App Logo

No.1 PSC Learning App

1M+ Downloads
image.png

Atan A + tan B = 0

Btan A = tan B

C2tanA+tanB1tanAtanB=1\frac{2tan A+ tan B}{1-tan A tan B}=1

D2tanA+tanB1+tanAtanB=0\frac{2tan A+ tan B}{1+ tan A tan B}=0

Answer:

A. tan A + tan B = 0

Read Explanation:

$tan (A+B) = \frac{tan A + tan B}{1-tanAtanB}$

so tan (A+B)=0

the numerator becomes zero


Related Questions:

In the given figure ABC=ABD,BC=BDthenCAB=\angle{ABC} = \angle{ABD}, BC = BD then \triangle{CAB} =\triangle___________

image.png

Find the value of tan60tan151+tan60tan15\dfrac{\tan 60^\circ - \tan 15^\circ}{1 + \tan 60^\circ \tan 15^\circ}

Find x if 2sin2x - 1 = 0

If 4θ is an acute angle, and cot 4θ = tan (θ - 5°) , then what is the value of θ?

Cos1o.cos2o.cos3o.......................cos100o is equal to