Trigonometry
NCERT Exercise 8.2
Question – 1 - Evaluate the following:
(i) sin 60o cos 30o + sin 30o cos 60o
Answer:
`sin60°cos30°+sin30°cos60°`
`=sqrt3/2xxsqrt3/2+1/2xx1/2`
`=3/4+1/4=1`
(ii) 2 tan2 45o + cos2 30o - sin2 60o
Answer:
`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°)`
Answer:
`(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°)`
Answer:
`(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°)`
Answer:
`(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
Answer: (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
Answer: (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
Answer: (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
Answer: (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.
Answer: 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
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.