Q. 294.3( 11 Votes )

# In a Δ ABC, ∠ B = 90° and tanA = 1/√3. Prove that

i. sinA cosC + cosA sinC = 1

ii. cosA cosC - sinA sinC = 0

Answer :

We have, tanA = k/(k√3) = BC/AB

Δ ABC is a right angled triangle

By Pythagoras theorem, (hypotenuse)^{2} = (perpendicular)^{2} + (base)^{2}

∴AC^{2} = BC^{2} + AB^{2}

= AC^{2} = (k)^{2} + (k√3)^{2}

= AC^{2} = k^{2} + 3k^{2}

= AC^{2} = 4k^{2}

→ 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

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