# NCERT Solutions For Class 10 Maths Chapter 8

## NCERT Solutions Class 10 Maths Introduction to 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 NCERT questions of Linear Equations in two variables. Maths is a subject which requires students understanding and reasoning skills. Along with this, it also requires students to practice maths on a regular basis. Students of Class 10 are suggested to solve NCERT questions in order to practice the questions that are usually asked in the examination. From all the chapters, NCERT Solutions Class 10 Maths Introduction to Trigonometry is an important topic for the students, and one needs to practice thoroughly to score well in the examination. BYJU’S provides NCERT Solutions Class 10 Maths Introduction to Trigonometry in the pdf format. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for maths class 10 chapter 8.

NCERT solutions for maths class 10 chapter 8 pdf is prepared by our subject experts under the guidelines of NCERT to assists students in their board exam preparations.

Class 10 Maths NCERT Solutions for chapter 8 includes the topic – Introduction. Trigonometric Ratios: Examples, Exercise. Trigonometric Ratios of Some Specific Angles : Trigonometric Ratios of 45°, Trigonometric Ratios of 30° and 60°, Trigonometric Ratios of 0° and 90°,  Examples, Exercise. Trigonometric Ratios of Complementary Angles: Examples, Exercise. Trigonometric Identities: Examples, Exercise. NCERT Solutions Class 10 Maths Introduction to Trigonometry is provided here for better understanding and clarification of the chapter.

### NCERT Solutions Class 10 Maths Chapter 8 Exercises

Exercise 8.1

Q1) In $$\bigtriangleup ABC$$ , $$90^{\circ}$$ at B, AB=24cm, BC = 7cm.

Determine:

(i)sin(A), cos(A)

(ii) sin(C), cos(C)

Ans.) In $$\bigtriangleup ABC$$ , $$\angle B=90^{\circ}$$

By Applying Pythagoras theorem, we get

$$AC^{2}=AB^{2}+BC^{2}$$

$$(24)^{2}+7^{2}$$ =(576+49)

$$AC^{2}$$ = $$625cm^{2}$$

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,

$$13^{2}=QR^{2}+12^{2}$$

$$169=QR^{2}+144$$

$$QR^{2}=169-144=25$$

$$QR=\sqrt{25}=5$$

tan(P) = $$\frac{opposite\;side}{adjacent\;side}=\frac{QR}{PQ}=\frac{5}{12}$$

cot(P) = $$\frac{adjacent\;side}{opposite\;side}$$ = $$\frac{PQ}{QR}$$ = $$\frac{5}{12}$$

tan(P) – cot(R) = $$\frac{5}{12}-\frac{5}{12}=0$$

Therefore ,tan(P) – cot(R) = 0

Q3) If sin(A) = 3/4, calculate cos(A) and tan(A)

Ans.) Let $$\bigtriangleup ABC$$ , 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,

$$AC^{2}=AB^{2}+BC^{2}$$

$$(4k)^{2}=AB^{2}+(3k)^{2}$$

$$16k^{2}-9k^{2}=AB^{2}$$

$$AB^{2}=7k^{2}$$

$$AB=\sqrt{7}k$$

cos(A) = AB/AC = $$\sqrt{7}k/4k=\sqrt{7}/4$$

tan(A) = BC/AB =$$3k/\sqrt{7}=3/\sqrt{7}$$

Q4) In question given below 15cot(A) = 8 ,find sin A and sec A.

Ans.)  Let $$\bigtriangleup ABC$$ 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,

$$AC^{2}=AB^{2}+BC^{2}$$

$$AC^{2}=(8k)^{2}+(15k)^{2}$$

$$AC^{2}=64k^{2}+225k^{2}$$

$$AC^{2}=289k^{2}$$

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  $$\bigtriangleup ABC$$ 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,

$$OP^{2}=OM^{2}+MP^{2}$$

$$(13k)^{2}=(12k)^{2}+MP^{2}$$

$$169(k)^{2}-144(k)^{2}=MP^{2}$$

$$MP^{2}=25k^{2}$$

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 $$\angle A$$ and $$\angle B$$ are acute angles such that

cos(A) = cos(B), then show $$\angle A$$ =$$\angle B$$ .

