Chapter 3:Pair of Linear Equations in Two Variables
Chapter 5: Arithmetic Progression
Chapter 8: Introduction to Trigonometry
Chapter 9:Some Applications of Trigonometry
Chapter 12:Area Related to Circles
Introduction to Trigonometry
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan2 45° + cos2 30° – sin2 60
(i) sin 60° cos 30° + sin 30° cos 60°
sin 30° = 1/2
cos 30° = √3/2
sin 60° = 3/2
cos 60°= 1/2
on putting value(ii) 2 tan2 45° + cos2 30° – sin2 60
We know that
sin 60° = √3/2
cos 30° = √3/2
tan 45° = 1
On putting value2 tan2 45° + cos2 30° – sin2 60
= 2(1)2 + (√3/2)2-(√3/2)2
= 2 + 0
= 2
(iii)
We know that,
cos 45° = 1/√2
sec 30° = 2/√3
cosec 30° = 2
On putting value we get
sin 30° = 1/2
tan 45° = 1
cosec 60° = 2/√3
sec 30° = 2/√3
cos 60° = 1/2
cot 45° = 1
Substitute the values in the given problem, we get
We know that,
cos 60° = 1/2
sec 30° = 2/√3
tan 45° = 1
sin 30° = 1/2
cos 30° = √3/2
Now, substitute the values in the given problem, we get
(5cos260° + 4sec230° – tan245°)/(sin2 30° + cos2 30°)
(iv)
(i) (A) is correct.
= 6/4√3 = √3/2 = sin 60°
(ii) (D) is correct.
tan 45° = 1
1-tan245°/1+tan245°
= (1-12)/(1+12)
= 0/2 = 0
The solution of the above equation is 0.
(iii) (A) is correct.
As sin 2A = 2sin A cos A
=2sin A cos A = 2 sin A
= 2cos A = 2 = cos A = 1
here cos 0 =1Therefore, A = 0°
(iv) (C) is correct.
tan 30° = 1/√3
2tan30°/1-tan230° = 2(1/√3)/1-(1/√3)2
= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°
3. If tan (A + B) = √3 and tan (A – B) = 1/√3 ,0° < A + B ≤ 90°; A > B, find A and B.
Solution:
tan (A + B) = √3
here √3 = tan 60°
On compare
⇒ tan (A + B) = tan 60°
(A + B) = 60° … (i)
Similarly
tan (A – B) = 1/√3
Here 1/√3 = tan 30°
⇒ tan (A – B) = tan 30°
(A – B) = 30° … equation (ii)
add the equation (i) and (ii), we get
A + B + A – B = 60° + 30°
2A = 90°
A= 45°
Now, substitute the value of A in equation (i)
45° + B = 60°
B = 60° – 45°
B = 15°
Therefore A = 45° and B = 15°
4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(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) cot A is not defined for A = 0°.
Solution:
(i) False.
Justify:
Let us take A = 30° and B = 60°, then
Substitute the values in the sin (A + B) formula, we get
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°
= 1/2 + √3/2 = 1+√3/2
Since the values obtained are not equal, the solution is false.
(ii) True.
Justify:
As we know
sin 0° = 0, sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2
sin 90° = 1
As we saw on θ increases sin θ is also increases as. So the statement is true
(iii) False.
As we know
cos 0° = 1,
cos 30° = √3/2,
cos 45° = 1/√2,
cos 60° = 1/2,
cos 90° = 0
Thus, the value of cos θ decreases as θ increases. So, the statement is false.
(iv) False
sin θ = cos θ,let θ=30°
So sin 30° = 1/2, cos 30° = √3/2, that are not equal
(v) True.
cot A = cos A/sin A
here A = 0°
cot 0° = cos 0°/sin 0° = 1/0 = undefined.
Hence, it is true