Question

If the median AD of ∆ABC be perpendicular to side AB then prove that tanA+2tanB=0.

Answer 1

Dear nishat

 

In Triangle ACD 

 use sine rule

a/2 sinB /sin(180-A-B)   = a/2 /sin(A-90)

sinB/sin(A+B)  = -1/cosA

 sinB cosA = -sin(A+B)

 sinBcosA  =-[sinAcosB + cosA sinB]

 tanA +2tanB =0

Answer 2

tan A - 2tan B = 0

Step-by-step explanation:

The image of the answer is attached below.

Given ABC is a triangle.

AD is the median of the triangle.

θ is an external angle of ΔABD.

⇒ θ = 90° + B (from the figure)

Using the rule m - n in ΔABC.

Cot θ = cot(90° + B)

$(1+1) \cot (90+B)=1 . \cot 90^{\circ}-1 . \cot \left(A-90^{\circ}\right)$

$\Rightarrow-2 \tan B=\cot \left(90^{\circ}-A\right)$

$\Rightarrow-2 \tan B=\tan A$   (since tan A = cot (90 - A))

Add tan A on both sides, we get

⇒ tan A - 2tan B = 0

Hence proved.

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