Question

in right triangle ABC , right angled at B, if tan A = 1 , then varify 2cos A sin A = 1​

Answer 1

Answer:

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Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

In right angled Δ ABC, right angled at B, if tan A = 1

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure of right angled triangle according to the question.

As we know that,

$\boxed{\bf{tan\:\theta = \frac{Perpendicular}{Base} }}$

A/q

$\mapsto\tt{tan\:A = \dfrac{BC}{AC} = \dfrac{1}{1} }$

$\underline{\mathcal{BY\:\:PYTHAGORAS\:\:THEOREM\::}}$

→ (Hypotenuse)² = (Base)² + (Perpendicular)²

→ (AB)² = (AC)² + (BC)²

→ (AB)² = (1)² + (1)²

→ (AB)² = 1 + 1

→ (AB)² = 2

→ AB = √2

Now,

V E R I F I C A T I O N :

Taking L.H.S :

$\mapsto\tt{2 cos A .sin A}$

$\mapsto\tt{2\times \dfrac{Base}{Hypotenuse} \times \dfrac{Perpendicular}{Hypotenuse}}$

$\mapsto\tt{2\times \dfrac{AC }{AB} \times \dfrac{BC}{AB}}$

$\mapsto\tt{2\times \dfrac{1 }{\sqrt{2} } \times \dfrac{1}{\sqrt{2} }}$

$\mapsto\tt{2\times \dfrac{1}{2} }$

$\mapsto\tt{\cancel{\dfrac{2}{2} }}$

$\mapsto\bf{ 1}$

Thus,

L.H.S = R.H.S

Verified .