Get here NCERT Solutions Class 10 Maths Chapter 8. These NCERT Solutions for Class 10 of Maths subject includes detailed answers of all the questions in Chapter 8 – Trigonometry provided in NCERT Book which is prescribed for class 10 in schools.
Resource: National Council of Educational Research and Training (NCERT) Solutions
Class: 10th Class
Subject: Maths
Chapter: Chapter 8 – Trigonometry
NCERT Solutions Class 10 Maths Chapter 8 – Free Download PDF
NCERT Solutions Class 10 Maths Chapter 8 – Trigonometry
NCERT Solutions for Class 10 Maths Chapter 8 is provided here so that students can check for the solutions whenever they are facing difficulty while solving the questions of Trigonometry. 10th Class Maths is a tricky subject which demands students understanding and reasoning skills. Along with this, it also requires students to practice mathematics on a regular basis in order to excel in flying colours. Students of Class 10 are suggested to solve these NCERT Questions in order to practice the exemplary problems that are usually asked in the competitive examinations.
Class 10 Chapter 8 Introduction to Trigonometry:
Trigonometry is a branch of mathematics which involves study of the relationships between the lengths and angles of triangles.
Examin provides NCERT Solutions Class 10 Maths chapter 8 Introduction to Trigonometry in the pdf format which you can download for absolutely free. The use of trigonometry class 10 formulas are provided in a detailed manner, where one can find step-by-step solutions to all the questions of chapter 8. The major topics covered in 10th class maths chapter 8 are:
- Trigonometric Ratios
- Trigonometric Ratios of some specific angles
- Trigonometric Ratios of Complementary angles
- Trigonometric Identities
The solutions to all these topics given-above are covered with a brief explanation and related formulas, in trigonometry class 10 questions. The most commonly used angles in Trigonometry, such as 0°, 30°, 45°, 60° and 90° have been tabulated for all the trigonometry ratios in this chapter. Also, the proofs of trigonometric identities are explained in a step by step manner, to make it easier to understand.
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercises
Here are the trigonometry class 10 formulas, trigonometric functions & trigonometric identities with solved ncert solutions exercise-wise:
- NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.1
- NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.2
- NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3
- NCERT Solutions Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.4
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.1
Q1) In △ABC , 90∘ at B, AB=24cm, BC = 7cm.
Determine:
(i)sin(A), cos(A)
(ii) sin(C), cos(C)
Ans.) In △ABC , ∠B=90∘
By Applying Pythagoras theorem, we get
AC2=AB2+BC2 (24)2+72 =(576+49)AC2 = 625cm2
à AC = 25cm
(i) sin(A) = BC/AC = 7/25
Cos(A) = AB/AC = 24/25
(ii) sin(C) = AB/AC =24/25
cos(C) = BC/AC = 7/25
Q2) In the given figure find tan(P) – cot(R)
Ans.) PR = 13cm,PQ = 12cm and QR = 5cm
According to Pythagorean theorem,
132=QR2+122 169=QR2+144 QR2=169−144=25 QR=25−−√=5
tan(P) = oppositesideadjacentside=QRPQ=512
cot(P) = adjacentsideoppositeside = PQQR = 512
tan(P) – cot(R) = 512−512=0
Therefore ,tan(P) – cot(R) = 0
Q3) If sin(A) = 3/4, calculate cos(A) and tan(A)
Ans.) Let △ABC , be a right-angled triangle, right-angled at B.
We know that sin(A) = BC/AC = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
AC2=AB2+BC2
(4k)2=AB2+(3k)2
16k2−9k2=AB2
AB2=7k2
AB=7–√k
cos(A) = AB/AC = 7–√k/4k=7–√/4
tan(A) = BC/AB =3k/7–√=3/7–√
Q4) In question given below 15cot(A) = 8 ,find sin A and sec A.
Ans.) Let △ABC be a right angled triangle, right-angled at B.
We know that cot(A) = AB/BC = 8/15
Given
Let AB side be 8k and BC side 15k
Where k is positive real number
By Pythagoras theorem we get,
AC2=AB2+BC2
AC2=(8k)2+(15k)2
AC2=64k2+225k2
AC2=289k2
AC = 17k
sin(A) = BC/AC = 15k/17k = 15/7
sec(A) =AC/AB =17k/8k = 17/8
Q5) Given sec Ѳ =13/12, calculate all other trigonometric ratios.
