Class 10 Mathematics

Trigonometry

Exercise 8.2 (NCERT)

Question – 1 - Evaluate the following:

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

Solution:

`sin60°cos30°+sin30°cos60°`

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

`=3/4+1/4=1`


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

Solution:

`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°)`

Solution:

`(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°)`

Solution:

`(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°)`

Solution:

`(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

Solution: (A) sin 60o

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

Solution: (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

Solution: (A) 0o

`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

Solution: (C) tan 60o

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.

Solution: We know;

`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

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

    Solution: 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.

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

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

    Solution: True, refer to values for trigonometric ratios.

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Exercise 1(part I)

Exercise 1(part II)

Exercise 3

Exercise 4(part I)

Exercise 4(Part II)

Application(part I)

Application(part II)