We know that, A is in second quadrant, B is in first quadrant.
In the second quadrant, sine function is positive. cosine and tan functions are negative.
In first quadrant, all functions are positive.
By using the formulas,
cos A = – √(1 – sin2 A) and sin B = √(1 – cos2 B)
So let us find the value of cos A and sin B
cos A = – √(1 – sin2 A)
= – √(1 – (1/2)2)
= – √(1- 1/4)
= – √((4-1)/4)
= – √(3/4)
= -√3/2
sin B = √(1 – cos2 B)
= √(1-(√3/2)2)
= √(1- 3/4)
= √((4-3)/4)
= √(1/4)
= 1/2
We know, tan A = sin A / cos A and tan B = sin B / cos B
tan A = (1/2)/( -√3/2) = -1/√3 and
tan B = (1/2)/(√3/2) = 1/√3
(i) tan (A + B) = (tan A + tan B) / (1 – tan A tan B)
= (-1/√3 + 1/√3) / (1 – (-1/√3) × 1/√3)
= 0 / (1 + 1/3)
= 0
(ii) tan (A – B) = (tan A – tan B) / (1 + tan A tan B)
= ((-1/√3) – (1/√3)) / (1 + (-1/√3) × (1/√3))
= ((-2/√3) / (1 – 1/3)
= ((-2/√3) / (3-1)/3)
= ((-2/√3) / 2/3)
= – √3
Question - 7 : - Evaluatethe following:
(i) sin 780 cos180 – cos 780 sin 180
(ii) cos470 cos 130 – sin 470 sin 130
(iii) sin 360 cos90 + cos 360 sin 90
(iv) cos800 cos 200 + sin 800 sin 200
Answer - 7 : -
(i) sin 780 cos 180 –cos 780 sin 180
We know that sin (A – B) = sin A cos B – cos A sin B
sin 780 cos 180 – cos780 sin 180 = sin(78 – 18) °
= sin 60°
= √3/2
(ii) cos 470 cos 130 –sin 470 sin 130
We know that cos A cos B – sin A sin B = cos (A + B)
cos 470 cos 130 – sin470 sin 130 = cos (47 + 13) °
= cos 60°
= 1/2
(iii) sin 360 cos 90 +cos 360 sin 90
We know that sin (A +B) = sin A cos B + cos A sin B
sin 360 cos 90 + cos360 sin 90 = sin (36 + 9) °
= sin 45°
= 1/√2
(iv) cos 800 cos 200 +sin 800 sin 200
We know that cos A cos B + sin A sin B = cos (A – B)
cos 800 cos 200 + sin800 sin 200 = cos (80 – 20) °
= cos 60°
= ½
Question - 8 : - If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following:
(i) sin (A + B) (ii) cos (A + B) (iii) tan (A + B)
Answer - 8 : -
Given:
cos A = -12/13 and cot B = 24/7
We know that, A lies in second quadrant, B in the third quadrant.
In the second quadrant sine function is positive.
In the third quadrant, both sine and cosine functions are negative.
By using the formulas,
sin A = √(1 – cos2 A), sin B = – 1/√(1 + cot2 B) and cos B = -√(1 – sin2 B),
So let us find the value of sin A and sin B
sin A = √(1 – cos2 A)
= √(1 – (-12/13)2)
= √(1 – 144/169)
= √((169-144)/169)
= √(25/169)
= 5/13
sin B = – 1/√(1 + cot2 B)
= – 1/√(1 + (24/7)2)
= – 1/√(1 + 576/49)
= -1/√((49+576)/49)
= -1/√(625/49)
= -1/(25/7)
= -7/25
cos B = -√(1 – sin2 B)
= -√(1-(-7/25)2)
= -√(1-(49/625))
= -√((625-49)/625)
= -√(576/625)
= -24/25
So, now let us find
(i) sin (A + B)
We know that sin (A + B) = sin A cos B + cos A sin B
So,
sin (A + B) = sin A cos B + cos A sin B
= 5/13 × (-24/25) + (-12/13) × (-7/25)
= -120/325 + 84/325
= -36/325
(ii) cos (A + B)
We know that cos (A + B) = cos A cos B – sin A sin B
So,
cos (A + B) = cos A cos B – sin A sin B
= -12/13 × (-24/25) – (5/13) × (-7/25)
= 288/325 + 35/325
= 323/325
(iii) tan (A + B)
We know that tan (A + B) = sin (A+B) / cos (A+B)
= (-36/325) / (323/325)
= -36/323
Question - 9 : - Prove that: cos 7π/12 + cos π/12 = sin 5π/12 – sin π/12
Answer - 9 : -
We know that, 7π/12 = 105°, π/12 = 15°; 5π/12 = 75°
Let us consider LHS: cos 105° + cos 15°
cos (90° + 15°) + sin (90° – 75°)
-sin 15° + sin 75°
sin 75° – sin 15°
= RHS
∴ LHS = RHS
Hence proved.
Question - 10 : - Prove that: (tan A + tan B) / (tan A – tan B) = sin (A + B) / sin (A – B)
Answer - 10 : - Let us consider LHS: (tan A + tan B) / (tan A – tan B)
= RHS
∴ LHS = RHS
Hence proved.