In a right angle triangle ABC, right angle is at B, if 
then find the value of
A. sin A cosC + cos A sin C
We have
Given that, tan A = √3/1
And tan A is given by,
⇒ 
⇒ perpendicular = √3x and base = x
Then, we can use Pythagoras theorem in ∆ABC,
(hypotenuse)2 = (perpendicular)2 + (base)2
⇒ (hypotenuse)2 = (√3x)2 + (x)2
⇒ (hypotenuse)2 = 3x2 + x2 = 4x2
⇒ hypotenuse = √(4x2) = 2x
We have, AB = √3x, BC = x and AC = 2x.
Using these values,
⇒ 
⇒ 
⇒ 
…(i)
Similarly, 
⇒ 
⇒ 
…(ii)
Also,
⇒ 
⇒ 
⇒ 
…(iii)
Similarly, 
⇒ 
⇒ 
…(iv)
We have to solve: sin A cos C + cos A sin C.
Substituting equations (i), (ii), (iii) & (iv) in above,
sin A cos C + cos A sin C = 
⇒ sin A cos C + cos A sin C = 1/4 + 3/4
⇒ sin A cos C + cos A sin C = 4/4 = 1
Thus, sin A cos C + cos A sin C = 1.
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