A = {{SinA} \over {CosA}} = \frac {1}{3}$
$sec A = \frac {1}{cos A} = \frac {\sqrt {10}}{3}$
$cot A = {{cosA} \over {sinA}} = 3$
Question 6. In a right triangle ABC right angled at B if $sin A =\frac {3}{5}$ . find all the six trigonometric ratios of C. Solution
Given $sin A= \frac{3}{5}$
Or $\frac {P}{H}=\frac{3}{5}$
Let P=3k and H=5k
Now By Pythagoras theorem
$P^2 + B^2 =H^2$
$9k^2 + B^2 =25k^2$
Or $ B =4k$
Now $CosA = {B \over H} = \frac {4}{5}$
$tan A = {{SinA} \over {CosA}} = \frac {3}{4}$
$cot A = {{cosA} \over {sinA}} = \frac {4}{3}$
$cosec A = {{1} \over {sinA}} = \frac {5}{3}$
$sec A = {{1} \over {cos A}} = \frac {5}{4}$
Now A + B + C=180
C=90 -A
$ sin C = sin (90 -A) = cos A = \frac {4}{5}$
$ cos C = cos (90 -A) = sin A = \frac {3}{5}$
$ tan C = tan (90 -A) = cot A = \frac {4}{3}$
$ cosec C = cosec (90 -A) = sec A = \frac {5}{4}$
$ sec C = sec (90 -A) = cosec A = \frac {5}{3}$
$ tan C = tan (90 -A) = cot A = \frac {4}{3}$
Question 7. The value of $(sin30 ^{\circ} + cos30^{\circ}) - (sin60^{\circ} + cos60^{\circ})$ is Solution
Question 8. True and False statement
a. The value of $sin \theta + cos \theta$ is always greater than 1
b. $tan \theta$ increases faster than $sin \theta$ as θ increase
c. The value of the expression $(cos^2 23^{\circ} - sin^2 67^{\circ})$ is positive.
d. The value of the expression $(sin 80^{\circ} - cos80^{\circ})$ is negative.
e. If $cosA + cos^2 A = 1$, then $sin^2 A + sin^4 A = 1$.
f. $(tan \theta + 2) (2 tan \theta + 1) = 5 tan \theta + sec ^2 \theta$.
g. If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.
h. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.
i. If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains unchanged
j. $cos^4 A - cos^2 A = sin^4 A - sin^2 A$
k. $cot^4 A - 1 = cosec^4 A - 2cosec^2 A$
l. $sin^4 A + cos^4 A = 1 - 2 sin^2 A cos^2 A$
m. $sin^4 A - cos^4 A = sin^2 A - cos^2 A = 2sin^2 A - 1 = 1 - 2 cos^2 A$ Solution
a. false
$sin 90^{\circ} + cos 90^{\circ}= 1+ 0 =1$
b. true
We know that $tan \theta = \frac {sin \theta}{cos \theta}$
Now as θ increase, sin θ increase and coθ decrease. So numerator increases and denominator decreases
In case of $ sin \theta = \frac {sin \theta}{1}$
Now as θ increase, sin θ increase and denominator remains constant at 1
Hence $tan \theta$ increases faster than $sin \theta$ as θ increase
The value of (sin 80° - cos80°) will be positive. The value of sin80° will be very close to that of sin90° (which is 1) and similarly, the value of cos80° will be very close to that of cos90° (which is zero). Clearly, the result will be positive decimal number and should be fairly close to 0.85.
e) true
$cos A+cos^2A=1$
$cosA = 1 - cos^2A = sin^2A$
Now
$sin^2A + sin^4A$
$= Cos A+Cos^2A$
=1
f) false
g) false
h) false
i) true
j) true
$cos^4 A - cos^2 A = sin^4 A - sin^2 A$
LHS
\[cos^4 A - cos^2 A\\
=cos^2 A(cos^2 A -1)\\
=-sin^2 A cos^2A\\
\]
RHS
\[sin^4 A - sin^2 A\\
=sin^2 A(sin^2 A-1)\\
=-sin^2 A cos^2A\\
\]
k) true
LHS
$=cot^4A-1$
$=(cot^2A)^2-1^2$
$=(cot^2A+1)(cot^2A-1)$
Now $1+cot^2A=cosec^2A$ which implies $cot^2A=cosec^2A-1$
$=cosec^2A(cosec^2A-1-1)$
$=cosec^2A(cosec^2A-2)$
$=cosec^4A-2cosec^2A$
=RHS
l) true
LHS
$sin^4 A + cos^4 A 2sin^2 A cos^2 A$
$=sin^4 A + cos^4 A + 2sin^2 A cos^2 A - 2sin^2 A cos^2 A$
$=(sin^2 A + cos^2 A)^2 - 2sin^2 A cos^2 A$
$=1- 2sin^2 A cos^2 A$
m) true
$sin^4 A - cos^4 A$
=$(sin^2 A + cos^2A)(sin^2 A - cos^2 A)$
Now as $sin^2 A + cos^2 A =1$
$=sin^2 A - cos^2 A$
$=sin^2 A - (1- sin^2 A)$
$=(2sin^2 A - 1 )$
Question 9. If $sin B =\frac {1}{2}$ , show that $3 cos B - 4 cos^3 B = 0$ Solution
$sin B =\frac {1}{2}$
So $\angle B =30 ^{\circ}$
$cos B = cos 30 =\frac {\sqrt {3}}{2}$
Now
$3 cos B - 4 cos^3 B$
$=3 \frac {\sqrt {3}}{2} - 4 (\frac {\sqrt {3}}{2})^3$
$=\frac {3\sqrt {3}}{2} -\frac {3\sqrt {3}}{2}$
=0
Question 10. If $tanA + \frac {1}{tanA}= 2$, find the value of $tan^2 A + \frac {1}{tan^2 A}$ Solution
$tanA + \frac {1}{tanA}= 2$
Squaring both the sides
$tan^2 A + \frac {1}{tan^2 A} + 2 = 4$
$tan^2 A + \frac {1}{tan^2 A}=2$
Question 16.
True and False
a. $cos A =\frac {4}{3}$ for some angle A.
b. $tan A = \frac {sin A}{cos A}$
c. $secA =\frac {1}{cosA}$ , for an acute angle
d. $sin 60^{\circ} = 2 sin 30^{\circ}$
e. $Sin A + Cos A =1$
f. $1 + cot^2 A =cosec^2 A$
g. $sin A \times cot A= cos A$
h. $cosec 50^{\circ} = sec 40^{\circ}$ Solution
a. false
b. true
c. true
d. False
e. false
f. true
g. true
h. true
Question 17
Fill in the blanks:
a. $5 cos 0^{\circ} + sin 90^{\circ}$ = __________
b. $tan 0^{\circ} =$ _____________
c. $tan 90^{\circ}$ is _____________
d. If $sin A= 1$, then A = ________________
e. $2 sin^2 60^{\circ}$ = ___________
f. $2 cos^2 45^{\circ}$ = __________________
g. $sin^2 A + cos^2 A=$ _______________
h. $(1 + tan^2 A ) (1 + sinA ) (1 - sin A) =$ ________. Solution
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