Ans.) Let △ABC be right-angled triangle, right-angled at B.
We know that sec Ѳ =OP/OM =13/12(Given)
Let side OP be 13k and side OM will be 12k where k is positive real number.
By Pythagoras theorem we get,
OP2=OM2+MP2
(13k)2=(12k)2+MP2
169(k)2−144(k)2=MP2
MP2=25k2
MP = 5
Now,
sin Ѳ = MP/OP = 5k/13k =5/13
cos Ѳ = OM/OP = 12k/13k = 12/13
tan Ѳ = MP/OM = 5k/12k = 5/12
cot Ѳ = OM/MP = 12k/5k = 12/5
cosec Ѳ = OP/MP = 13k/5k = 13/5
Q6) If ∠A and ∠B are acute angles such that
cos(A) = cos(B), then show ∠A =∠B .
Ans.) Let △ABC in which CD⊥AB .
A/q,
cos(A) = cos(B)
à AD/AC = BD/BC
à AD/BD = AC/BC
Let AD/BD =AC/BC =k
AD =kBD …. (i)
AC=kBC …. (ii)
By applying Pythagoras theorem in △CAD and △CBD we get,
CD2=AC2−AD2 ….(iv)
From the equations (iii) and (iv) we get,
AC2−AD2=BC2−BD2 AC2−AD2=BC2−BD2 k2(BC2−BD2)=BC2−BD2 k2=1
Putting this value in equation (ii) , we obtain
AC = BC
∠A=∠B (Angles opposite to equal side are equal-isosceles triangle)
Q7) If cot Ѳ = 7/8, evaluate :
(i) (1+sin Ѳ)(1-sin Ѳ) / (1+cos Ѳ)(1-cos Ѳ)
(ii) cot2Θ
Ans.) Let △ABC in which ∠B=90∘
and ∠C=Θ
A/q,
cot Ѳ =BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number
According to Pythagoras theorem in △ABC we get.
AC2=AB2+BC2
AC2=(8k)2+(7k)2
AC2=64k2+49k2
AC2=113k2
AC=113−−−√k
sin Ѳ = AB/AC = 8k/113−−−√k=8/113−−−√
and cos Ѳ = BC/AC = 7k/113−−−√k=7/113−−−√
(i) (1+sin Ѳ)(1-sinѲ)/(1+cosѲ)(1-cos Ѳ) = (1−sin2Θ)/(1−cos2Θ)
= 1−(8/113−−−√)2/1−(7/113−−−√)2
= {1-(64/113)}/{1-(49/113)} = {(113-64)/113}/{(113-49)/113} = 49/64
(ii) cot2Θ=(7/8)2=49/64
Q8) If 3cot(A) = 4/3, check whether (1−tan2A)/(1+tan2A)=cos2A−sin2A or not.
Ans.) Let △ABC in which ∠B=90∘
A/q,
cot(A) = AB/BC = 4/3
Let AB = 4k an BC =3k, where k is a positive real number.
AC2=AB2+BC2
AC2=(4k)2+(3k)2
AC2=16k2+9k2
AC2=25k2
AC=5k
tan(A) = BC/AB = 3/4
sin(A) = BC/AC = 3/5
cos(A) = AB/AC = 4/5
L.H.S. = (1−tan2A)(1+tan2A)=1−(3/4)2/1+(3/4)2=(1−9/16)/(1+9/16)=(16−9)/(16+9)=7/25
R.H.S. =cos2A−sin2A=(4/5)2−(3/4)2=(16/25)−(9/25)=7/25
R.H.S. =L.H.S.