Ans.) Let  $$\bigtriangleup ABC$$ in which $$CD\perp AB$$ .

A/q,

cos(A) = cos(B)

AC=kBC  …. (ii)

By applying Pythagoras theorem in $$\bigtriangleup CAD$$ and $$\bigtriangleup CBD$$ we get,

$$CD^{2}=AC^{2}-AD^{2}$$ ….(iv)

From the equations (iii) and (iv) we get,

$$AC^{2}-AD^{2}=BC^{2}-BD^{2}$$

$$AC^{2}-AD^{2}=BC^{2}-BD^{2}$$

$$k^{2}(BC^{2}-BD^{2})=BC^{2}-BD^{2}$$

$$k^{2}=1$$

Putting this value in equation (ii) , we obtain

AC = BC

$$\angle A=\angle B$$ (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) $$cot^{2}\Theta$$

Ans.) Let $$\bigtriangleup ABC$$ in which  $$\angle B=90^{\circ}$$

and $$\angle C=\Theta$$

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 $$\bigtriangleup ABC$$ we get.

$$AC^{2}=AB^{2}+BC^{2}$$

$$AC^{2}=(8k)^{2}+(7k)^{2}$$

$$AC^{2}=64k^{2}+49k^{2}$$

$$AC^{2}=113k^{2}$$

$$AC=\sqrt{113}k$$

sin Ѳ = AB/AC = $$8k/\sqrt{113}k= 8/\sqrt{113}$$

and cos Ѳ = BC/AC = $$7k/\sqrt{113}k= 7/\sqrt{113}$$

(i) (1+sin Ѳ)(1-sinѲ)/(1+cosѲ)(1-cos Ѳ) = $$(1-sin^{2}\Theta)/(1-cos^{2}\Theta )$$

= $${1-(8/\sqrt{113})^{2}}/{1-(7/\sqrt{113})^{2}}$$

= {1-(64/113)}/{1-(49/113)} = {(113-64)/113}/{(113-49)/113} = 49/64

(ii) $$cot^{2}\Theta =(7/8)^{2}=49/64$$

Q8) If 3cot(A) = 4/3, check whether $$(1-tan^{2}A)/(1+tan^{2}A)= cos^{2}A-sin^{2}A$$ or not.

Ans.) Let $$\bigtriangleup ABC$$ in which $$\angle B=90^{\circ}$$

A/q,

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

$$AC^{2}=AB^{2}+BC^{2}$$

$$AC^{2}=(4k)^{2}+(3k)^{2}$$

$$AC^{2}=16k^{2}+9k^{2}$$

$$AC^{2}=25k^{2}$$

$$AC=5k$$

tan(A) = BC/AB = 3/4

sin(A) = BC/AC = 3/5

cos(A) = AB/AC = 4/5

L.H.S. = $$(1-tan^{2}A)(1+tan^{2}A)=1-(3/4)^{2}/1+(3/4)^{2}=(1-9/16)/(1+9/16)=(16-9)/(16+9)=7/25$$

R.H.S. =$$cos^{2}A-sin^{2}A=(4/5)^{2}-(3/4)^{2}=(16/25)-(9/25)=7/25$$

R.H.S. =L.H.S.

Hence, $$(1-tan^{2}A)/(1+tan^{2}A)=cos^{2}A-sin^{2}A$$

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

LetΔEFG in which $$\angle F=90^{\circ}$$, E/q

$$tan E=\frac{FC}{EF}$$

$$tan E=\frac{FC}{EF}=\frac{1}{\sqrt{3}}$$

Where k is the positive real number of the problem

By Pythagoras theorem in ΔEFG we get:

