Q9 of 268 Page 4

If A + B = 90o, sin A = a, sin B = b, then prove that

(a) a2 + b2 = 1


(b)

(a) LHS = a2 +b2

= (sin A)2 + (sin B)2


= sin2 A + sin2 B


= sin2 A + sin2 (90° - A) [∵ cos θ = sin (90° - θ)]


= sin2 A + cos2 A


= 1 [∵ sin2 θ + cos2 θ = 1]


=RHS


Hence Proved


(b) LHS = tan A


Now, taking RHS


⇒


{given, A +B = 90°)


[∵ cos θ = sin (90° - θ)]


⇒ tan A


=LHS


∴ LHS = RHS


Hence Proved


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