Trigonometry Class 10

Ex 8.1, 1 teackhoo
In A ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cosA A
Step1 : Finding sides of triangle
25cm
24cm
In right triangle ABC,
Using Pythagoras theorem B c
7cm
(Hypotenuse)? = (Height)? + (Base)?
AC? = AB? + BC?
AC? = 24? + 7?
AC? = 576 +49
AC? = 625
AC=V625
AC= 254
AC=25

Ex 8.1, 2 teackoo
In figure , find tan P — cot R.
P
Finding sides of triangle
In right A PQR,
12 cm 13 cm

Using Pythagoras theorem

(Hypotenuse)? = (Height)? + (Base)? Q R

PR? = PQ? + QR?

13? = 12? + QR?

169 = 144 + QR?

169 — 144 = QR?

25 = QR?

QR? = 25

QR=v25

Ex 8.1, 3 teachoo
If sin A= :, calculate cos A and tan A.
A
Given
4x
sinA= 3
4
side opposite to ZA _ 3
Hypotenuse “4 B 3x Cc

Be _3

AC 4
Let

BC = 3x

AC = 4x
We find AB using Pythagoras Theorem

teackhoo
Ex 8.1, 4
Given 15 cot A = 8, find sin A and sec A.
Given
15cotA=8
8
cot A= is
Now,
tanA=——
cota
1
ae
15
15
“8

Ex 8.1, 5 teackhoo
Given sec 8 =, calculate all other trigonometric ratios.
Finding cos 8
1
cos @ = sec® A
12
cos 9 = —
13 13x
12x
Now, L
B Cc
2
cos@= —
13
side adjacent to 40 _ 12
Hypotenuse ~ 43
AB _ 12
AC” 13

Ex 8.1, 6 teackoo
If 2 A and Z B are acute angles such that cos A = cos B, then
show that 2 A= ZB.
A
cos A= side adjacent A
Hypotenuse
AC
Cc B
Similarly,
cosB = side adjacent B
Hypotenuse
BC
== (2)

Ex 8.1, 7 teackoo
_7 _ py A+ sin 0) — sin 0)
If cot 6 = a? evaluate : (i) (bcos O\(1 cos 8)
We will first calculate the value of sin 8 & cos 8
Now,
1
tan 8 = core
tano=2
7
We can write
tang =2
7
side opposite 40 __ 8
side adjacent 40 7

teackhoo
Ex 8.1, 8
If 3 cot A = 4, check whether 2-2" 4) = cos? A-sinA or not.
(1 + tan*A)
Given
3cotA=4
cot A= 4
3
So,
tanA= —
cotA
1
tanA= =
@)
tanA= 3
4

Ex8.1,9 teackhoo
In triangle ABC, right-angled at B, if tan A= nat find the value of
(i) sin AcosC+cosAsinC
A
Given
1
tanA= 7 V3x
Side opposite to A _1 B Cc
Side adjacent toA “3 x
Bo
AB V3
Let BC =x
& AB = V3 x

Ex 8.1, 10 teachoo
In A PQR, right-angled at Q, PR + QR = 25 cm and PQ=S cm.
Determine the values of sin P, cos P and tan P.
Given
PR+QR=25cm P
Thus, a R
PR+QR=25cm
PR=25-—OQR
PR=25-x
In right triangle PQR,
Using Pythagoras theorem

Ex 8.1, 11 teackoo
State whether the following are true or false. Justify your answer.
(i) The value of tan Ais always less than 1.
A
tan A= eee
= Be
AB
B Cc
Assuming BC > AB
Example:
Let AB= 10 cm and BC =15cm
Hence,
tanA= =
=i
10

Ex 8.2, 1 teackoo
Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
P| os | 30" | s+ co" | 20")
sin 0 1 z v3 1
We know that, 2 22
ve oi 12
B cos 1 - 73 3 ie}
sin 60° = 7 1 a Not
tn 0 Be 1 NB Get
cos 30° = v3
2
sin 30° = +
2
cos 60° ==
2
Putting all values
sin 60° cos 30° + sin 30° cos 60°

