Question

prove geometrically: tanA(180-A)=-tanA need urgently​

Answer 1

Appropriate Question:

• To prove : tan(180-A) = -tanA

Solution:

We know,

$\red\bigstar\underline{\boxed{\sf\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}}}$

$\underline{\mathfrak{\blue{Putting\; the\; values}}}$

$\implies\sf\tan (180^\circ-A)=\dfrac{\tan 180^\circ-\tan A}{1+\tan180^\circ\tan A}$

tan 180° = 0

$\implies\sf\tan (180^\circ-A)=\sf\dfrac{0-\tan A}{1-0.\tan A}\\\\\implies\sf\tan (180^\circ-A)=\sf\dfrac{-\tan A}{1-0}\\\\\implies\sf\tan (180^\circ-A)=\sf\dfrac{-\tan A}{1}\\\\\implies\sf\tan (180^\circ-A)=\sf-\tan A$

$\quad\large\underline{\green{\mathfrak{Hence\;verified\displaystyle !\,}}}$

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$\dag\underline{\red{\sf Trigonometric\:table}}:-$

$\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}$

Answer 2

Step-by-step explanation:

Formula: tan(a-b)=tan(a)-tan(b)/1+tan(a)tan(b)

here,a=180 &b=A

putting the values in formula..

-->tan(180-A)= tan(180)-tan(A)/1+tan(180)tan(A)

*tan(180)=0 (by calculator)

-->tan(180-A) = (0) - tanA/1+ 0[tan(A)]

-->tan(180 -A) = -tanA /1+0

--> tan(180 -A) = -tan A / 1 = -tanA