Question

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 1

Given as

cos A = -12/13 and cot B = 24/7

As we know that, A lies in second quadrant, B in the third quadrant.

Since, in the second quadrant sine function is positive.

Since, in the third quadrant, both sine and cosine functions are negative.

On using the formulas,

sin A = √(1 – cos² A), sin B = – 1/√(1 + cot² B) and cos B = -√(1 – sin² B),

Therefore 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

Therefore, now let us find

(i) sin (A + B)

As we know that sin (A + B) = sin A cos B + cos A sin B

Therefore,

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)

As we know that cos (A + B) = cos A cos B – sin A sin B

Therefore,

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)

As we know that tan (A + B) = sin (A + B)/cos (A + B)

= (-36/325)/(323/325)

= -36/323