$$EG^{2}=EF^{2}+FG^{2}$$

$$EG^{2}=(\sqrt{3k}^{2}))+K^{2}$$

$$EG^{2}=3k^{2}+K^{2}$$

$$EG^{2}=4k^{2}$$

$$EG=2K$$

sinE = FG/EG = 1/2

cosE = EF/EG =  $$\frac{\sqrt{3}}{2}$$  ,
sin G = EF/EG = $$\frac{\sqrt{3}}{2}$$ cosE = FG/EG = 1/2
(i) sin EcosG + cosE sin G = (1/2\ast1/2) + ($$\frac{\sqrt{3}}{2}\ast\frac{\sqrt{3}}{2}$$)= 1/4+3/4 = 4/4 = 1
(ii) cosEcosG – sin E sin C = $$(\frac{\sqrt{3}}{2}\ast\frac{1}{2})-(\frac{\sqrt{3}}{2}\ast\frac{1}{2})$$= $$(\frac{\sqrt{3}}{4})-(\frac{\sqrt{3}}{4})$$= 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.

Given that, MO + NO = 25 , MN = 5
Let MO be x.  ∴ NO = 25 – x

By Pythagoras theorem ,
$$MO^{2}=MN^{2}+NO^{2}$$
$$X^{2}=5^{2}+(25-x)^{2}$$
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.

(i) False.

In ΔMNC in which $$\angle N$$ = $$90^{\circ}$$,

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.

$$MC^{2}=MN^{2}+NC^{2}$$
$$5^{2}=3^{2}+4^{2}$$
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,
$$MC^{2}=MN^{2}+NC^{2}$$
$$(12k)^{2}=(5k)^{2}+NC^{2}$$
$$NC^{2}+25k^{2}=144K^{2}$$
$$NC^{2}=119k^{2}$$

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 $$\angle M$$.
(v) False.

$$sin\Theta$$ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ $$sin\Theta$$will always less than 1 and it can never be 4/3 for any value of $$\Theta$$.

Exercise 8.2

1) Calculate the following:

• $$sin60^{\circ}cos30^{\circ}+sin30^{\circ}cos60^{\circ}$$

• $$2tan^{2}45^{\circ}+co^{2}30^{\circ}-sin^{2}60^{\circ}$$

• $$\frac{cos45^{\circ}}{\left ( sec30^{\circ}+cosec30^{\circ} \right )}$$

• $$\frac{\left ( sin30^{\circ}+tan45^{\circ}-cosec60^{\circ} \right )}{\left ( sec30^{\circ}+cos60^{\circ}+cot45^{\circ} \right )}$$

• $$\frac{\left ( 5cos^{2}60^{\circ}+4sec^{2}30^{\circ}-tan^{2} 45^{\circ}\right )}{\left ( sin^{2} 30^{\circ}+cos^{2}30^{\circ}\right )}$$

Ans.- (i) $$sin60^{\circ}cos30^{\circ}+sin30^{\circ}cos60^{\circ}$$

= $$\left ( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\right )+\left ( \frac{1}{2}\times \frac{1}{2} \right )=\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1$$

(ii) $$2tan^{2}45^{\circ}+co^{2}30^{\circ}-sin^{2}60^{\circ}$$

=$$2\times \left ( 1 \right )^{2}+\left ( \frac{\sqrt{3}}{2} \right )^{2}-\left( \frac{\sqrt{3}}{2} \right )^{2}=2$$

(iii) $$\frac{cos45^{\circ}}{\left ( sec30^{\circ}+cosec30^{\circ} \right )}$$

= $$\frac{1}{\frac{\sqrt{2}}{2\sqrt{3}+2}}=\frac{1}{\frac{\sqrt{2}}{\frac{\left ( 2+2\sqrt{3} \right )}{\sqrt{3}}}}$$

= $$\frac{\sqrt{3}}{\sqrt{2}\times \left ( 2+2\sqrt{3} \right )}=\frac{\sqrt{3}}{2\sqrt{2}+2\sqrt{6}}$$

= $$\frac{\sqrt{} 3\left ( 2\sqrt{6} -2\sqrt{2}\right )}{\left ( 2\sqrt{6}+2\sqrt{2} \right )\left ( 2\sqrt{6} -2\sqrt{2}\right )}$$

