Class 10 Maths

# Trigonometry

## NCERT Exercise 8.1

### Part 2

Question – 7 - If text(cot θ)=7/8, evaluate the following:

Solution: text(cot θ)=7/8=b/p

This means, b = 7 and p = 8.

We can calculate h by using Pythagoras theorem;

h^2 = p^2 + b^2

Or, h^2 = 8^2 + 7^2

= 64 + 49 = 113

Or, h = sqrt(113)

(a) ((1+text(sin θ))(1-text(sin θ)))/((1+text(cos θ))(1-text(cos θ))

Solution:

((1+text(sin θ))(1-text(sin θ)))/((1+text(cos θ))(1-text(cos θ))

=(1-text(sin)^2θ)/(1-text(cos)^2θ)

=(1-(8/sqrt(113))^2)/(1-(7/sqrt(113))^2)

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

=(113-64)/(113-49)=(49)/(64)

(b) cot2 θ

Solution:

text(cot)^2θ=(7/8)^2=(49)/(64)

Question – 8 - If 3 cot A = 4, check whether (1-text(tan)^2A)/(1+text(tan)^2A)=text(cos)^2A-text(sin)^2A or not.

Solution: 3 cot A = 4 means cot A = 4/3 = b/p

Hence, p = 3 and b = 4.

We can calculate h by using Pythagoras theorem;

h^2 = p^2 + b^2

Or, h^2 = 3^2 + 4^2

= 9 + 16 = 25

Or, h = 5

Now; the equation can be checked as follows:

(1-text(tan)^2A)/(1+text(tan)^2A)

=(1-(3/4)^2)/(1+(3/4)^2

=(1-(9)/(16))/(1+(9)/(16))=(7)/(25)

RHS:

text(cos)^2A-text(sin)^2A

=(4/5)^2-(3/5)^2

=(16)/(25)-(9)/(25)=(7)/(25)

It is clear that LHS = RHS.

Question – 9 - In triangle ABC, right-angled at B, if text(tan A)=1/sqrt3 find the value of:

Solution:

text(tan A)=1/sqrt3=p/b

This means, p = 1 and b = sqrt3

We can calculate h by using Pythagoras theorem;

h^2 = p^2 + b^2

Or, h^2 = 1^2 + (sqrt3)^2

= 1 + 3 = 4

Or, h = sqrt2=2

(a) Sin A Cos C + Cos A Sin C

Solution: Sin A Cos C + Cos A Sin C

=1/2xx1/2+sqrt3/2xxsqrt3/2

=1/4+3/4=1

(Note: side opposite to angle A = side adjacent to angle C and side adjacent to angle A = side opposite to angle C)

(b) Cos A Cos C – Sin A Sin C

Solution: Cos A Cos C – Sin A Sin C

=sqrt3/2xx1/2-1/2xxsqrt3/2=0

Question – 10 - In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution: Given; PR + QR = 25 cm and PQ = 5 cm.

Hence, PR = 25 – QR

We can calculate PR and QR by using Pythagoras theorem;

PQ^2 = PR^2 – OR^2

Or, 5^2 = (25 – OR)^2 – OR^2

Or, 25 = 625 + OR^2 – 50OR – OR^2

Or, 25 = 625 – 50QR

Or, 50 QR = 625 – 25 = 600

Or, QR = 12

Hence, PR = 25 – 12 = 13

Now;

text(sin P)=(QR)/(PR)=(12)/(13)

text(cos P)=(QP)/(PR)=(5)/(13)

text(tan P)=(QR)/(QP)=(12)/(5)

Question – 11 - State whether the following are true or false. Justify your answer.

1. The value of tan A is always less than 1.

Solution: False; value of tan begins from zero and goes on to become more than 1.
2. sec A = 12/5 for some value of angle A.

Solution: True, value of cos is always more than 1.
3. cos A is the abbreviation used for the cosecant of angle A.

Solution: False, cos is the abbreviation of cosine.
4. cot A is the product of cot and A.

Solution: False, cot A means cotangent of angle A.
5. sin A = 4/3 for some angle A.

Solution: False, value of sin is less than or equal to 1, while this value is more than 1.