Question

Solve this one plzzzzz

Answer 1

Heya......✋
Answer is in attachment
hope it helps...:)

Answer 2

Hey Friend!!

cos¢ = 2t / 1 + t² = b/h (base) / (height )

by Applying Pythagoras theorem .

P (perpendicular )=
$\sqrt{h {}^{2} \ - b {}^{2} }$

=> √ ( 1 + t) ² - (2t)²

=> √( 1 + t² + 2t - 4t²

=> √-3t² + 2t + 1

=> √-3t² + 3t - t + 1

=> √ - 3t ( t - 1 ) - 1 ( 1 - 1 )

=> √( -3t - 1 )( t-1 )

P= √(-3t - 1 ) ( t - 1 )

tan¢ = P/b

putting the value of P and b

tan¢ = √(-3t - 1 )(t-1 ) / ( 1 + t²)

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Hope it helps you !!!

@Rajukumar111