= $$\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{\left ( 2\sqrt{6}^{2}\ -\left ( 2\sqrt{2} \right )^{2}\right )}$$

$$\frac{2\sqrt{3}\left ( \sqrt{6} -\sqrt{2}\right )}{24-8}=\frac{2\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{16}$$

$$\frac{\sqrt{3}\left ( \sqrt{6}-\sqrt{2} \right )}{8}=\frac{\left ( \sqrt{18}-\sqrt{6} \right )}{8}=\frac{\left ( 3\sqrt{2}-\sqrt{6} \right )}{8}$$

(iv)  (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

= $$\left (\frac{ \frac{1}{2} + 1 – \frac{2}{\sqrt{3}}}{2\sqrt{3} + \frac{1}{2} + 1} \right )$$

= $$\left ( \frac{\frac{3}{2} – \frac{2}{\sqrt{3}}}{\frac{3}{2} + \frac{2}{\sqrt{3}}} \right )$$

= $$\frac{\left ( 3\sqrt{3} – 4 \right )^{2}}{\left ( 3\sqrt{3} \right )^{2} – 4^{2}}$$

= $$\frac{\left ( 27 + 16 – 24\sqrt{3} \right )}{\left ( 27 – 16 \right )}$$

= $$\frac{\left ( 43 – 24\sqrt{3} \right )}{11}$$

(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)

= $$5\left ( \frac{1}{2} \right )^{2} + 4 \left ( \frac{2}{\sqrt{3}} \right )^{2} – \frac{1^{2}}{\left ( \frac{1}{2} \right )^{2} + \left ( \frac{\sqrt{3}}{2} \right )^{2} }$$

= $$\frac{\left ( \frac{5}{4} + \frac{16}{3 – 1} \right )}{\left ( \frac{1}{4} + \frac{3}{4} \right )}$$

= $$\frac{\frac{\left ( 15 + 64 – 12 \right )}{12}}{\frac{4}{4}}$$

=$$\frac{67}{12}$$

(i)  $$\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}$$ =

(A) sin $$60^{\circ}$$ (B) cos $$60^{\circ}$$ (C) tan $$60^{\circ}$$ (D)        sin $$30^{\circ}$$

(ii) $$\frac{1-tan^{2}45^{\circ}}{1+tan^{2}30^{\circ}}$$ =

tan $$90^{\circ}$$ (B) 1  (C) sin $$45^{\circ}$$  (D) 0

(iii) sin 2P = 2 sin P is true when P =

$$0^{\circ}$$ (B)  $$30^{\circ}$$    (C)  $$45^{\circ}$$   (D)  $$60^{\circ}$$

(iv)    $$\frac{2tan30^{\circ}}{1-tan^{2}30^{\circ}}$$ =

cos $$60^{\circ}$$ (B)  sin $$60^{\circ}$$   (C)  tan $$60^{\circ}$$     (D)  sin $$30^{\circ}$$

Ans.-

(i)  (A) IS correct.

$$\frac{2tan30^{\circ}}{1+tan^{2}30^{\circ}}$$ = $$2\frac{\left ( 1 \right )}{\frac{\sqrt{3}}{1}}+\left ( \frac{1}{\sqrt{3}} \right )^{2}$$

$$\frac{\left ( 2\sqrt{3} \right )}{1+\frac{1}{3}} = \frac{\left (\frac{2}{\sqrt{3}}\right )}{\frac{4}{3}}$$

$$= \frac{6}{4\sqrt{3}} = \frac{\sqrt{3}}{2} = sin 60^{\circ}$$

(ii)(D) is correct

$$\frac{1-tan^{2}45^{\circ}}{1+tan^{2}30^{\circ}}$$

= $$\frac{\left ( 1 – 1^{2} \right )}{\left ( 1 + 1^{2} \right )} = \frac{0}{2} = 0$$

(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

$$\frac{2tan30^{\circ}}{ 1 – tan^{2}30^{\circ}} = 2\left ( \frac{\frac{1}{\sqrt{3}}}{1 – \left ( \frac{1}{\sqrt{3}} \right )^{2}} \right )$$

$$\frac{\left ( 2\sqrt{3} \right )}{ 1 – \frac{1}{3}} = \frac{2\sqrt{3}}{\frac{2}{3}} = \sqrt{3} = tan 60^{\circ}$$

