Question

182. If tan A = 1 and tan B = 13; evaluate:
(i) cos A cos B - sin A sin B.
(ii) sin A cos B + cos A sin B.​

Answer 1

Answer:⤵️

(1) -6/√85

(2) 7/√85

Explanation:⤵️

tan A = 1

=> tan A = tan 45°

=> A = 45°

tan B = 13/1

we know that,

tan B = h/b = 13/1

then let,

h= 13x

b= x

then we know that,

p^2 = h^2 + b^2

=> p^2 = 169x^2 + x^2

=> p^2 = 170x^2

=> p = (√170)x

now,

cos A = cos45° = 1/√2

cos B = b/p = x/(√170)x

= 1/√170

sin A = Sin 45° = 1/√2

sin B = h/p = 13x/(√170)x

= 13/√170

NOW,

(1) =>

cosA cosB - sinA sinB

=>cos (A+B)

=(1/√2)(1/√170)-(1/√2)(13/√170)

= -12/√340

= -12/2√85

= -6/√85

(2) =>

sinA cosB + sinB cosA

=> sin(A+B)

=(1/√2)(1/√170)+(13/√170)(1/√2)

= 14/√340

= 14/2√85

= 7/√85