Hence, (1−tan2A)/(1+tan2A)=cos2A−sin2A
Q9) In triangle EFG, right-angled at F, if tan E =1/√3 find the value of:
(i) sin EcosG + cosE sin G
(ii) cosEcosG – sin E sin G
Answer
LetΔEFG in which ∠F=90∘, E/q
tanE=FCEF tanE=FCEF=13√
Where k is the positive real number of the problem
By Pythagoras theorem in ΔEFG we get:
EG2=EF2+FG2 EG2=(3k−−√2))+K2 EG2=3k2+K2 EG2=4k2 EG=2K
sinE = FG/EG = 1/2
cosE = EF/EG = 3√2 ,
sin G = EF/EG = 3√2 cosE = FG/EG = 1/2
(i) sin EcosG + cosE sin G = (1/2\ast1/2) + (3√2∗3√2)= 1/4+3/4 = 4/4 = 1
(ii) cosEcosG – sin E sin C = (3√2∗12)−(3√2∗12)= (3√4)−(3√4)= 0
Q10)In Δ MNO, right-angled at N, MO + NO = 25 cm and MN = 5 cm. Determine the values of sin M, cos M and tan M.
Answer
Given that, MO + NO = 25 , MN = 5
Let MO be x. ∴ NO = 25 – x
By Pythagoras theorem ,
MO2=MN2+NO2 X2=52+(25−x)2 50x = 650
x = 13
∴ MO = 13 cm
NO = (25 – 13) cm = 12 cm
sinM = NO/MO = 12/13
cosM = MN/MO = 5/13
tanM = NO/MN = 12/5
Q11) State whether the following are true or false. Justify your answer.
(i) The value of tan M is always less than 1.
(ii) secM = 12/5 for some value of angle M.
(iii) cosM is the abbreviation used for the cosecant of angle M.
(iv) cot M is the product of cot and M.
(v) sin θ = 4/3 for some angle θ.
Answer
(i) False.
In ΔMNC in which ∠N = 90∘,
MN = 3, NC = 4 and MC = 5
Value of tan M = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as
it will follow the Pythagoras theorem.
MC2=MN2+NC2 52=32+42 25 = 9 + 16
25 = 25
(ii) True.
Let a ΔMNC in which ∠N = 90º,MC be 12k and MB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
MC2=MN2+NC2 (12k)2=(5k)2+NC2 NC2+25k2=144K2 NC2=119k2
Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) False.
Abbreviation used for cosecant of angle M is cosec M.cosM is the abbreviation used for cosine of angle M.
(iv) False.
cotM is not the product of cot and M. It is the cotangent of ∠M.
(v) False.
sinΘ = Height/Hypotenuse
We know that in a right angled triangle, Hypotenuse is the longest side.
∴ sinΘwill always less than 1 and it can never be 4/3 for any value of Θ.
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.2
1) Calculate the following:
- sin60∘cos30∘+sin30∘cos60∘
- 2tan245∘+co230∘−sin260∘
- cos45∘(sec30∘+cosec30∘)
- (sin30∘+tan45∘−cosec60∘)(sec30∘+cos60∘+cot45∘)
- (5cos260∘+4sec230∘−tan245∘)(sin230∘+cos230∘)
Ans.- (i) sin60∘cos30∘+sin30∘cos60∘
= (3√2×3√2)+(12×12)=34+14=44=1
(ii) 2tan245∘+co230∘−sin260∘
=2×(1)2+(3√2)2−(3√2)2=2
(iii) cos45∘(sec30∘+cosec30∘)
= 12√23√+2=12√(2+23√)3√
= 3√2√×(2+23√)=3√22√+26√
= √3(26√−22√)(26√+22√)(26√−22√)
= 23√(6√−2√)(26√2 −(22√)2)
23√(6√−2√)24−8=23√(6√−2√)16
3√(6√−2√)8=(18√−6√)8=(32√−6√)8
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
= (12+1–23√23√+12+1)
= (32–23√32+23√)
= (33√–4)2(33√)2–42
= (27+16–243√)(27–16)
= (43–243√)11
(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)
= 5(12)2+4(23√)2–12(12)2+(3√2)2
= (54+163–1)(14+34)
= (15+64–12)1244
=6712
2) Find the correct answer and explain your choice:
(i) 2tan30∘1+tan230∘ =
(A) sin 60∘ (B) cos 60∘ (C) tan 60∘ (D) sin 30∘
(ii) 1−tan245∘1+tan230∘ =
tan 90∘ (B) 1 (C) sin 45∘ (D) 0
(iii) sin 2P = 2 sin P is true when P =
0∘ (B) 30∘ (C) 45∘ (D) 60∘
(iv) 2tan30∘1−tan230∘ =
cos 60∘ (B) sin 60∘ (C) tan 60∘ (D) sin 30∘
Ans.-
(i) (A) IS correct.
