Question

In a right angle triangle ABC, right angle is at B, if tan A= root3 then find the value of
(i) sin A cos C+ cos A sin C
(ii) cos A cos C-sin A sin C​

Answer 1

Answer:

1)5/4 2) 0

Step-by-step explanation:

I have attached solution in the picture

Answer 2

Answer:

(i) 1

(ii) 0

Step-By-Step Explanation:

(i) sin A cos C+ cos A sin C

Given: tan A = √3

⇒ Tanθ = Opposite/Adjacent = BC/AB = $\frac{\sqrt{3} }{1}$

Let: BC = √3k, AB = 1k

∴ By Pythagoras Theorem,

= (AC)² = (AB)² + (BC)²

= (AC)² = (1k)² + (√3k)²

= (AC)² = 1k² + 3k²

= AC² = 4k²

= AC = √4k² (√,² gets Cancelled)

= AC = 2k

∴ $\frac{(\sqrt{3})^2 }{4} + \frac{1}{4}$

= $\frac{3+1}{4} = \frac{4}{4} = 1$

(ii) cos A cos C-sin A sin C​

= $\frac{1k}{2k}$ x $\frac{\sqrt{3k} }{2k}$ - $\frac{\sqrt{3k} }{2k}$ x $\frac{1k}{2k}$  (All [k's] will get Cancelled)

= $\frac{\sqrt{3} }{4} -\frac{\sqrt{3} }{4}=\frac{0}{4}$ = 0