Given below are the Class 10 Maths trigonometry worksheet with answers.
a. Match the following
b. True and False
c. Fill in the blanks
d. Short Answer Type
e. Long Answer Type
Question 2. 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 3. 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 $sin^2 A + cos^2 A =1$
$=sin^2 A - cos^2 A$
$=sin^2 A - (1- sin^2 A)$
$=(2sin^2 A - 1 )$
Fill in the blanks
Question 4
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
a. 6
b. 0
c. undefined
d. 90
e. 3
f. 1
g. 1
h. 1
Short Answer type
Question 5. 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 6. 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 7. 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 8. 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 9. 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 10. 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 11. The value of $(sin30 ^{\circ} + cos30^{\circ}) - (sin60^{\circ} + cos60^{\circ})$ is Solution
Now as $cos 90^{\circ}=0$
$cos 1^{\circ}.cos 2^{\circ} .cos3^{\circ} .... cos 180^{\circ}=0$
Long Answer Type
Question 18
Prove the following identity
$(sin A - cosec A)^2 + (cos A - sec A)^2 = tan^2 A+cot^2 A-1$ Question 19
Prove the following identity
(i) $cos^6 A + sin^6 A = 1 - 3 sin^2 A Cos^2 A$
(ii)$\frac {1}{cot A + tan A} = sin A cos A$ Question 20
Prove the following
(i) $(sin A + cos A)(cot A + tan A)=sec A + cosec A$
(ii)$sin 30 cos 60 + sin 60 cos 30=1$
Summary
This Class 10 trigonometry worksheet with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.You can download in PDF form also using the below links