2tan30∘1+tan230∘ = 2(1)3√1+(13√)2 (23√)1+13=(23√)43 =643√=3√2=sin60∘
(ii)(D) is correct
1−tan245∘1+tan230∘
= (1–12)(1+12)=02=0
(iii) (A) is correct
sin 2P = 2 sin P is true when
P = sin 2P = sin 0° = 0
2 sin P = 2sin 0° = 2×0 = 0
or,
sin 2P = 2sin PcosP
=>2sin PcosP = 2 sin P
=>2cos P = 2 =>cosP = 1
=>P = 0°
(iv) (C) is correct
2tan30∘1–tan230∘=2(13√1–(13√)2)
(23√)1–13=23√23=3–√=tan60∘
3) If tan (P + Q) = 3–√ and tan ( P – Q) = 13√;00<P+Q<=90∘;P>Q , calculate P and Q
Ans:- tan (P + Q) = 3–√
=>tan (P + Q) = tan 60°
=> (P + Q) = 60° … (i)
=>tan (P – Q) = 13√
=>tan (P – Q) = 30°
=> (P – Q) = 30° … (ii)
Adding (i) and (ii), we get
P + Q + P – Q = 60° + 30°
2P = 90°
=> P = 45°
Putting the value of P in equation (i)
45° + Q = 60°
=> Q = 60° – 45° = 15°
Hence, P = 45° and Q = 15°
4) Check whether the given statements are true or false, also give a reason for your answer:
(i) sin (P + Q) = sin P + sin Q.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cotP is not defined for P = 0°.
Ans:-
(i) False
Let P = 30° and Q = 60°, then
sin (P + Q) = sin (30° + 60°) = sin 90° = 1 and,
sin P + sin Q = sin 30° + sin 60°
= 12+3√2=1+3√2
(ii) True
Sin 0° = 0
Sin 30° = 12
Sin 45° = 12√
Sin 60° = 3√2
Sin 90° = 1
Thus, the value of sinθ increases as θ increases
(iii) False
Cos 0° = 1
Cos 30° = 3√2
Cos 45° = 12√
Cos 60° = 12
Cos 90° = 0
Thus, the value of Cosθ decreases as θ increases.
(iv) True
cotP=cosPSinP cot0∘=cos0∘Sin0∘=10=notdefined
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.3
1) Calculate:
(i) sin18∘cos72∘
(ii) tan26∘cot64∘
(iii) cos 48° – sin 42°
(iv) cosec 31° – sec 59°
Ans:-
(i) sin18∘cos72∘
= sin(90∘–18∘)cos72∘
= cos72∘cos72∘=1
(ii) tan26∘cot64∘
= tan(90∘–36∘)cot64∘
= cot64∘cot64∘=1
(iii) cos 48° – sin42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59°
= sec 59° – sec 59° = 0
2) Show that :
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Ans:-
(i)tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1
(ii) cos 38° cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52° = 0
3) We have 2P = cot ( P – 18 ° ), where 2P is an acute angle, calculate the value of P.
Ans:- According to question,
tan 2P = cot (P- 18°)
=>cot (90° – 2P) = cot (P -18°)
Equating angles,
=>90° – 2P = P- 18°
=>108° = 3P
=> P = 36
4) If tan P = cot Q, prove that P + Q = 90°.
AnswerAccording to question,
tanP = cot Q
=>tan P = tan (90° – Q)
=>P = 90° – Q
=>P + Q = 90°
5) If the value of sec 4P = cosec (P – 20°), in which 4P is an acute angle, find the value of P.
Ans:-According to question
sec 4P = cosec (P – 20°)
=> cosec (90° – 4P) = cosec (P – 20°)
Equating angles,
=> 90° – 4P= P- 20°
=> 110° = 5P
=> P = 22°
Q6) If X,Y and Z are interior angles of a triangle XYZ, then show that
sin (Y+Z/2) = cos X2
Answer
In a triangle, sum of all the interior angles
X + Y + Z = 180∘ ⇒ Y + Z = 180∘ – X
⇒ Y+Z2 = (180∘−X)2 ⇒ Y+Z2 = (90∘−X2) ⇒ sin (Y+Z2) = sin (90∘−X2) ⇒ sin (Y+Z2) = cosX2
Q7) Express sin 67∘ + cos 75∘ in terms of trigonometric ratios of angles between 0∘ and 45∘.
