Class 10 Maths

# Trigonometry

## NCERT Exercise 8.2

Question – 1 - Evaluate the following:

(i) sin 60o cos 30o + sin 30o cos 60o

sin60°cos30°+sin30°cos60°

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

=3/4+1/4=1

(ii) 2 tan2 45o + cos2 30o - sin2 60o

tan^2\45°+cos^2\30°-sin^2\60°

=2xx1^2+(sqrt3/2)^2-(sqrt3/2)^2

=2+0=2

(iii) (cos 45°)/(sec 30°+text(cosec) 30°)

(cos 45°)/(sec 30°+text(cosec) 30°)

=(1/sqrt2)/(2/sqrt3+2)=(1/sqrt2)/((2+2sqrt3)/(sqrt3))

=1/sqrt2xx(sqrt3)/(2(sqrt3+1))

=(sqrt3)/(2sqrt2(sqrt3+1))

=(sqrt3)/(2sqrt2(sqrt3+1))xx(2sqrt2(sqrt3-1))/ (2sqrt2(sqrt3+1))

=(sqrt3xx2sqrt2(sqrt3-1))/(8(3-1))

=(6sqrt2-2sqrt6)/(16)

=(2(3sqrt2-sqrt6))/(16)

=(3sqrt2-sqrt6)/(8)

(iv) (sin30°+tan45°-text(cosec) 60°)/(Sec30°+cos60°+cot45°)

(sin30°+tan45°-text(cosec) 60°)/(Sec30°+cos60°+cot45°)

=(1/2+1-2/sqrt3)/(2/sqrt2+1/2+1)

=(3/2-2/sqrt3)/(3/2+2/sqrt3)

=(3/2-2/sqrt3)/(3/2+2/sqrt3)xx(3/2-2/sqrt3)/(3/2-2/sqrt3)

=(9/4+4/3-2xx3/2xx2/sqrt3)/(9/4-4/3)

=((27+16)/(12)-2sqrt3)/((27-16)/(12))

=(43-24sqrt3)/(11)

(v) (5cos^2\60°+4sec^2\30°-tan^2\45°)/(sin^2\30°+cos^2\30°)

(5cos^2\60°+4sec^2\30°-tan^2\45°)/(sin^2\30°+cos^2\30°)

=(5xx(1/2)^2+4xx(2/sqrt3)^2-1^2)/((1/2)^2+(sqrt3/2)^2)

=(5/4+16/3-1)/(1/4+3/4)

=(5/4+16/3-1)/(1)

=(15+64-12)/(12)=(67)/12

Question – 2 - Choose the correct option and justify your choice:

(i) (2tan30°)/(1+tan^2\30°)=?

(A) sin 60o, (B) cos 60o, (C) tan 60o, (D) sin 30o

Explanation:

(2tan30°)/(1+tan^2\30°)

=(2xx1/sqrt3)/(1+(1/sqrt3)^2)

=(2/sqrt3)/(1+1/3)=2/sqrt3xx3/4

=sqrt3/2=sin60°

(ii) (1-tan^2\45°)/(1+tan^2\45°)=?

(A) tan 90o, (B) 1, (C) sin 45o, (D) 0

Explanation:

(1-tan^2\45°)/(1+tan^2\45°)

=(1-1)/(1+1)=0/2=0

(iii) Sin 2A = 2 Sin A is true when A = ?

(A) 0o, (B) 30o, (C) 45o, (D) 60o

Sin 0° = 0 and Sin 2 xx 0° = Sin 0° = 0

Sin 30° =1/4 But Sin 2 xx 30° = Sin 60° is not equal to Sin 30°. The same holds true for sin 45°.

(iv) (2tan30°)/(1-tan^2\30°)=?

(A) cos 60o, (B) sin 60o, (C) tan 60o, (D) sin 30o

Explanation:

(2tan30°)/(1-tan^2\30°)

=(2xx1/sqrt3)/(1-(1/sqrt3)^2)=(2/sqrt3)/(1-1/3)

=(2/sqrt3)/((3-1)/(3))=2/sqrt3xx3/2

=sqrt3=tan60°

Question 3: If tan (A + B) = sqrt3 and tan(A-B)=1/sqrt3, 0°< A + B ≤ 90°, A > B; find A and B.

tan 60° = sqrt3

Hence, A + B = 60°

Similarly;

tan30°=1/sqrt3

Hence, A – B = 30°

Adding equation (1) and (2), we get;

A+B+A-B=60°+30°

Or, 2A=90°

Or, A=45°

So, B=60°-45°=15°

Question – 4 - State whether the following are true or false. Justify your answer.

• Sin (A+B) = Sin A + Sin B

Answer: False; this can be proved by assuming different values for A and B.
• The value of sin θ increases as θ increases.

Answer: True, the value of sin θ increases from zero for zero degree and goes up to 1 for 90 degree.
• The value of cos θ increases are θ increases.

Answer: False, the value of cos decreases from one to zero.
• Sin θ = Cos θ for all values of θ.

Answer: False, values of sin and cos are equal only for 45 degree.
• Cot A is not defined for A = 0°

Answer: True, refer to values for trigonometric ratios.