Evaluate the following :
(ii) 2 tan2 45° + cos2 30° – sin2 60°

Evaluate the following :
(iii) "cos 45°" /"sec 30° + cosec 30°"

Evaluate the following :
"sin 30° + tan 45° – cosec 60°" /"sec 30° + cos 60° + cot 45°"

Evaluate the following :
"5 cos2 60° + 4 sec2 30° – tan2 45°" /"sin2 30° + cos2 30°"

Ex 8.2, 2 teackhoo
Choose the correct option and justify your choice :
+ Z2tan 30°
{i) 1+ tan? 30°
(A) sin60° (B)cos60° (C)tan60° (D)sin 30°
J 0° | 0745" | co | 50"
: 1 1 v3
We know that, sn O + & = 1
V3 1 1
cos 1 = = = ie}
tan 30° = a 2 ve 2
3 1 Not
tan 0 1 NB gee
So,
1
2tan30° _ 2x (3)
1+ tan?30° 1\2
2
— _v3
7 1
1 +3

Ex 8.2, 2 teackhoo
Choose the correct option and justify your choice :
i 1 — tan? 45°
ware Po a0 aso ao
° . ° . 1 1
(A) tan 90 (B)1 = (C) sin 45 sin 0 2 2 OF 1
ve oi 12
cos 1 - 73 3 ie}
We know that, a Not
tan 0 1 NB eg
tan 45° =1
So,
1-tan? 45° 1-(1)?
1+ tan2 45° 14 (1)2
_ 1-1
~ 444
-?2
“2

Ex 8.2, 2 teackoo
Choose the correct option and justify your choice :
(iii) sin 2A = 2 sin Ais true when A=
(A) 0° (B)30° (C)45° = (D) 60°
Given
sin 2A=2sinA
Here, we substitute the value of option in the equation and
whichever satisfies the question would be solution.

Ex 8.2, 2 teachoo
Choose the correct option and justify your choice :
(i } 2 tan 30°
Mv 1— tan? 30°
(A) cos 60° (B)sin 60° (C)tan60° (D) sin 30°
J | 0: | 0" [45°] so" | 90"
We know that, sn o 1 +4 WB 4
2 v2 2
eo 1 vei 2
tan 30 = cos 1 > 8 3 fe)
1 Not
tn 0 = 1 V3 det
So,
1
2tan30° 2x(%)
1—tan?30° 1\2
1-(%)

Ex 8.2, 3 teackhoo
1
If tan (A+B) = V3 and tan (A-B) = 7g 0°<A+B S90"; A>B,
find A and B.
Given that
1
tan (A+B) =V3 tan(A-B)=—
But we know that But we know that
tan 60° = V3 ot
tan 30° = a
Thus,
Thus
t A+B) = tan 60°
an ( )=tan tan (A—- B) = tan 30°
~ A+B=60° (1)
*A-B=30° (2)

Ex 8.2, 4 teackhoo
State whether the following are true or false. Justify your answer.
(i) sin (A+B) =sinA+sinB.
Po a0 aso ao
We have to prove sin 0 1 = 8 1
sin (A+ B)=sinA+sinB
cos 1 8 4 1 ie}
2 vz 2
Assuming tan 0 z 1 v3 ie
A= 60° & B = 30°
L.H.S R.H.S
sin (A+B) sinA+sinB
= sin (60° + 30°) = sin 60° + sin 30°
= sin 90° -t,t
2
=1
_ V3+1
~ 2

State whether the following are true or false. Justify your answer.
(ii) The value of sin θ increases as θ increases.

State whether the following are true or false. Justify your answer.
(iii) The value of cos θ increases as θ increases.

State whether the following are true or false. Justify your answer.
(iv) sin θ = cos θ for all values of θ.

State whether the following are true or false. Justify your answer.
(v) cot A is not defined for A = 0°.

