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प्रश्न
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
उत्तर
LHS = `sin(90^circ - A).cos(90^circ - A)`
⇒ cosA.sinA
RHS = `tanA/(1 + tan^2A) = tanA/sec^2A = (sinA/cosA)/(1/cos^2A)`
⇒ RHS = `sinA/cosA . cos^2A = cosA.sinA`
Thus , LHS = RHS
⇒ `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
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