If theta is an acute angle and 3sinetheta=4 costheta
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Squaring both sides : 9 sin^2 A = 16 cos^2A
{ here A is theta } ( cos^2A = 1- Sin^2 A )
So 9 sin^2 A = 16 - 16 sin^2 A
9+ 16(sin^2A)= 16
25sin^2 A = 16
sin^2A = 16/25
SinA = 4/5
So 3/4 Sin A = 3/5 = cos A
{ here A is theta } ( cos^2A = 1- Sin^2 A )
So 9 sin^2 A = 16 - 16 sin^2 A
9+ 16(sin^2A)= 16
25sin^2 A = 16
sin^2A = 16/25
SinA = 4/5
So 3/4 Sin A = 3/5 = cos A
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