Question

Q4.If acos⁡Z+bsin⁡Z=4,a sin⁡Z-b cos⁡Z=3 then a^2+b^2
a)7
b)12
c)25
d)none

Answer 1

Answer:

Given, a cosx + b sinx = 4 ...( i )

a sinx - b cosx = 3 ...( ii )

Square on both sides on ( i )

⇒ ( a cosx + b sinx )^2 = 4^2

From the identities of factorization, we know ( a + b )^2 = a^2 + b^2 + 2ab

⇒ ( a cosx )^2 + ( b sinx )^2 + 2( a cox . b sinx ) = 16

⇒ a^2 cos^2 x + b^2 sin^2 x + 2ab.cosx.sinx = 16 ...( iii )

Square on both sides on ( ii )

⇒ ( a sinx - b cosx )^2 = 3^2

From the identities of factorization, we know ( a - b )^2 = a^2 + b^2 - 2ab

⇒ ( a sinx )^2 + ( b cosx )^2 - 2( b cox . a sinx ) = 16

⇒ a^2 sin^2 x + b^2 cos^2 x - 2ab.cosx.sinx = 16 ...( iv )

Now adding ( iii ) and ( iv ) to each other : -

⇒ a^2 cos^2 x + b^2 sin^2 x + 2ab.cosx.sinx + a^2 sin^2 x + b^2 cos^2 x - 2ab.cosx.sinx = 16 + 9

⇒ a^2 cos^2 x + a^2 sin^2 x + b^2 sin^2 x + b^2 cos^2 x = 25

⇒ a^2( cos^2 x + sin^2 x ) + b^2( sin^2 x + cos^2 x ) = 25

From the identities of trigonometry we know, sin^2 A + cos^2 A = 1

∴ a^2( 1 ) + b^2( 1 ) = 25

⇒ a^2 + b^2 = 25

Hence the value of a^2 + b^2 is 25.

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i think it may be x not z