3) If tan (P + Q) = $$\sqrt{3}$$ and tan ( P – Q) = $$\frac{1}{\sqrt{3} }$$;$$0^{0} < P + Q <= 90^{\circ}; P>Q$$
, calculate P and Q

Ans:-     tan (P + Q) = $$\sqrt{3}$$

=>tan (P + Q) = tan 60°

=> (P + Q) =  60°     … (i)

=>tan (P – Q) = $$\frac{1}{\sqrt{3}}$$

=>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°

= $$\frac{1}{2} + \frac{\sqrt{3}}{2} = 1 + \frac{\sqrt{3}}{2}$$

(ii) True

Sin 0° = 0

Sin 30° = $$\frac{1}{2}$$

Sin 45° = $$\frac{1}{\sqrt{2}}$$

Sin 60° = $$\frac{\sqrt{3}}{2}$$

Sin 90° = 1

Thus, the value of $$sin \theta$$ increases as $$\theta$$ increases

(iii) False

Cos 0° = 1

Cos 30° = $$\frac{\sqrt{3}}{2}$$

Cos 45° = $$\frac{1}{\sqrt{2}}$$

Cos 60° = $$\frac{1}{2}$$

Cos 90° = 0

Thus, the value of $$Cos \theta$$ decreases as $$\theta$$ increases.

(iv) True

$$cot P = \frac{cos P}{ Sin P}$$

$$cot 0^{\circ} = \frac{cos 0^{\circ}}{ Sin 0^{\circ}} = \frac{1}{0} = not \: defined$$

Exercise 8.3

1) Calculate:

(i) $$\frac{sin 18 ^{\circ}}{cos 72 ^{\circ}}$$

(ii) $$\frac{tan 26 ^{\circ}}{cot 64 ^{\circ}}$$

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°

Ans:-

(i) $$\frac{sin 18 ^{\circ}}{cos 72 ^{\circ}}$$

= $$\frac{{sin }\left ( 90 ^{\circ} – 18^{\circ} \right )}{cos 72^{\circ}}$$

= $$\frac{cos 72^{\circ}}{cos 72^{\circ}} = 1$$

(ii) $$\frac{tan 26 ^{\circ}}{cot 64 ^{\circ}}$$

= $$\frac{{{tan }\left ( 90 ^{\circ} – 36^{\circ} \right )} }{cot 64^{\circ}}$$

=  $$\frac{cot 64^{\circ}}{cot 64^{\circ}} = 1$$

(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°.

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 $$\frac{X}{2}$$

In a triangle, sum of all the interior angles

X + Y + Z = $$180^{\circ}$$

$$\Rightarrow$$ Y + Z = $$180^{\circ}$$ – X

$$\Rightarrow$$ $$\frac{Y+Z}{2}$$ = $$\frac{(180^{\circ}-X)}{2}$$

$$\Rightarrow$$ $$\frac{Y+Z}{2}$$ = $$\left ( 90^{\circ}-\frac{X}{2} \right )$$

$$\Rightarrow$$ sin $$\left ( \frac{Y+Z}{2} \right )$$ = sin $$\left ( 90^{\circ}-\frac{X}{2} \right )$$

$$\Rightarrow$$ sin $$\left ( \frac{Y+Z}{2} \right )$$ = cos$$\frac{X}{2}$$

Q7) Express sin $$67^{\circ}$$ + cos $$75^{\circ}$$ in terms of trigonometric ratios of angles between $$0^{\circ}$$ and $$45^{\circ}$$.

sin $$67^{\circ}$$ + cos $$75^{\circ}$$

= sin ($$90^{\circ}$$ – $$23^{\circ}$$) + cos ($$90^{\circ}$$ – $$15^{\circ}$$)
= cos $$23^{\circ}$$ + sin $$15^{\circ}$$

