Question

Q1) If asin2(x)+bcos2(y)=c , bsin2(y)+acos2(y)=d and

atan(x)=btan(y)

Then prove that : a2/b2=(d−a)(c−a)/(b−c)(b−d​

Answer 1

asin

2

x+bcos

2

x=c⇒asin

2

x+b(1−sinx)=c

⇒asin

2

x+b−bsin

2

x=c⇒(a−b)sin

2

x=c−b

⇒sin

2

x=

a−b

c−b

asin

2

x+bcos

2

x=c⇒a(1−cos

2

x)+bcos

2

x=c

⇒a−acos

2

x+bcos

2

x=c⇒(b−a)cos

2

x=c−a

⇒cos

2

x=

a−b

c−b

∴tan

2

x=

a−c

c−b

bsin

2

y+acos

2

y=d⇒bsin

2

y+a(1−sin

2

y)=d

⇒bsin

2

y+a−asin

2

y=d⇒(b−a)sin

2

y=d−a

⇒sin

2

y=

b−a

d−a

bsin

2

y+acos

2

y=d⇒b(1−cos

2

y)+acos

2

y=d

⇒b−bcos

2

y+acos

2

y=d⇒(a−d)cos

2

y=d−b

⇒cos

2

y=

a−b

d−b

∴tan

2

y=

a−d

d−b

atanx=btany⇒

tany

tanx

=

a

b

⇒

tan

2

y

tan

2

x

=

a

2

b

2

⇒

b

2

a

2

=

(b−c)(b−d)

(d−a)(c−a)