Question

plzz kindly solve this ques ​

Answer 1

Answer:

answer for the given problem is given

Answer 2

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In triangle ABC, angle ABC = 90°

$therefore \: \tan(a) = \frac{bc}{ab}$

Since tan A = 1 (Given)

$\frac{bc}{ab } = 1$

therefore BC = AB

Let AB = BC = K , where k is a positive number

$\underline\bold\blue {Now: -}$

${ac}^{2} = {ab}^{2} + {bc}^{2}$

$therefore \: ac \: = \sqrt{ {ab}^{2} } + {bc}^{2} = \sqrt{ {k}^{2} } + {k}^{2}$

$therefore \: ac \: = k \sqrt{2}$

$therefore \sin(a) = \frac{bc}{ac} = \frac{k}{k \sqrt{2} } = \frac{1}{ \sqrt{2} }$

$\cos(a) = \frac{ab}{ac} = \frac{k}{k \sqrt{2} } = \frac{1}{ \sqrt{2} }$

$2 \sin(a) \cos(a) = 2 ( \frac{1}{ \sqrt{2} } )( \frac{1}{ \sqrt{2} }) = 1$

$therefore \: 2 \sin(a ) \cos(a) = 1$