Ex 8.3,1 teackhoo
Express the trigonometric ratios sin A, sec A and tan A in terms
of cot A.
tanA
We know that
1

tanA= cota
cosecA
We know that

1+ cot? A=cosec?A

cosec*A=1+cot?A

cosecA=+V1+cot7A

Ex 8.3, 2 teackoo
Write all the other trigonometric ratios of 2 Ain terms of sec A.
cosA
We know that

cos A= —

secA

tanA
We know that

1+tan?A=sec?A

tan*A=sec2A —1

tanA=+vsec?A —1
Here, A is acute angle (i.e. less than 90°}

Ex 8.3, 3 teackhoo
Choose the correct option. Justify your choice.
{i) 9sec*A-9tan2A
{A) 1 {B)9 {C) 8 (D)O
9 sec? A-9 tan?A
As sec? A=1+tan’A
=9(1+tan? A)-9tan?A
=9+9tan?A-9tan2A
=9+0
=9
So, the correct answer is (b)

Choose the correct option. Justify your choice.
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) –1

Choose the correct option. Justify your choice.
(iii) (sec A + tan A) (1 – sin A)
sec A (B) sin A (C) cosec A (D) cos A

Choose the correct option. Justify your choice.
(iv) (1 + 𝑡𝑎𝑛2𝐴)/(1 + 𝑐𝑜𝑡2𝐴)
sec2 A (B) –1 (C) cot2 A (D) tan2 A

Ex 8.3, 4 teackoo
Prove the following identities, where the angles involved are
acute angles for which the expressions are defined.
. _ 2_1-cosd
(i) (cosec 8 — cot 8)? = Te cos8
Solving L.H.S
(cosec 6 — cot 6)?
We need to make it in terms of cos 8 & sin ®
_ ( 1 cos 2"
~ \sin @ sin @
_ ¢ — cos 2"
~ sin @
_ (1~cos 6° We know that
sin? @ cos? 8+ sin?6=1
sin? 8@=1-cosz20

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(ii) "cos A" /"1 + sin A" +"1 + sin A" /"cos A" =2 sec A

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iii)tan⁡θ/(〖1 − cot〗⁡θ " " )+cot⁡θ/(1 − tan⁡θ ) =1+ sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
("1 + sec" A)/"sec A" ="sin2 A" /"1 – cos A" "[Hint : Simplify LHS and RHS separately]"
(1 + sec⁡𝐴)/sec⁡〖 𝐴〗
= 1/sec⁡〖 𝐴〗 +sec⁡〖 𝐴〗/sec⁡〖 𝐴〗
= 1/sec⁡〖 𝐴〗 +1
= cos A + 1
(𝒔𝒊𝒏𝟐 𝑨)/(1 − cos⁡𝐴 )
= (𝟏 − 𝒄𝒐𝒔𝟐 𝑨)/(1 − cos⁡𝐴 )
= (12 − 𝑐𝑜𝑠2 𝐴)/(1 − cos⁡𝐴 )
= ((1 − 𝑐𝑜𝑠 𝐴)(1+ cos⁡〖𝐴)〗)/(1 − cos⁡𝐴 )
= 1 + cos A
∴ L.H.S = R.H.S
Hence proved

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
"cos A – sin A + 1" /"cos A + sin A – 1" = cosec A + cot A,
using the identity cosec2 A = 1 + cot2 A.