Answer
sin 67∘ + cos 75∘
= sin (90∘ – 23∘) + cos (90∘ – 15∘)
= cos 23∘ + sin 15∘
NCERT Solutions for Class 10 Maths Chapter 8 – Introduction to Trigonometry Exercise 8.4
Q1) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer
cosec2A−cot2A=1 ⇒ cosec2A = 1 + cot2A ⇒ 1sin2A = 1 + cot2A ⇒ sin2A = 1/(1+cot2A)
⇒ sin A= ±11+cot2A√ Now,
sin2A=11+cot2A ⇒1−cos2A=11+cot2A ⇒ cos2A = 1−11+cot2A ⇒cos2A = (1−1+cot2A)(1+cot2A) ⇒1sec2A = (cot2A)(1+cotA) ⇒secA = (1+cotA)(cot2A)
⇒secA=±1+cot2A√cotA
also,
tan A = sinAcosAand cot A = cosAsinA ⇒ tan A = 1cotA
Q2) Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer
We know that,
sec A = 1cosA ⇒cos A = 1secA also,
cos2A + sin2A = 1
⇒ sin2A = 1 – cos2A ⇒ sin2A = 1 – (1sec2A)
⇒ sin2A = (sec2A−1)sec2A ⇒ sin A=±sec2A−1√secA
also,
sin A = 1cosecA ⇒ cosec A = 1sinA ⇒cosec A=±secAsec2A−1√ Now,
sec2A – tan2A = 1
⇒ tan2A = sec2A + 1
⇒ tan A=sec2A+1−−−−−−−−√ also,
tan A = 1cotA ⇒ cot A = 1tanA ⇒ cot A=±1sec2A+1√
Q3 Evaluate :
(i) (sin263∘+sin227∘)(cos217∘+cos273∘) (ii) sin25∘cos65+∘+cos25∘sin65∘
Answer
(i) (sin263∘+sin227∘)(cos217∘+cos273∘)
= [sin2(90∘–27∘)+sin227∘][cos2(90∘–73∘)+cos273∘] =(cos227∘+sin227∘)(sin227∘+cos273∘) = 11 =1 ( becausesin2A+cos2A=1)
(ii) sin25∘cos65+∘+cos25∘sin65∘ =sin(90∘−25∘)cos65∘+cos(90∘−65∘)sin65∘
=cos65∘cos65∘+sin65∘sin65∘
= cos65∘+sin65∘=1
4) Choose the correct option. Justify your choice.
(i) 9 sec2A – 9 tan2A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan Θ + sec Θ) (1 + cot Θ – cosec Θ)
(A) 0 (B) 1 (C) 2 (D) – 1
(iii) (secA + tanA) (1 – sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan2A1+cot2A=
(A) sec2A
(B) -1
(C) cot2A
(D) tan2A
Answer
(i) (B) is correct.
9 sec2A– 9 tan2A
= 9 (sec2A– tan2A )
= 9×1 = 9 ( because sec2A– tan2A = 1)
(ii) (C) is correct
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ – 1/sin θ)
= (cosθ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ
= (cosθ+sin θ)2-12/(cos θ sin θ)
= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)
= (1+ 2cos θ sin θ -1)/(cos θ sin θ)
= (2cos θ sin θ)/(cos θ sin θ) = 2
(iii) (D) is correct.
(secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1+sin A/cos A) (1 – sinA)
= (1 – sin2A)/cos A
= cos2A/cos A = cos A
(iv) (D) is correct.
1+tan2A/1+cot2A
= (1+1/cot2A)/1+cot2A
= (cot2A+1/cot2A)×(1/1+cot2A)
= 1/cot2A = tan2A
Q5) Prove the following identities, where the angles involved are acute angles for which theexpressions are defined.
