請數學高手入來

1) 若x+y=90，化簡(sin^2y(sinxcosy+sinycosx))/(1-sin^2x)

[sin^2y(sinxcosy+sinycosx)]/cos^2x

=[sin^2y sin(x+y)]/cos^2x

=[sin^2y sin90]/cos^2x

=sin^2y/cos^2x

------------------------------------------------------------------------
2)已知cosx=7/9及x是一個銳角
(a) 求sinx和tanx的值。

∵sin^2x+cos^2x=1

cosx=7/9

x是一個銳角

∴sin^2x+(7/9)^2=1

sin^2x=1-(7/9)^2

sin^2x=32/81

sinx=(4√2)/9


∵sinx=(4√2)/9

cosx=7/9

∴tanx=sinx/cosx

=[(4√2)/9]/( 7/9)

=(4√2)/7


(b)由此，求(63(cos^2x-sin^2x))/(16tan^2(90-x))
以根式表示。

[63(cos^2x-sin^2x)]/[16tan^2(90-x)]

=[63(cos2x)]/[16*(1/tanx)^2]

=[63(2cos^2x-1)]/(16/tan^2x)

={63[2*(7/9)^2-1]}/{16/[(4√2)/7]^2}

=(119/9)/(49/2)

=34/63

1) =cos^2 x(sin^2 x+cos^2x)/cos^2x=1
2(a)sinx=4開方(2/9) tanx= 4開方2/7
(b)306/343