If asin2 x+bcos2 â¡ x=c, bsin2 â¡ y+acos2 â¡ y=d and atan x=btan â¡ y, then (a2/b2) is equal to (Here a, b, c and d are distinct)

Question

If asin2 x+bcos2 â¡ x=c, bsin2 â¡ y+acos2 â¡ y=d and atan x=btan â¡ y, then (a2/b2) is equal to (Here a, b, c and d are distinct) (A) ((b - c) (d - b)/(a - d) (a - b)) (B) ((a - d) (c - a)/(b - c) (d - b)) (C) ((d - a) (c - a)/(b - c) (d - b)) (D) ((b - c) (b - d)/(a - c) (a - d))

Tardigrade Admin · Accepted Answer

Dividing the given equations by cos2 x cos2â¡y respectively. We get, tan2x=(c - b/a - c) and tan2y=(d - a/b - d) ⇒ (a2/b2)=((tan)2 y/(tan)2 x)=((. a - d .) (. c - a .)/(. b - c .) (. d - b .))

Q. If $a s i n^{2} x + b co s^{2} x = c, b s i n^{2} y + a co s^{2} y = d$ and $a t an x = b t an y,$ then $\frac{a ^{2}}{b ^{2}}$ is equal to (Here $a, b, c$ and $d$ are distinct)

$\frac{( b - c ) ( d - b )}{( a - d ) ( a - b )}$

$\frac{( a - d ) ( c - a )}{( b - c ) ( d - b )}$

$\frac{( d - a ) ( c - a )}{( b - c ) ( d - b )}$

$\frac{( b - c ) ( b - d )}{( a - c ) ( a - d )}$

Dividing the given equations by $co s^{2} x & co s^{2} y$ respectively. We get,
$t a n^{2} x = \frac{c - b}{a - c}$ and $t a n^{2} y = \frac{d - a}{b - d}$
$\Rightarrow \frac{a ^{2}}{b ^{2}} = \frac{( t an ) ^{2} y}{( t an ) ^{2} x} = \frac{( a - d ) ( c - a )}{( b - c ) ( d - b )}$

Q. If asin2x+bcos2⁡x=c,bsin2⁡y+acos2⁡y=d and atanx=btan⁡y, then b2a2​ is equal to (Here a,b,c and d are distinct)

(a−d)(a−b)(b−c)(d−b)​

(b−c)(d−b)(a−d)(c−a)​

(b−c)(d−b)(d−a)(c−a)​

(a−c)(a−d)(b−c)(b−d)​