# In a Δ ABC, ∠ B = 90° and tanA = 1/√3. Prove thati. sinA cosC + cosA sinC = 1ii. cosA cosC - sinA sinC = 0

We have, tanA = k/(k√3) = BC/AB

Δ ABC is a right angled triangle By Pythagoras theorem, (hypotenuse)2 = (perpendicular)2 + (base)2

AC2 = BC2 + AB2

= AC2 = (k)2 + (k√3)2

= AC2 = k2 + 3k2

= AC2 = 4k2

AC = 2k

sinA = BC/AC = k/(2k) = 1/2

cosA = AB/AC = (k√3)/(2k) = √3/2

sinC = AB/AC = (k√3)/(2k) = √3/2

cosC = = BC/AC = k/(2k) = 1/2

i. sinA cosC + cosA sinC = (1/2)(1/2) + ( √3/2)( √3/2)

= = 4/4

= 1

RHS = LHS

HENCE PROVED

ii. cosA cosC - sinA sinC = (1/2)(√3/2) - (1/2)(√3/2)

= (√3/4) - (√3/4)

= 0

RHS = LHS

HENCE PROVED

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos  Quiz | Task on Trigonometric Ratios46 mins  NCERT I Trigonometric Ratios50 mins  Basic Concepts of Trigonometry45 mins  Trigonometric Identities33 mins  Quiz on Trigonometric Ratios31 mins  Applying the Trigonometric Identities52 mins  Champ Quiz | Trigonometric Identities33 mins  NCERT | Trigonometric Identities52 mins  Smart Revision | Trigonometric Identities40 mins  T- Ratios of Specified Angles58 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 