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vi) √((1 + sin⁡𝐴 )/(1 −〖 sin〗⁡𝐴 )) = sec A + tan A

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vii) (sin θ − 2 sin3 θ)/(2 cos3 θ − cos θ)=tan θ

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(ix) (cosec A – sin A)(sec A – cos A) = 1/(𝑡𝑎𝑛 𝐴 +cot⁡ 𝐴)
[Hint : Simplify LHS and RHS separately]

Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
((1 +𝑡𝑎𝑛2 𝐴)/(1 + 𝑐𝑜𝑡2 𝐴))=((1 −tan⁡〖 𝐴〗)/(1 −cot⁡ 𝐴))^2=𝑡𝑎𝑛2 𝐴

teachoo.com
Ex 8.4, 3
Evaluate :
(i) sin’ 63° + sin? 27°
cos’ 17° +cos* 73° We know that
63+27=90
Hence 63 = 90-27
sin’ 63° + sin? 27°
cos’ 17° + cos* 73° Similarly
17+ 73=90
_ sin? (90 - 27) + sin? 27° Hence 17 = 90 — 73
~ cos? (90 — 73) + cos? 73°
cos? 27° + sin? 27° cos (90-8) =sin®
~ sin? 73° + cos? 73° & sin (90 — 8) = cos 8
1 +
=7 (As cos? A + sin? A= 1)
=1
sin? 65° + sin? 27°
Hence, —~_-_s 1
sin’ 17° + sin® 73°

Evaluate :
sin 25 cos 65 + cos 25 sin 65

teachoo
Example 1
Given tan A= :, find the other trigonometric ratios of the angle A
Given, A
tanA= 4
3
3x
side opposite toangle A _ 4
side adjacent to angle A “3
c 4x B
Bo 4
AB 3
Let
BC = 4x
AB = 3x
We find AC using Pythagoras Theorem

Example 2 teachoo
If 2 B and Z Q are acute angles such that sin B = sin Q, then prove
that2B=2Q.
Pp
A .

Given: sin B=sinQ
To prove: 2B=2Q c BOR Q
Proof:
Let’s take two right angle triangles ABC & POR

. _ side opposite to B . _ side opposite to Q

sin B= Hypotenuse | sinQ= Hypotenuse

. PR

sin B= « sinQ= PQ

Since,
sinB=sinQ

Example 3 teackoo
Consider A ACB, right-angled at C, in which AB = 29 units, BC = 21
units and 2 ABC = 6 (see fig.). Determine the values of
(i) cos? @ + sin? 8, A
29

Step 1 : Finding sides of triangle
In right triangle ABC, DS
Using Pythagoras theorem
(Hypotenuse)? = (Height)? + (Base)?

AB = AC? + BC?

AC? = AB? — BC?

AC? = (29)? — (21)
Using a? — b? = (a +b} (a—b)

AC? = (29-21) (29 +21)

Example 4 teachoo
In a right triangle ABC, right-angled at B, if tan A = 1, then verify
that 2 sin Acos A= 1.
In a right angle triangle ABC
A
tanA=1
side opposite toangle A _ 1 k
Side adjacent toangle A ~
B c
Bead k
AB
AB=BC
Let
AB = BC =k
Where k is a positive number.

Example 5 teackoo
In A OPQ, right-angled at P, OP = 7 cm and OQ—-PQ=1cm
(see Fig.). Determine the values of sin Q and cos Q.
Q
Given in A OPQ,
0Q-PQ=1¢cm
0Q=1+PQ
Using Pythagoras theorem
(Hypotenuse)? = (Height) + (Base)? Te °
OQ? = PQ? + OP?
(1+ PQ)? = PQ? + (7)
1+ PQ2+2PQ = PQ?+49
1+ PQ2+2PQ-PQ? - 49=0
2PQ—48 =0
2PQ=48

Example 6 teachoo
In A ABC, right-angled at B, AB=5 cm and Z ACB = 30° (see Fig.).
Determine the lengths of the sides BC and AC.
A
10
Given em
Semi
AB=5cm & Z ACB = 30° 30
B Cc
5v3 cm
According to diagram,
_ side opposite to angle sinC= side opposite to angle C
tan side adjacent to angle Hypotenuse
. o _ AB
tan 30° == sin 30° =7
1_ 5 15
Va BC 2° AC
BC=5 V3 cm AC= 10cm
Hence, AC=10 cm& BC=5 v3 cm

