Class 10 Maths trigonometry worksheet

Self explanatory

a. false
b. true
c. true
d. False
e. false
f. true
g. true
h. true

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

c. false
$(cos^2 23^{\circ} - sin^2 67^{\circ})$
$=(cos^2 23^{\circ} - sin^2 (90^{\circ} -23^{\circ}))$
$=(cos^2 23^{\circ} - cos^2 23^{\circ})$
=0

d. false

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 $sin^2 A + cos^2 A =1$
$=sin^2 A - cos^2 A$
$=sin^2 A - (1- sin^2 A)$
$=(2sin^2 A - 1 )$

a. 6
b. 0
c. undefined
d. 90
e. 3
f. 1
g. 1
h. 1

$\cos 0^{\circ}= 1$
$\cos 45^{\circ}= \frac {1}{\sqrt {2}}$
$\cos 60^{\circ}= \frac {1}{2}$
$\cos 90^{\circ}= 0$
We can see that values of cos decreases as angle increases from 0° to 90°

$\sin 0^{\circ}= 0$
$\sin 30^{\circ}= \frac {1}{2}$
$\sin 45^{\circ}= \frac {1}{\sqrt {2}}$
$\sin 60^{\circ}= \frac {\sqrt {3}}{2}$
$\sin 90^{\circ}= 1$
We can see that values of sin increases as angle increases from 0° to 90°

$\tan 0^{\circ}= 0$
$\tan 30^{\circ}= \frac {1}{\sqrt {3}}$
$\tan 45^{\circ}= 1$
$\tan 60^{\circ}= \sqrt {3}$
$\tan 90^{\circ}= undefined$
We can see that values of tan increases as angle increases from 0° to 90°

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 $\Cos A = {B \over H} = \frac {4}{5}$
$\tan A = {{\sin A} \over {\cos A}} = \frac {3}{4}$

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 $sinA = {P \over H} = \frac {1}{\sqrt {10}}$
$cosA = {B \over H} = \frac {3}{\sqrt {10}}$
$tan A = {{SinA} \over {CosA}} = \frac {1}{3}$
$sec A = \frac {1}{cos A} = \frac {\sqrt {10}}{3}$
$cot A = {{cosA} \over {sinA}} = 3$

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}$

$(sin30 ^{\circ} + cos30^{\circ}) - (sin60^{\circ} + cos60^{\circ})$
$=(\frac {1}{2} + \frac {\sqrt {3}}{2}) - (\frac {\sqrt {3}}{2} + \frac {1}{2})$
=0

$sin B =\frac {1}{2}$
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

$tanA + \frac {1}{tanA}= 2$
Squaring both the sides
$tan^2 A + \frac {1}{tan^2 A} + 2 = 4$
$tan^2 A + \frac {1}{tan^2 A}=2$

$2 sin^2 30^{\circ} - 3cos^2 45^{\circ} + tan^2 60^{\circ}$
$=2 (\frac {1}{2})^2 - 3 (\frac {1}{\sqrt {2}})^2 + (\sqrt {3})^2$
$=\frac {1}{2} - \frac {3}{2} + 3 $
=2

$sin^2 30^{\circ} cos^2 45^{\circ} + 4tan^2 30^{\circ} + \frac {1}{2} sin^2 90^{\circ} - 2 cos^2 90^{\circ} + \frac {1}{24}cos^2 0^{\circ}$
$= \frac {1}{4} \times \frac {1}{2} + 4 \times \frac {1}{3} + \frac {1}{2} -0 + \frac {1}{24}$
$= \frac {1}{8} + \frac {4}{3} + \frac {1}{2}+ \frac {1}{24}$
$=\frac {3+ 32 + 12 +1}{24}$
$=2$

$cot^2 30^{\circ} - 2 cos^2 60^{\circ} - \frac {3}{4} sec^2 45^{\circ} - 4sec^2 30^{\circ}$
$= 3 - \frac {1}{2} -3 -4 \times \frac {4}{3}$
$= \frac {-13}{3}$

Now as $cos 90^{\circ}=0$
$cos 1^{\circ}.cos 2^{\circ} .cos3^{\circ} .... cos 180^{\circ}=0$