Question

answer the 7th and 8th questions . please quickly

Answer 1

(7) cot θ = 7/8 = adjacent side/opposite side
Hypotenuse² = opp² + adj²
hyp² = 7²+8²
hyp² = 49+64
hyp² = 113
hyp = √113

sin θ = 8/√113
Cos θ = 7/√113

(i)(1+sinθ)(1-sinθ)/(1-cosθ)(1+cosθ)
= (1²-sin²θ)/(1²-cos²θ)
= (1-sin²θ)/(1-cos²θ)
= cos²θ/sin²θ
= cot²θ
= (7/8)² = 49/64

(ii) (1+sinθ)/cosθ
= (1+8/√113)/7/√113
= (√113+8)/7

(8) tan A = √3
tan A = tan 60°
A = 60°
B = 90°
C = 30°

(i)sin A.sin C + cos A.sin C
= sin 60°.cos 30° + cos 60°.sin 30°
= √3/2 × √3/2 + ½ + ½
= 3/4 + 1/4
= 4/4 = 1 = sin 90°

(ii) Cos A.Cos C - Sin A.sin C
= cos 60°.cos30° - cos 60°.sin 30°
= ½ × √3/2 - ½ × √3/2
= √3/4 - √3/4
= 0 = cos 90°

Hope it helps