Example 7 teachoo
In A PQR, right-angled at Q (see Fig. }, PQ = 3 cm and PR=6cm.
Determine Z QPR and Z PRQ.
P
“ side opposite t0 an DS.
3cm
+ _ side opposite to an.
sin R= Hypotenuse Q a a R
irR= 22
sir R= PR
ree? en
=e . 1 3
6 sin 0 3 Eg 1
. 1 vao4 1
sin R= 5 cos 1 > 82 0
1 Not
sin R = sin 30° tm 0 = 1 VB def
R= 30°
So, Z PRQ = 30°

Example 8 teackoo
If sin (A-B) =>, cos (A+ B)=>,0°2A+ZB<90°, A> B,
Por 130" | 45°) 60" | 90"|
find A and B. . 1 1 3
sin ie} 3 Boo 1
v3 i 1
cos 1 - ei 3 ie}
Given data, 1 Not
tan 0 1 NB gee
. 1 A 1
sin (A-B) => cos (A+B)=>
But we know that But we know that
1 o_ 1
sin 30° =— cos 60° =-—
2 2
Thus, Thus
sin (A—B) = sin 30° cos {A+ B) = cos 60°
A-B=30° (1) A+B=60° ...(2)

Example 9 teachoo
Express the ratios cos A, tanA and sec A in terms of sin A.
cosA
Since,
cos? A+sin? A =1
cos?A=1-sin?A
cosA=+V1 —sin?A
Here, A is acute (i.e. less than 90°)
& cos Ais positive when A is acute
So, cosA=V1 — sin?A

Example 10 teachoo
Prove that sec A (1— sin A)(sec A + tan A) = 1.
Solving L.H.S
sec A (1-sin A) (secA + tan A)
Writing everything in terms of sin A and cos A
1 . 1 sin A

~ cos q(i-sin A) ( cos A + cos ‘)

_ (-=sin A) 1+sin A

~ cos A ( cos A )

_ (—sin A) (1+sin A)

~ cos AXcos A
Since (a — b) (a + b) = a? — b?

_ @?-sin? A)

~ cos? A

Example 11 teackoo
cotA—cosA cosecA—1
Prove that —————- = ———_—_
cotA+cosA cosecA+1
Taking L.H.S
cotA—cosA
cotA+cosA
Writing everything in terms of sin A and cos A
cos A
- sin.A 7 cos A
cos A +cos A
sin A
_cos A—cos A sin A
~ sin A
cos A +cos A sin A
sin A
_ (cos A- cos Asin A)
~ (cos A+ cos Asin A)

Example 12 teachoo
sin@-cos@4+1 1 . . .
Prove that SinO4cos0—-1 = SecO—tane’ using the identity
sec?8 = 1+ tan76.
Solving L.H.S
sin 8 -—cos04+1
sin 8+ cos0-1
Dividing the numerator & denominator by cos @
=< (sin 6 — cos 6 +1)
= cos a
ws gesin 8+ cos 6-1)
sin 8 cos 8 1
= (a a) - ( a) + (Ss a)
~ ¢sin 8 cos 8 1
(a 7) + (G a) ~ (as a)
_ (tan ®@- 14sec 0)
~ (tan 8 + 1 — cos 0)

feachoo.com
Example 9
tan 65°
Evaluate ———— .
cot 25°
tan 65°
cor?s We know
_ tan (90°— 25°) 25 +65 =90
cot 25° Hence 65 = 90-25
cot 25°
~ cot 25° (tan (90 — 8) = cot 8)
=1

feachoo.com
Example 10
If sin 3A = cos (A— 26°), where 3A is an acute angle, find the
value of A.
Given that,
sin 3A = cos {(A— 26°)
cos (90° — 3A) = cos (A - 26°) (sin 8 = cos (90 - 6))
Comparing angles
90-3A=A- 26°
- 3A—-A=-— 26-90
-4A=-116
Aarts
4
A=29
Hence, A = 29°

feachoo.com
Example 11
Express cot 85° + cos 75° in terms of trigonometric ratios of
angles between 0° and 45°.
We know that
5+85=90
Hence 85 =90-5
cot 85° + cos75° And
15+75=90
= cot (90° - 5) + cos (90° - 15°) Hence 75 = 90-15
=tan5°+sin 15° cos (90-9) =sin®@
& cot (90-6) = tan 8
Hence required value is
tan 5° + sin 15°

