Q. 183.9( 12 Votes )

If tanθ = 4/3, sh

Answer :

We have, tanθ = (4k)/(3k) = BC/AC (For some value of k)



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


AB2 = BC2 + AC2


= AB2 = (4k)2 + (3k)2


= AB2 = 16k2 + 9k2


= AB2 = 25k2


= (5k)2


AB = 5k


sinθ = BC/AB = (4k)/(5k) = 4/5


cosθ = AC/AB = (3k)/(5k) = 3/5


consider LHS = sinθ + cosθ =


= 7/5


= RHS


HENCE PROVED


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