Question

a sin^2tha+bcos^2tha=c then prove that tan tha=sqrt c-b/a-c

Answer 1

$asin^2(x) +bcos^2(x) = c asin^2(x) +b(1 - sin^2(x)) = c sin^2(x) = \frac{c-b}{a-b} cos^2(x) = 1- \frac{c-b}{a-b} = \frac{a-c}{a-b} tan^2(x) = \frac{sin^2(x)}{cos^2(x)} = \frac{(c-b)/(a-b)}{(a-c)/(a-b)} = \frac{c-b}{a-c} Hence, tan(x) = \sqrt{\frac{c-b}{a-c} }$

Answer 2

Step-by-step explanation:

asin

2

(x)+bcos

2

(x)=casin

2

(x)+b(1−sin

2

(x))=csin

2

(x)=

a−b

c−b

cos

2

(x)=1−

a−b

c−b

=

a−b

a−c

tan

2

(x)=

cos

2

(x)

sin

2

(x)

=

(a−c)/(a−b)

(c−b)/(a−b)

=

a−c

c−b

Hence,tan(x)=

a−c

c−b