Question 'Skysails' is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively. Based on the following figure related to sky sailing answer the questions: Question 1 (i) In the given figure, if tan 𝜃 = cot (30° + 𝜃), where θ and 30° + 𝜃 are acute angles, then the value of 𝜃 is: (a) 45° (b) 30° (c) 60° (d) None of these Given,
tan θ = cot(30° + 𝜃)
tan θ = tan[90° – (30° + 𝜽)]
tan θ = tan(90° – 30° – 𝜃)
tan θ = tan(60° – 𝜃)

Question Authority wants to construct a slide in a city park for children. The slide was to be constructed for children below the age of 12 years. Authority prefers the top of the slide at a height of 4 m above the ground and inclined at an angle of 30° to the ground. Based on the following figure related to the slide answer the questions: Question 1 The distance of AB is: (a) 8 m (b) 6 m (c) 5 m (d) 10 m In Δ ABC
sin 30° = 𝐴𝐶/𝐴𝐵
1/2 = 4/𝐴𝐵
AB = 8 m

Question A circus artist is climbing from the ground along a rope stretched from the top of a vertical pole and tied at the ground. The height of the pole is 12 m and the angle made by the rope with ground level is 30°. Give answer to the following questionsQuestion 1 The distance covered by the artist in climbing the top of the pole is: (a) 24 m (b) 36 m (c) 28 m (d) 22 m We need to find AC

The value of (sin 30° + cos 30°) – (sin 60°+ cos 60°) is
(A) – 1 (B) 0 (C) 1 (D) 2
Now,
(sin 30° + cos 30°) – (sin 60°+ cos 60°)
= (𝟏/𝟐+√𝟑/𝟐)−(√𝟑/𝟐+𝟏/𝟐)
= 1/2−1/2+√3/2−√3/2
= 0

The value of tan⁡〖30°〗/cot⁡〖60°〗 is
(A) 1/√2 (B) 1/√3 (C) √3 (D) 1
Now,
tan⁡〖30°〗/cot⁡〖60°〗 = tan 30° × 𝟏/𝒄𝒐𝒕⁡〖𝟔𝟎°〗
= tan 30° × tan 60°
= 1/√3 × √3
= 1

The value of (sin 45° + cos 45°) is
(A) 1/√2 (B) √2 (C) √3/2 (D) 1

If cos A = 4/5 , then the value of tan A is
(A) 3/5 (B) 3/4 (C) 4/3 (D) 5/3

If sin A = 1/2 , then the value of cot A is
(A)√3 (B) 1/√3 (C) √3/2 (D) 1
Since sin A = 1/2
∴ cosec A = 2

The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1 (B) 0 (C) 1 (D) 3/2
cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)

Given that sin θ = 𝑎/𝑏, then cos θ is equal to
(A) 𝑏/√(𝑏^2 − 𝑎^2 ) (B) 𝑏/𝑎
(C) √(𝑏^2 − 𝑎^2 )/𝑏 (D) 𝑎/√(𝑏^2 − 𝑎^2 )

If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α
Given that
cos (α + β) = 0
cos (α + β) = cos 90°