Excercise 8.4

Q1) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

$$cosec^{2}A-cot^{2}A=1$$
$$\Rightarrow$$ $$cosec^{2}A$$ = 1 + $$cot^{2}A$$
$$\Rightarrow$$ $$\frac{1}{sin^{2}A}$$ = 1 + $$cot^{2}A$$
$$\Rightarrow$$ $$sin^{2}A$$ = 1/(1+$$cot^{2}A$$)
$$\Rightarrow$$ sin A= $$\frac{\pm 1}{\sqrt{1+cot^{2}A}}$$
Now,
$$sin^{2}A=\frac{1}{1+cot^{2}A}$$
$$\Rightarrow$$$$1-cos^{2}A=\frac{1}{1+cot^{2}A}$$
$$\Rightarrow$$ $$cos^{2}A$$ = $$1-\frac{1}{1+cot^{2}A}$$
$$\Rightarrow$$cos2A = $$\frac{(1-1+cot^{2}A)}{(1+cot^{2}A)}$$
$$\Rightarrow$$$$\frac{1}{sec^{2}A}$$ = $$\frac{(cot^{2}A)}{(1+cot^{A})}$$
$$\Rightarrow$$secA = $$\frac{(1+cot^{A})}{(cot^{2}A)}$$

$$\Rightarrow$$secA=$$\frac{\pm \sqrt{1+cot^{2}A}}{cotA}$$

also,
tan A = $$\frac{sin A}{cos A}$$and cot A = $$\frac{cos A}{sin A}$$

$$\Rightarrow$$ tan A = $$\frac{1}{cot A}$$

Q2) Write all the other trigonometric ratios of $$\angle A$$ in terms of sec A.

We know that,
sec A = $$\frac{1}{cos A}$$
$$\Rightarrow$$cos A = $$\frac{1}{sec A}$$
also,
$$cos^{2}A$$ + $$sin^{2}A$$ = 1
$$\Rightarrow$$  $$sin^{2}A$$ = 1 – $$cos^{2}A$$
$$\Rightarrow$$  $$sin^{2}A$$ = 1 – ($$\frac{1}{sec^{2}A}$$)
$$\Rightarrow$$  $$sin^{2}A$$ = $$\frac{\left ( sec^{2}A-1 \right )}{sec^{2}A}$$

$$\Rightarrow$$  sin A=$$\frac{\pm \sqrt{sec^{2}A-1}}{sec A}$$

also,
sin A = $$\frac{1}{cosec A}$$
$$\Rightarrow$$ cosec A = $$\frac{1}{sin A}$$

$$\Rightarrow$$cosec A=$$\frac{\pm sec A}{\sqrt{sec^{2}A-1}}$$
Now,
$$sec^{2}A$$ – $$tan^{2}A$$ = 1
$$\Rightarrow$$ $$tan^{2}A$$ = $$sec^{2}A$$ + 1

$$\Rightarrow$$ tan A=$$\sqrt{sec^{2}A+1}$$
also,
tan A = $$\frac{1}{cot A}$$
$$\Rightarrow$$ cot A = $$\frac{1}{tan A}$$

$$\Rightarrow$$  cot A=$$\frac{\pm 1}{\sqrt{sec^{2}A+1}}$$

Q3 Evaluate :

(i) $$\frac{\left ( sin^{2}63^{\circ}+sin^{2}27^{\circ} \right )}{\left ( cos^{2}17^{\circ}+cos^{2}73^{\circ} \right )}$$
(ii)  $$sin 25^{\circ} cos 65+^{\circ}+cos 25^{\circ}sin 65^{\circ}$$

(i) $$\frac{\left ( sin^{2}63^{\circ}+sin^{2}27^{\circ} \right )}{\left ( cos^{2}17^{\circ}+cos^{2}73^{\circ} \right )}$$

= $$\frac{[sin^{2}\left ( 90^{\circ} – 27^{\circ}\right )+sin^{2}27^{\circ}]}{[cos^{2}\left ( 90^{\circ} – 73^{\circ}\right )+cos^{2}73^{\circ}]}$$
=$$\frac{(cos^{2}27^{\circ}+sin^{2}27^{\circ})}{(sin^{2}27^{\circ}+cos^{2}73^{\circ})}$$
= $$\frac{1}{1}$$ =1          $$\left ( \ because sin^{2}A+cos^{2}A=1 \right )$$