(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(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]
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cosA+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
(vi)1+sinA1−sinA−−−−−√=secA+tanA
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately] (x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
Answer
(i) (cosecΘ−cotΘ)2 = (1-cos θ)/(1+cos θ)
L.H.S. = (cosecΘ−cotΘ)2
=(cosec2Θ+cot2Θ−2cosecΘcotΘ)
=(1sin2Θ+cos2Θsin2Θ−2cosΘsin2Θ)
= (1 + cos2Θ – 2cos θ)/(1 – cos2Θ)
= (1−cosΘ)2 /(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
= [cos2A +(1+sinA)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
= (1 + 1 + 2sin A)/(1+sin A)cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)] = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cosθ-sin θ)/cos θ] = sin2Θ /[cos θ(sin θ-cos θ)] + cos2Θ /[sin θ(cos θ-sin θ)] = sin2Θ /[cos θ(sin θ-cos θ)] – cos2Θ /[sin θ(sin θ-cos θ)] = 1/(sin θ-cos θ) [(sin2Θ /cos θ) – (cos2Θ /sin θ)] = 1/(sin θ-cos θ) × [(sin3Θ – cos3Θ)/sin θ cos θ] = [(sin θ-cos θ)(sin2Θ +cos2Θ +sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ] = (1 + sin θ cos θ)/sin θ cos θ)
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.
(iv) (1 + sec A)/sec A = sin2Θ /(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
= cos A + 1
R.H.S. = sin2Θ /(1-cos A)
= (1 –cos2Θ)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
= cos A + 1
L.H.S. = R.H.S.
(v) (cos A–sin A+1)/(cosA+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cosA+sin A–1)
Dividing Numerator and Denominator by sin A,
= (cos A–sin A+1)/sin A/(cosA+sin A–1)/sin A
= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1)
= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
= (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
= cot A + cosec A = R.H.S.
(vi)1+sinA1−sinA−−−−−√=secA+tanA
Dividing Numerator and Denominator of L.H.S. by cos A,
= 1cosA+sinAcosA√1cosA−sinAcosA√
= secA+tanA√secA−tanA√
= secA+tanA√secA−tanA√XsecA+tanA√secA+tanA√
=(secA+tanA)2√sec2A−tan2A√
=secA+tanA1
= sec A + tan A = R.H.S.
(vii) (sin θ – 2sin3Θ)/(2cos3Θ -cos θ) = tan θ
L.H.S. = (sin θ – 2sin3Θ)/(2cos3Θ – cos θ)
= [sin θ(1 – 2sin2Θ)]/[cos θ(2cos2Θ – 1)] = sin θ[1 – 2(1-cos2Θ)]/[cosθ(2cos2Θ-1)] = [sin θ(2cos2Θ -1)]/[cos θ(2cos2Θ -1)] = tan θ = R.H.S.
(viii) (sinA+cosecA)2 + (cosA+secA)2 = 7+tan2A +cot2A L.H.S. = (sinA+cosecA)2 + (cosA+secA)2 = (sin2A + cosec2A + 2 sin A cosec A) + (tcos2A + sec2A + 2 cos A sec A)
= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A = 1 + 2 + 2 + 2 + tan2A + cot2A = 7+tan2A+cot2A = R.H.S.
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin2A)/sin A][(1-cos2A)/cos A] = (cos2A/sin A)×(sin2A/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin2A+cos2A)/sin A cos A] = cos A sin A
L.H.S. = R.H.S.
(x) (1+tan2A/1+cot2A) = (1−tanA1−cotA)2 =tan2A L.H.S. = (1+tan2A/1+cot2A)
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A] = tan2A
After learning the Trigonometry chapter of Class 10, now you must have understood the following concepts; say,
In a right-angle triangle ABC, which is right angled at B,
- Sina A = (side opp to angle A) / hypotenuse
- Cos A = (side adjacent to angle A) / hypotenuse
- Tan A = (side opp to angle A) / (side adjacent to angle A)
- Cosec A = 1 / Sin A
- Sec A= 1 / Cos A
- Tan A= 1 / Cot A
- Tan A = Sin A / Cos A
- Sin (90° – A) = cos A
- Cos (90° – A) = sin A
- Tan (90° – A) = cot A
- Cot (90° – A) = tan A
- Sec (90° – A) = cosec A
- Cosec (90° – A) = sec A