The value of (tan 1° tan 2° tan 3° ... tan 89°) is
(A) 0 (B) 1 (C) 2 (D)
tan 1° tan 2° tan 3° ... tan 89°
= tan 1° tan 2° tan 3° … tan 44° tan 45° tan 46° … tan 87° tan 88° tan 89°
Using tan θ = cot (90° − θ)
= tan 1° tan 2° tan 3° … tan 44° tan 45° cot (90° − 46°) … cot (90° − 87°) cot (90° − 88°) cot (90° − 89°)
= tan 1° tan 2° tan 3° … tan 44° tan 45° cot 44° … cot 3° cot 2° cot 1°
= tan 1° cot 1° tan 2° cot 2° tan 3° cot 3° … tan 44° cot 44° × tan 45°
= tan 1° × 𝟏/(𝒕𝒂𝒏 𝟏°) . tan 2° × 𝟏/(𝒕𝒂𝒏 𝟐°) . tan 3° × 𝟏/(𝒕𝒂𝒏 𝟑°) … tan 44° × 𝟏/(𝒕𝒂𝒏 𝟒𝟒°) × tan 45°
= 1 . 1 . 1 … 1 × tan 45°
= tan 45°
= 1
So, the correct answer is (B)

If cos 9α = sin α and 9α < 90° , then the value of tan 5α is
(A) 1/√3 (B) √3 (C) 1 (D) 0
Given
cos 9α = sin α
cos 9α = cos (90° − α)

If ∆ABC is right angled at C, then the value of cos (A + B) is
(A) 0 (B) 1 (C) 1/2 (D) √3/2
Given
∠ C = 90°

If sin A + sin2 A = 1, then the value of the expression (cos2 A + cos4 A) is
(A) 1 (B) 1/2 (C) 2 (D) 3
Given
sin A + sin2 A = 1
sin A = 1 − sin2 A
sin A = cos2 A

Given that sin α = 1/2 and cos β = 1/2 , then the value of (α + β) is
(A) 0° (B) 30° (C) 60° (D) 90°
Now,
sin α = 𝟏/𝟐
sin α = sin 30°
∴ α = 30°
cos β = 𝟏/𝟐
cos β = cos 60°
∴ β = 60°

The value of the expression [(sin^2⁡22"°" + sin^2⁡68"°" )/(cos^2⁡〖22"°" 〗+ cos^2⁡68"°" )+sin^2⁡〖63"°" +cos⁡〖63"°" sin⁡27"°" 〗 〗 ] is
(A) 3 (B) 2 (C) 1 (D) 0
(sin^2⁡22"°" + sin^2⁡68"°" )/(cos^2⁡〖22"°" 〗+ cos^2⁡68"°" )+sin^2⁡〖63"°" +cos⁡63"°" 𝒔𝒊𝒏⁡𝟐𝟕"°" 〗
Using sin θ = cos (90° − θ)
= (sin^2⁡22"°" + sin^2⁡68"°" )/(cos^2⁡〖22"°" 〗+ cos^2⁡68"°" )+sin^2⁡〖63"°" +cos⁡〖63"°" 𝐜𝐨𝐬⁡〖(𝟗𝟎°−𝟐𝟕"°)" 〗 〗 〗
= (sin^2⁡22"°" + sin^2⁡68"°" )/(cos^2⁡〖22"°" 〗+ cos^2⁡68"°" )+sin^2⁡〖63"°" +cos⁡〖63"°" 𝒄𝒐𝒔⁡〖𝟔𝟑°〗 〗 〗
= (sin^2⁡22"°" + sin^2⁡68"°" )/(cos^2⁡〖22"°" 〗+ cos^2⁡68"°" )+〖𝒔𝒊𝒏〗^𝟐⁡〖𝟔𝟑"°" +〖𝒄𝒐𝒔〗^𝟐⁡𝟔𝟑"°" 〗
Using cos2 θ + sin2 θ = 1
(sin^2⁡22"°" + 〖𝒔𝒊𝒏〗^𝟐⁡𝟔𝟖"°" )/(cos^2⁡〖22"°" 〗+ 〖𝒄𝒐𝒔〗^𝟐⁡𝟔𝟖"°" )+1

If 4 tan θ = 3, then ((4 sin⁡〖𝜃 − cos⁡𝜃 〗)/(4 sin⁡〖𝜃 + cos⁡𝜃 〗 )) is equal to
(A) 2/3 (B) 1/3 (C) 1/2 (C) 3/4
Given
4 tan θ = 3
tan θ = 𝟑/𝟒