(ii) $$sin 25^{\circ} cos 65+^{\circ}+cos 25^{\circ}sin 65^{\circ}$$
=$$sin(90^{\circ}-25^{\circ})cos 65^{\circ}+cos(90^{\circ}-65^{\circ})sin65^{\circ}$$

=$$cos 65^{\circ}cos 65^{\circ}+sin 65^{\circ}sin 65^{\circ}$$

= $$cos 65^{\circ}+sin 65^{\circ}$$=1

4) Choose the correct option. Justify your choice.
(i) 9 $$sec^{2}A$$ – 9 $$tan^{2}A$$ =
(A) 1                 (B) 9              (C) 8                (D) 0
(ii) (1 + tan $$\Theta$$ + sec $$\Theta$$) (1 + cot $$\Theta$$ – cosec $$\Theta$$)
(A) 0                 (B) 1              (C) 2                (D) – 1
(iii) (secA + tanA) (1 – sinA) =
(A) secA           (B) sinA        (C) cosecA      (D) cosA

(iv) $$\frac{1+tan^{2}A}{1+cot^{2}A}$$=

(A) $$sec^{2}A$$

(B) -1

(C) $$cot^{2}A$$

(D) $$tan^{2}A$$

(i) (B) is correct.

9 $$sec^{2}A$$- 9 $$tan^{2}A$$

= 9 ($$sec^{2}A$$- $$tan^{2}A$$                 )
= 9×1 = 9             ($$\ because$$  $$sec^{2}A$$- $$tan^{2}A$$ = 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)= (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)$$\sqrt{\frac{1+sin A}{1-sin A}}=sec A+tan A$$

(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)+ (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

(i) $$\left ( cosec\Theta -cot\Theta \right )^{2}$$ = (1-cos θ)/(1+cos θ)
L.H.S. =  $$\left ( cosec\Theta -cot\Theta \right )^{2}$$

=$$\left ( cosec^{2}\Theta +cot^{2}\Theta-2cosec\Theta cot\Theta \right )$$

=$$\left ( \frac{1}{sin^{2}\Theta }+\frac{cos^{2}\Theta }{sin^{2}\Theta }-2 \frac{cos\Theta }{sin^{2}\Theta } \right )$$

= (1 + $$cos^{2}\Theta$$ – 2cos θ)/(1 – $$cos^{2}\Theta$$)
= $$\left ( 1-cos \Theta\right )^{2}$$ /(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
= [$$cos^{2}A$$ +$$\left ( 1+sin A\right )^{2}$$]/(1+sin A)cos A
= ($$cos^{2}A$$ + $$sin^{2}A$$ + 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]
= $$sin^{2}\Theta$$ /[cos θ(sin θ-cos θ)] + $$cos^{2}\Theta$$ /[sin θ(cos θ-sin θ)]
= $$sin^{2}\Theta$$ /[cos θ(sin θ-cos θ)] – $$cos^{2}\Theta$$ /[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [($$sin^{2}\Theta$$ /cos θ) – ($$cos^{2}\Theta$$ /sin θ)]
= 1/(sin θ-cos θ) × [($$sin^{3}\Theta$$ – $$cos^{3}\Theta$$)/sin θ cos θ]
= [(sin θ-cos θ)($$sin^{2}\Theta$$ +$$cos^{2}\Theta$$ +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 = $$sin^{2}\Theta$$ /(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. = $$sin^{2}\Theta$$ /(1-cos A)
= (1 -$$cos^{2}\Theta$$)/(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 $$cosec^{2}A$$ = 1+$$cot^{2}A$$.
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 – $$cosec^{2}A$$ + $$cot^{2}A$$ + cosec A)/(cot A+ 1 – cosec A) (using $$cosec^{2}A$$ – $$cot^{2}A$$ = 1)
= [(cot A + cosec A) – ($$cosec^{2}A$$ – $$cot^{2}A$$)]/(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)$$\sqrt{\frac{1+sin A}{1-sin A}}=sec A+tan A$$

Dividing Numerator and Denominator of L.H.S. by cos A,

= $$\frac{\sqrt{\frac{1}{cos A}+\frac{sin A}{cos A}}}{\sqrt{\frac{1}{cos A}-\frac{sin A}{cos A}}}$$

= $$\frac{\sqrt{sec A+tan A}}{\sqrt{sec A- tan A}}$$

= $$\frac{\sqrt{sec A+tan A}}{\sqrt{sec A- tan A}}X\frac{\sqrt{sec A+tan A}}{\sqrt{sec A+ tan A}}$$

=$$\frac{\sqrt{\left ( sec A+tan A \right )^{2}}}{\sqrt{sec^{2}A-tan^{2}A} }$$

=$$\frac{sec A+tan A}{1}$$

= sec A + tan A = R.H.S.

