Question

you have to prove that sin A cos A = 1/2 root 2​

Answer 1

Answer:

Step-by-step explanation:

Given :

To prove :

$sinAcosA=\frac{\sqrt{2} }{4}$ if $tanA=\sqrt{2} -1$

Solution :

$tanA=\sqrt{2} -1$

By multiplying & dividing by $\sqrt{2}+1$

We get,

$tanA=\sqrt{2} -1 \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{(\sqrt{2}-1)(\sqrt{2}+1)}{\sqrt{2}+1} = \frac{(\sqrt{2})^2-(1)^2}{\sqrt{2}+1} = \frac{2 -1}{\sqrt{2}+1} = \frac{1}{\sqrt{2}+1}$

⇒ $tanA= \frac{1}{\sqrt{2}+1}$

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We know that,

$tanA=\frac{sinA}{cosA} = \frac{1}{\sqrt{2} + 1}$,

Then,

$sinA =\frac{Opposite(or)Perpendicular}{Hypotenuse} = \frac{1k}{Hypotenuse}$

$cosA =\frac{Adjcent(or)Base}{Hypotenuse} = \frac{(\sqrt{2}+1)k}{Hypotenuse}$

⇒ $(Hypotenuse)^2 = (Perpendicular)^2+(Base)^2$

⇒ $(Hypotenuse)^2 = (1k)^2 + [(\sqrt{2} +1)k]^2$

⇒ $(Hypotenuse)^2 = k^2 + [(2+1+2\sqrt{2})k]$

⇒ $(Hypotenuse)^2 = k^2 + (3k^2+2k^2\sqrt{2})$

⇒ $(Hypotenuse)^2 =4k^2+2k^2\sqrt{2}$

⇒ $Hypotenuse =\sqrt{4k^2+2k^2\sqrt{2}} = k\sqrt{4+2\sqrt{2}}$

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$sinAcosA = \frac{perpendicular}{Hypotenuse} \times \frac{Base}{Hypotenuse} = \frac{1k}{k\sqrt{4+2\sqrt{2}}} \times \frac{(\sqrt{2}+1)k}{k\sqrt{4+2\sqrt{2}}} = \frac{\sqrt{2}+1 }{4+2\sqrt{2} } = \frac{\sqrt{2}+1}{2\sqrt{2}(\sqrt{2}+1)} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$

Hence proved,.

Answer 2

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