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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


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figure shows a triangle and its circumcircle what is the radius of the circle

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AC=5 cm, angle ABC =60°

If tanθ + cotθ = 2 and θ is acute, then the value of tan100θ +cot100θ is equal to:

Conert Radian to Degree :

9π3\frac{9\pi}{3}

Conert Radian to Degree :

4π3\frac{4\pi}{3}