Question

In triangle ABC, angle B=90• if tanA=1/√3 then show that sinA. cosC+cosA.sinC=1

Answer 1

Answer:

Sins, Triangle of ABC whose angle B=90° and

TanA=1/√3

Therefore, TanA=1/√3=BC/AB. [From to angle A]

Therefore,Let, BC=k and AB=√3k

According to pithagoraous rule,

AC²=AB²+BC²

»AC²=(√3k)²+(k)²

»AC²=3k²+k²

»AC²=4k²

Therefore,AC=2k

And sinA=BC/AC=k/2k=1/2

cosA=AB/AC=√3k/2k=√3/2

sinC=AB/AC=√3k/2k=√3/2

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

:. sinA.cosC+cosA.sinC

=1/2×1/2+√3/2×√3/2

=1/4+3/4

=1+3/4

=4/4

=1. [proved]