(vii) (sin θ – 2$$sin^{3}\Theta$$)/(2$$cos^{3}\Theta$$ -cos θ) = tan θ
L.H.S. = (sin θ – 2$$sin^{3}\Theta$$)/(2$$cos^{3}\Theta$$ – cos θ)
= [sin θ(1 – 2$$sin^{2}\Theta$$)]/[cos θ(2$$cos^{2}\Theta$$ – 1)]
= sin θ[1 – 2(1-$$cos^{2}\Theta$$)]/[cosθ(2$$cos^{2}\Theta$$-1)]
= [sin θ(2$$cos^{2}\Theta$$ -1)]/[cos θ(2$$cos^{2}\Theta$$ -1)]
= tan θ = R.H.S.

(viii) $$\left ( sin A+cosec A \right )^{2}$$ + $$\left (cosA+sec A \right )^{2}$$ = 7+$$tan^{2}A$$ +$$cot^{2}A$$
L.H.S. =  $$\left ( sin A+cosec A \right )^{2}$$ + $$\left (cosA+sec A \right )^{2}$$
= ($$sin^{2}A$$ + $$cosec^{2}A$$ + 2 sin A cosec A) + ($$tcos^{2}A$$ + $$sec^{2}A$$ + 2 cos A sec A)
= ($$sin^{2}A$$ + $$cos^{2}A$$) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + $$tan^{2}A$$ + 1 + $$cot^{2}A$$
= 1 + 2 + 2 + 2 + $$tan^{2}A$$ + $$cot^{2}A$$
= 7+$$tan^{2}A$$+$$cot^{2}A$$ = 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-$$sin^{2}A$$)/sin A][(1-$$cos^{2}A$$)/cos A]
= ($$cos^{2}A$$/sin A)×($$sin^{2}A$$/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[($$sin^{2}A$$+$$cos^{2}A$$)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.

(x)  (1+$$tan^{2}A$$/1+$$cot^{2}A$$) = $$\left ( \frac{1-tan A}{1-cot A} \right )^{2}$$ =$$tan^{2}A$$
L.H.S. = (1+$$tan^{2}A$$/1+$$cot^{2}A$$)
= (1+$$tan^{2}A$$/1+1/$$tan^{2}A$$)
= 1+$$tan^{2}A$$/[(1+$$tan^{2}A$$)/$$tan^{2}A$$]
= $$tan^{2}A$$

Downloadable files of ebooks, notes, pdfs are available at BYJU’S for students to start their exam preparation without delay. Once you have registered with BYJU’S access to online course is very simple. After studying this chapter you know the following concepts: In a right angle triangle ABC where the right angle is at B:

1. Sina A = (side opp to angle A) / hypotenuse
2. Cos A = (side adjacent to angle A) /  hypotenuse
3. Tan A = (side opp to angle A) / (side adjacent to angle A)
4. Cosec A = 1 / Sin A
5. Sec A= 1 / Cos A
6. Tan A= 1 / Cot A
7. Tan A = Sin A / Cos A
8. Sin (90° – A) = cos A
9. Cos (90° – A) = sin A
10. Tan (90° – A) = cot A
11. Cot (90° – A) = tan A
12. Sec (90° – A) = cosec A
13. Cosec (90° – A) = sec A

Sign up on BYJU’S the learning app to learn more. Apart from NCERT Solutions for Class 10 Maths chapter 8, get NCERT solutions for other maths chapters – NCERT Solutions for Class 10 Maths as well. Students can download worksheets, assignments, NCERT Class 10 Maths Solutions PDF and other study materials for exam preparation and score better marks.