If sin θ – cos θ = 0, then the value of (sin4 θ + cos4 θ) is
(A) 1 (B) 3/4 (C) 1/2 (D) 1/4
Given
sin θ – cos θ = 0
sin θ = cos θ

sin (45° + θ) – cos (45° – θ) is equal to
(A) 2 cos θ (B) 0 (C) 2 sin θ (D) 1
Now,
sin (45° + θ) – cos (45° – θ)
Using sin A = cos (90° − A)
= cos (90° − [45° + θ] ) – cos (45° – θ)
= cos (90° − 45° − θ) – cos (45° – θ)
= cos (45° − θ) – cos (45° – θ)
= 0

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Ex 8.3,1
Evaluate :
+, sin 18°
(i) cos 72°

sin 18°

cos 72°

; We know that

= — sini’ 18+72=90

cos (90° — 18°) Hence 72 = 90 - 18
— Sin 18° (cos (90-8) =sin@)

sin 18°
=1

Evaluate :
(ii) tan 26 /cot 64

Evaluate :
(iii) "cos 48 sin 42 "

Evaluate :
(iv) cosec 31 sec 59

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Ex 8.3, 2
Show that :
(i) tan 48° tan 23° tan 42° tan 67 =1 We know that
48 +42=90
Taking L.H.S Hence 42 = 90-48
tan 48° tan 42° tan 23° tan 67° Similarly,
23+67=90
= tan 48° tan(90° — 48°) tan 23° tan(90° — 23°) | Hence 67 = 90 — 23
= tan 48° cot 48° tan 23° cot 23° (tan (90 - 8) = cot 8)
= (tan 48° x ——) x (tan 23° x ——} 1
= (an tan 48° an tan 23° (cot 6 = ane!
=1x1
=1
= R.H.S
“ LHS = R.H.S
Hence proved

Show that :
(ii) cos 38 cos 52 sin 38 sin 52 = 0

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Ex 8.3, 3
If tan 2A = cot (A— 18°}, where 2A is an acute angle, find the
value of A.
tan 2A = cot (A— 18°)
cot (90 — 2A) = cot (A— 18°) (tan 9 = cot (90 - 8))
Comparing angles

90-2A=A-18

90+18=A+2A

108=3A

=A

36=A

A= 36°

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Ex 8.3, 4
If tan A = cot B, prove that A+B = 90°.
tan A=cotB
tan A = tan (90 —B) (cot @ = tan (90 - 8)
Comparing angles

A=90-B

A+B=90°

Hence proved.

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Ex 8.3 ,5
If sec 4A = cosec (A— 20°), where 4A is an acute angle, find the
value of A.
sec 4A = cosec (A— 20°)
cosec (90 — 4A) = cosec (A — 20°) (sec 8 = cosec (90 — 8))
Comparing angles
90-4A=A-—20
-4A-A=-20-90
—5A=-110
Ante
-5
A=22°

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Ex 8.3, 6
if A, Band C are interior angles of a triangle ABC, then show
that sin C=) = cos“ A
2 2 |
B Cc
In A ABC
Sum of angles of a triangle = 180° {Angle sum property of triangle,
A+B+C=180°
B+C=180°-A
Multiplying both sides by >
B+C _ 180°-A
2 2
B+c _ 180° A
2 2 2
B+ _gge_A AL)
2 2

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Ex 8.3, 7
Express sin 67° + cos 75° in terms of trigonometric ratios of
angles between 0° and 45°. We know that

67+23=90

Hence 67 = 90-23
sin 67° + cos 75° And

15+75=90
= sin (90 — 23) + cos (90-15) Hence 75 = 90-15
= cos 23° + sin 15° cos (90 - 8) = sin 8

& sin (90-8) = cos 6
Hence required value is
cos 23° + sin 15°